Basic Genetics

The basis for order in life lies in a very large molecule called deoxyribonucleic acid, mercifully abbreviated to DNA. A related molecule, ribonucleic acid (RNA) provides the genetic material for some microbes, and also helps read the DNA to make proteins.

Read?

Yes, read.

DNA has a shape rather like a corkscrewed ladder. The "rungs" of the ladder are of four different types. The information in DNA comes in how those types are ordered along the molecule, just as the information in Morse code comes in how the dashes and dots are ordered. The information in three adjacent rungs is "read" by a kind of RNA that hooks onto a particular triad of rungs at one end and grabs a particular amino acid at the other. Special triads say "start here" and "end here" and mark off regions of the DNA molecule we call discrete genes. The eventual result is a chain of amino acids that makes up a protein, with each amino acid corresponding to a set of three rungs along the DNA molecule. There are also genes that tell the cell when to turn on or turn off another gene. The proteins produced may be structural or they may be enzymes that facilitate chemical reactions in the body.

We now know that chromosomes are essentially DNA molecules. In an advanced (eukaryotic) cell, these chromosomes appear as threadlike structures packaged into a more or less central part of the cell, bound by a membrane and called the nucleus. What is more important is that the chromosomes in a body cell are arranged in pairs, one from the father and one from the mother. Further, the code for a particular protein is always on the same place on the same chromosome. This place, or location, is called a locus (plural loci.)

There are generally a number of slightly different genes that code for forms of the same protein, and fit into the same locus. Each of these genes is called an allele. Each locus, then, will have one allele from the mother and one from the father. How?

When an animal makes an egg or a sperm cell (gametes, collectively) the cells go through a special kind of division process, resulting in a gamete with only one copy of each chromosome. Unless two genes are very close together on the same chromosome, the selection of which allele winds up in a gamete is strictly random. Thus a dog who has one gene for black pigment and one for brown pigment may produce a gamete which has a gene for black pigment OR for brown pigment. If he's a male, 50% of the sperm cells he produces will be B (black) and 50% will be brown (b).

When the sperm cell and an egg cell get together, a new cell is created which once again has two of each chromosome in the nucleus. This implies two alleles at each locus (or, in less technical terms, two copies of each gene, one derived from the mother and one from the father,) in the offspring. The new cell will divide repeatedly and eventually create an animal ready for birth, the offspring of the two parents. How does this combination of alleles affect the offspring?

There are several ways alleles can interact. In the example above, we had two alleles, B for black and b for brown. If the animal has two copies of B, it will be black. If it has one copy of B and one of b, it will be just as black. Finally, if it has two copies of b, it will be brown, like a chocolate Labrador. In this case we refer to B as dominant to b and b as recessive to B. True dominance implies that the dog with one B and one b cannot be distinguished from the dog with two B alleles. Now, what happens when two black dogs are bred together?

We will use a diagram called a Punnett square. For our first few examples, we will stick with the B locus, in which case there are two possibilites for sperm (which we write across the top) and two for eggs (which we write along the left side. Each cell then gets the sum of the alleles in the egg and the sperm. To start out with a very simple case, assume both parents are black not carrying brown, that is, they each have two genes for black. We then have:

	B	B
B	BB (black)	BB (black)
B	BB (black)	BB (black)

All of the puppies are black if both parents are BB (pure for black.

Now suppose the sire is pure for black but the dam carries a recessive gene for brown. In this case she can produce either black or brown gametes, so

	B	B
B	BB (pure for black)	BB (pure for black)
b	Bb (black carrying brown)	Bb (black carrying brown)

This gives appoximately a 50% probability that any given puppy is pure for black, and a 50% probability that it is black carrying brown. All puppies appear black. We can get essentially the same diagram if the sire is black carrying brown and the dam is pure for black. Now suppose both parents are blacks carrying brown:

	B	b
B	BB (pure for black	Bb (black carrying brown)
b	Bb (black carrying brown)	bb (brown)

This time we get 25% probabilty of pure for black, 50% probability of black carrying brown, and - a possible surprise if you don't realize the brown gene is present in both parents - a 25% probability that a pup will be brown. Note that only way to distinguish the pure for blacks from the blacks carrying brown is test breeding or possibly DNA testing - they all look black.

Another possible mating would be pure for black with brown:

	B	B
b	Bb (black carrying brown)	Bb (black carrying brown)
b	Bb (black carrying brown)	Bb (black carrying brown)

In this case, all the puppies will be black carrying brown.

Suppose one parent is black carrying brown and the other is brown:

	B	b
b	Bb (black carrying brown)	bb (brown)
b	Bb (black carrying brown)	bb (brown)

In this case, there is a 50% probability that a puppy will be black carrying brown and a 50% probability that it will be brown.

Finally, look at what happens when brown is bred to brown:

	b	b
b	bb (brown)	bb (brown)
b	bb (brown)	bb (brown)

Recessive to recessive breeds true - all of the pups will be brown.

Note that a pure for black can come out of a mating with both parents carrying brown, and that such a pure for black is just as pure for black as one from ten generations of all black parentage. THERE IS NO MIXING OF GENES. They remain intact through their various combinations, and B, for instance, will be the same B no matter how often it has been paired with brown. This, not the dominant-recessive relationship, is the real heart of Mendelian genetics.

This type of dominant-recessive inheritance is common (and at times frustrating if you are trying to breed out a recessive trait, as you can't tell by looking which pups are pure for the dominant and which have one dominant and one recessive gene.) Note that dominant to dominant can produce recessive, but recessive to recessive can only produce recessive. The results of a dominant to recessive breeding depends on whether the dog that looks to be the dominant carries the recessive. A dog that has one parent expressing the recessive gene, or that produces a puppy that shows the recessive gene, has to be a carrier of the recessive gene. Otherwise, you really don't know whether or not you are dealing with a carrier, bar genetic testing or test breeding.

One more bit of terminology before we move on - an animal that has matching alleles (BB or bb) is called homozygous. An animal that has two different alleles at a locus (Bb) is called heterozygous.

A pure dominant-recessive relationship between alleles implies that the heterozygous state cannot be distinguished from the homozygous dominant state. This is by no means the only possibility, and in fact as DNA analysis advances, it may become rare. Even without such analysis, however, there are many loci where three phenotypes (appearances) come from two alleles. An example is merle in the dog. This is often treated as a dominant, but in fact it is a type of inheritance in which there is no clear dominant - recessive relationship. It is sometimes called overdominance, if the heterozyote is the desired state. I prefer incomplete dominance, recognising that in fact neither of the alleles is truly dominant or recessive relative to the other.

As an example, we will consider merle. Merle is a diluting gene, not really a color gene as such. If the major pigment is eumelanin, a dog with two non-merle genes (mm) is the expected color - black, liver, blue, tan-point, sable, recessive red. If the dog is Mm, it has a mosaic appearance, with random patches of the expected eumelanin pigment in full intensity against a background of diluted eumelanin. Phaeomelanin (tan) shows little visual effect, though there is a possibility that microscopic examination of the tan hair would show some effect of M. Thus a black or black tan-point dog is a blue merle, a brown or brown tan-point dog is red merle, and a sable dog is sable merle, though the last color, with phaeomelanin dominating, may be indistinguishable from sable in an adult. (The effect of merle on recessive red is unknown, and I can't think of a breed that has both genes.) What makes this different from the black-brown situation is that an MM dog is far more diluted than is an Mm dog. In those breeds with white markings in the full-color state the MM dog is often almost completely white with a few diluted patches, and has a considerable probablity of being deaf, blind, and/or sterile. Even in the daschund, which generally lacks white markings, the so-called double dapple (MM) has extensive white markings and may have reduced eye size. Photographs of Shelties with a number of combinations of merle with other genes are available on this site, but the gene also occurs in Australian Shepherds, Collies, Border Collies, Cardiganshire Welsh Corgis, Beaucerons (French herding breed), harlequin Great Danes, Catahoula leopard dogs, and Daschunds, at the least.

Note that both of the extremes - normal color and double merle white - breed true when mated to another of the same color, very much like the Punnett squares above for the mating of two browns or two pure for blacks. I will skip those two and go to the more interesting matings involving merles.

First, consider a merle to merle mating. Remember both parents are Mm, so we get:

	M	m
M	MM (sublethal double merle)	Mm (merle)
m	Mm (merle)	mm (non-merle)

Assuming that merle is the desired color, this predicts that each pup has a 25% probability of inheriting the sublethal (and in most cases undesirable by the breed standards) MM combination, only 50% will be the desired merle color, and 25% will be acceptable full-color individuals. (In fact there is some anecdotal evidence that MM puppies make up somewhat less than 25% of the offspring of merle to merle breedings, but we'll discuss that separately.) Merle, being a heterozygous color, cannot breed true.

Merle to double merle would produce 50% double merle and is almost never done intentionally. The Punnet square for this mating is:

	M	M
M	MM (sublethal double merle)	MM (sublethal double merle)
m	Mm (merle)	Mm (merle)

Merle to non-merle is the "safe" breeding, as it produces no MM individuals:

	m	m
M	Mm (merle)	Mm (merle)
m	mm (non-merle)	mm (non-merle)

We get exactly the same probability of merle as in the merle to merle breeding (50%) but all of the remaining pups are acceptable full-colored individuals.

There is one other way to breed merles, which is in fact the only way to get an all-merle litter. This is to breed a double merle (MM) to a non-merle (mm). This breeding does not a use a merle as either parent, but it produces all merle puppies. (The occasional exception will be discussed elsewhere.) In this case,

	M	M
m	Mm (merle)	Mm (merle)
m	Mm (merle)	Mm (merle

The problem with this breeding is that it requires the breeder to maintain a dog for breeding which in most cases cannot be shown and which may be deaf or blind. Further, in order to get that one MM dog who is fertile and of outstanding quality, a number of other MM pups will probably have been destroyed, as an MM dog, without testing for vision and hearing, is a poor prospect for a pet. In Shelties, the fact remains that several double merles have made a definite contribution to the breed. This does not change the fact that the safe breeding for a merle is to a nonmerle.

Thus far, we have concentrated on single locus genes, with two alleles to a locus. Even something as simple as coat color, however, normally involves more than one locus, and it is quite possible to have more than two alleles at a locus. What happens when two or more loci are involved in one coat color?

 

Basic Genetics II: Multiple Loci

Usually more than one gene locus is involved in coat color. We'll take one of the simplest, in which the two loci each have two alleles, with a simple dominant-recessive relationship. The model we will use is the Labrador Retriever. One locus we have already examined: the brown locus. We will now add a second locus, on a different chromosome, called E. An EE or Ee dog will show whatever eumelanin pigment is possible. An ee dog apparently can manufacture only phaeomelanin in the hair, though the skin and eye pigment still includes melanin (of whatever color is allowed by the B series).

A black Lab may be BBEE, BBEe, BbEE or BbEe - any combination that includes at least one B and one E gene.

A chocolate (brown) Lab may be bbEE or bbEe.

A yellow Lab with a black nose may be BBee or Bbee

A yellow Lab with a liver nose is bbee - but since ee dogs tend in many cases to lose nose pigment in winter, this may not be easy to distinguish from BBee or Bbee.

Suppose we mate two BbEe dogs, both blacks carrying brown and yellow:

	BE	Be	bE	be
BE	BBEE (pure for black)	BBEe (black carrying yellow)	BbEE (black carrying brown)	BbEe (black carrying brown and yellow)
Be	BBEe (black carrying yellow)	BBee (pure for yellow, black nose)	BbEe (black carrying brown and yellow)	Bbee (yellow carrying brown)
bE	BbEE (black carrying brown)	BbEe (black carrying brown and yellow)	bbEE (pure for brown)	bbEe (brown carrying yellow)
be	BbEe (black carrying brown and yellow)	Bbee (yellow carrying brown)	bbEe (brown carrying yellow)	bbee (brown-nosed yellow)

Each puppy has one chance in sixteen of having the combination shown in any section of the table above. In this mating between two black dogs both carrying brown and yellow, there is a 9/16 probability that a particular pup will be black, a 3/16 probability that the pup will be brown, a 3/16 probability that the pup will be a black-nosed yellow, and a 1/16 probability of a brown-nosed yellow. Since nose color does not come into registration, the registered colors would be 9 black:3 brown:4 yellow.

What happens if more than two loci are involved? The basic principle is the same - put all of the possible combinations in a sperm cell along the top and all of the possible combinations in an egg cell along the left side. The problem is that the number of possible combinations doubles for each additional locus. For a single locus, we had a 2 x 2 square with 4 cells. For two loci, we had a 4 x 4 square with 16 cells. With three loci, we have an 8 x 8 square with 64 cells. Besides, we've pretty well exhausted the acceptable colors for Labs.

Shetland Sheepdogs might be a good model for our three-locus model. For the moment we'll omit the recessive black, and consider that Sheltie color is determined by three loci.

At the A (agouti) locus, a^y is sable and a^t is tan-point (black and tan = referred to as tricolor if a white spotting gene is present.) An a^ya^t dog is sable, but generally somewhat darker than an a^ya^y dog. The difference is generally of the same order as the difference within a^ya^y or within a^ya^t, so it is not possible to be absolutely sure whether a^t is present by looking at the dog.

At the M locus, Shelties have both M and m, as discussed earlier. Mm produces blue merle in a^ta^t dogs and sable merle on a^ya^t and a^ya^y.

At the S (spotting) locus most correctly marked Shelties have two copies of sⁱ, Irish spotting. A sⁱsⁱ dog generally ranges from white on the chest and feet to high white stockings, white tail tip, and a full shawl collar. The probability of the full collar, as well as white stifles, seems to be somewhat enhanced if the dog is sⁱs^w, so s^w, color headed white, tends to be maintained in the breed as well. (I am keeping it simple by ignoring s^p, piebald, which may also occur in Shelties.) s^ws^w dogs are predominantly white with color on the head and perhaps a few body spots. While healthy, they cannot be shown.

Suppose we mate two white-factored, tri-factored sable merles (not a likely mating, but this is an illustration!) The genetic formula for each parent is a^ya^tMmsⁱs^w. There are eight possible gametes for each sex:

	ayMsi	ayMsw	aymsi	aymsw	atMsi	atMsw	atmsi	atmsw
ayMsi	DMS	DMS	SM	SM	DMS	DMS	StM	StM
ayMsw	DMS	DMS*	SM	WSM	DMS	DMS*	StM	WStM
aymsi	SM	SM	S	S	StM	StM	St	St
aymsw	SM	WSM	S	WS	StM	WStM	St	WSt
atMsi	DMS	DMS	StM	StM	DM	DM	BM	BM
atMsw	DMS	DMS*	StM	WStM	DM	DM*	BM	WBM
atmsi	StM	StM	St	St	BM	BM	T	T
atmsw	StM	WStM	St	WSt	BM	WBM	T	WT

There is no way I could fill in this chart with the detail I used in the 2-loci charts and still have it fit readably into a browser window, so I have used a shorthand to indicate the apparent color:

• S = pure for sable with Irish markings (3)
• St = tri-factored sable with Irish markings (6)
• T = tricolor with Irish markings (3)
• SM = pure for sable merle with Irish markings (6)
• StM = tri-factored sable merle with Irish markings (12)
• BM = blue merle with Irish markings (6)
• WS = white with pure for sable head (1)
• WSt = white with trifactored sable head (2)
• WT = white with tricolor head (1)
• WSM = white with pure for sable merle head (2)
• WStM = white with tri-factored sable merle head (4)
• WBM = white with blue merle head. (2)
• DMS = homozygous merle, dilute sable markings (12)
• DM = normal homozygous merle (4)

I have not distinguished white-factored from Irish dogs, and I have ignored the possibility that the MMs^ws^w pups (starred in chart) might not be viable. In practice such a breeding would probably never be made, as Sheltie breeders tend to avoid breeding merle to merle and white factor to white factor, but it does illustrate the variety that can be obtained with two alleles at each of three loci.

In this case, all three loci are visibly affecting the color. The only exception is the interaction between color-headed white and double merle, and this is frankly an unknown. There are times, however, when a particular gene combination at one locus can block expression of a gene combination at another locus. I will follow Searle on nomenclature and distinguish between a dominant-recessive relationship between alleles at a particular locus and an epistatic-hypostatic relationship between two loci.

The first example is very obvious, but only because the gene action is clear-cut. Consider Cocker Spaniels. They have two alleles at the S locus (S, fully colored, and s^p, piebald.) An SS dog is solid color, an Ss^p dog may have minor white marking (and is often unshowable) and an s^ps^p dog is a parti-color. The second gene is ticking. Ticking works by producing flecks of color in white areas. TT produces ticks of color in any white areas on the dog, tt has clear white areas, and Tt probably produces less ticking than TT, with considerable variation among breeds. s^ps^p and probably Ss^p dogs will show ticking if T is present, since they have white areas that are "available" for ticking, though if the base color is red, tan or cream the ticking may not be obvious. But if the dog is SS, there are no white spots for the ticking to show up on. SS is thus epistatic to ticking.

The final example involves the genes for dominant black, which may or may not (my feeling is probably not, as there are records of dominant black to tricolor producing both sables and tris) be the top dominant in the A series where it is generally placed. I will assume it is at a separate locus K, with K being dominant black, epistatic to anything at the A series, while kk allows the A series to show through. We also have the E series, in which E allows the A series to show through while ee allows only red-yellow pigment in the hair. Functionally we can consider that the A locus determines where eumelanin and phaeomelanin are produced, the K locus allows only eumelanin to be produced if E is at the E locus, but ee at the E locus overrides that to allow only phaeomelanin production. Sounds like a mess? You bet it does! K at the K locus is epistatic to the A locus, but ee (pure recessive at the E locus) is epistatic to both the A and the K locus. But it agrees with what is observed.

Let's look at a breed cross between two "red" dogs. We'll take an accidental breeding I know of between a Belgian Tervuren (a^ya^yEEkk) and a Golden Retriever (??eeKK). Note that ee is epistatic to the A series, so if dominant black is not at the A locus, we do not know what the normal A allele is in the Golden. The gametes are a^yEk for the Terv and ?eK for the Golden. Every puppy inherits a^y?EeKk and is black, as was in fact observed (to the initial astonishment of the owner.) If we mated two of these pups, we would get a 16/64 probability of ee which would be red regardless of what was at other loci. Of the other 48/64 (Ee and EE dogs), 75% would be Kk or KK, and hence black. so there is a 36/64 probability that a particular puppy will be black. The remaining 12/64 will show what is present at the A locus. Of the 12, we expect that 9 will have the a^y gene in at least one dose, and with dominant black moved to the K locus a^y is dominant over all other A alleles. So there is only a 3/64 chance that a given puppy will actually show what A allele is normal in a Golden - if in fact all Goldens have the same allele at A!

Note that in this particular case we can get identical results in the first generation by postulating a top dominant A^s dominant black at the A locus, with A^s- dogs having solid eumelanin pigmentation (unless overridden by ee.) In this case the parent gametes would be A^se and a^yE, giving A^sa^yEe black pups. In the next generation we would again get 4/16 ee red, 9/16 A^s-E- black, and 3/16 a^ya^yE- sables. If we could be absolutely sure that the Terv used was not a^ya^t, the appearance of a tricolor would be good evidence for the first hypothesis.

We still need to discuss penetrance, variable expression, and threshold traits, as well as linkage and crossing over (and their influence on the accuracy of DNA testing), test breeding, and testing whether a suspected allele is in fact at a particular locus. Some further comments about merle are also in the works.

 

Basic Genetics III: Linkage and Crossing  Over

Up until now we have assumed that all genes were inherited independently. However, we have also said that genes are arranged on chromosomes, which are essentially long strands of DNA residing in the nucleus of the cell. This certainly opens the possibility that two otherwise unrelated genes could reside on the same chromosome. Does independent inheritance hold for these genes?

To start with, we need to consider the rather complex process that forms gametes (egg and sperm cells, each with only one copy of each chromosome) from normal cells with two copies of each chromosome., one derived from each parent. I am not going to go into the details, beyond remarking that at one stage of this process, the maternally-derived chromosome lines up with the corresponding paternally-derived chromosome, and only one of the two goes to a specific gamete. If this were all there were to it dogs, having 39 chromosome pairs, would have only 39 "genes", each of which would code for a wide variety of traits. In fact, things are a little more complicated yet, because while the paternal and maternal chromosomes are lined up, they can and do exchange segments, so that at the time they actually separate, each of the two chromosomes will most likely contain material from both parents.

At this point we need to define a couple of terms. Two genes are linked if they are close together on the same chromosome and thus tend to be inherited together. Linkage in common usage, however, may apply to a single gene having more than one effect. An example which is not linkage in the sense used here is the association between deafness and extreme white spotting. White spotting is due to the melanocytes, the cells which produce pigment, not managing to migrate to all parts of the fetus. Now it turns out that in order for the inner ear to develop properly, it must have melanocytes. If the gene producing white spotting also prevents the precursors of the melanocytes from reaching the inner ear, the result will be deafness in that ear. In other words, the same gene could easily influence both processes. Thus deafness and white spotting are associated, but they are not linked. They are due to what is called pleiotropic (affecting the whole body) effects of a single gene.

In true linkage, there is always the possibility that linked genes can cross over. Imagine each chromosome as a piece of rope, with the genes marked by colored stripes. The matching of the maternal and paternal chromosomes is more or less controlled by the colored stripes, which tend to line up. But the chromosomes are flexible. They bend and twist around each other. They are also self healing, and when both the maternal and paternal chromosomes break, they may heal onto the paired chromosome. This happens often enough that genes far apart on long chromosomes appear to be inherited independently, but if genes are close together, a break is much less likely to form between them than at some other part of the paired chromosomes.

Such breaks, called "crossing over" do occur, and occur often enough that they are used to map where genes genes are located on specific chromosomes. In general, neither linkage nor crossing over is of much importance to the average dog breeder, though one should certainly keep in mind the possibility that the spread of an undesirable gene through a breed is due to the undesirable gene being linked to a gene valued in the breed ring. Crossing over is also important in the use of marker genes for testing whether a dog carries a specific gene, most often a gene producing a health problem.

There are two distinct ways of using DNA testing to identify dogs carrying specific, undesirable genes. The first (and preferable) is actually to sequence the undesirable gene and its normal allele. This allows determination of whether the dog is homozygous normal, a heterozygous carrier, or homozygous affected. Since the genes themselves are being looked at, the results should be unambiguous. (The breeding decisions based on these results are still going to depend on the priorities of the breeders.)

In some tests, however, a marker gene is found that appears to be associated with the trait of interest, but is not actually the gene producing that trait. Such a marker is tightly linked to the gene actually causing that trait. This does not work at all badly providing that the group on which the test was validated is closely related to the group to which the test was applied. Use of this type of test on humans usually requires that the test be validated on close relatives, and applied only to people closely related to the validation group.

It is true that dogs of a given breed tend to be closely related to each other. However, the breed-wide relationship is generally through more distant ancestors than most people can trace in their own genealogy. In Shetland Sheepdogs, for instance, almost all US show stock can be traced to dogs imported from the British Isles between 1929 and 1936, with only a tiny influence of imports after 1950. This means that a crossover appearing on one side of the Atlantic since 1950 (20 or so dog generations) might not show up on the other side. Marker tests that work on U.S. populations might not work at all on British dogs, or on a dog with recent British ancestry.

Even without physical separation here is always the possibility that at some point in the breed history a crossover occurred. Quite a large fraction of the breed may have the original relationship between the marker gene and the problem gene, but if a crossover occurred in an individual who later had a considerable influence on the breed, the breed may also contain individuals in which the marker gene is associated with the opposite form of the problem gene. Since the relationship between individuals of the same breed may go back 30 generations or more, and there is a chance of a crossover occurring in each generation, linked markers need to be used with caution and with constant checking that marker test results correlate with clinical results.

Let's look more closely at this.

Let our marker gene be ma, with ma^a being the gene associated with the healthy gene, and ma^b being the marker that seems to be associated with the defective gene, both being true for the test population. For the genes actually producing the problem, we will use H, with H^h being the normal, healthy gene and h^d being the recessive gene which causes the problem. In the original test population, ma^a was always on the same chromosome with H^h, and ma^b was on the same chromosome with h^d. In other words, chromosomes are either ma^aH^h or ma^bh^d, never ma^ah^d or ma^bH^h. If a dog has ma^a on both chromosomes, it is also H^h on both chromosomes, a genetic clear. If it has ma^a on one chromosome and ma^b on the other, it also has one H^h gene and one h^d gene, and is a carrier. If it has ma^b on both chromosomes, it has h^d on both chromosomes and is a genetic affected. At least, that is the assumption on which marker tests are based.

Now suppose that at some point a crossover occurred between the ma and H loci. The probability of a crossover may be very small in any individual breeding, but remember that there are a lot of breedings behind any particular dog. We can still assume that most of the chromosomes will still be of the ma^aH^h or ma^bh^d type, or the original validation of the marker test would have failed. But now suppose that a small fraction of the chromosomes are of types ma^ah^d and/or ma^bH^h. We now have four chromosome types, and sixteen possible combinations. Some of these will test the same, since the only difference is in which chromosome comes from the mother and which from the father, but there are still sixteen possible outcomes. In the table below both the marker results (upper) and the true results (lower) are shown for each possible combination:

	maaHh	mabhd	maahd	mabHh
maaHh	clear maamaa	carrier maamab	clear maamaa	carrier maamab
	clear HhHh	carrier Hhhd	carrier Hhhd	clear HhHh
mabhd	carrier maamab	affected mabmab	carrier maamab	affected mabmab
	carrier Hhhd	affected hdhd	affected hdhd	carrier Hhhd
maahd	clear maamaa	carrier maamab	clear maamaa	carrier maamab
	carrier Hhhd	affected hdhd	affected hdhd	carrier Hhhd
mabHh	carrier maamab	affected mabmab	carrier maamab	affected mabmab
	clear HhHh	carrier Hhhd	carrier Hhhd	clear HhHh

Note that in only six of the sixteen possible types is the marker indication of genotype correct. If the crossover genotypes are rare (as would normally be the case if the marker test verified at all) most of the population will be in the upper left quarter of the table, where the marker will correctly predict the true genotype. But if any of the chromosomes trace back to a crossover, a marker test may give a false sense of security (carrier or affected shows clear by marker testing) or result in discarding a healthy dog (carrier or clear shows affected or carrier by marker testing.)

If only three chromosome types are available, the two verifying types plus one crossover, then if the marker gene is associated at times with the healthy allele, (ma^bH^h) the result will include dogs which are affected or carriers by marker analysis which are genetically carriers or clears (false positives.) If the other chromosome type has the undesirable allele not always associated with the marker (ma^ah^d) the results will include dogs clear or carriers by marker analysis that are actually carriers or affected (false negatives.) However, the existance of one crossover chromosome type would make me suspicious that the other might also exist in the breed.

So are marker tests of any use at all?

Yes! In the first place, they demonstrate that the actual gene is on a relatively limited portion of a known chromosome. The marker gene can thus assist in finding and sequencing the gene actually causing the health problem.

In the second place, marker tests are accurate so long as neither parent of an individual has a crossover chromosome. In humans, such tests are most likely to be used when a problem runs in a particular family. The linkage of a marker with the genes actually producing the problem is generally based on studies of how the marker is linked to the genes in that particular family. With dogs, the verification is normally done on a breed basis, and the fact that breeds may actually be split into groups (color, size, country of origin) which interbreed rarely if ever is likely to be ignored. Dogs closely related via close common ancestors to the test population are the best candidates for marker testing. In general, keep up conventional testing side by side with the marker testing. If the marker testing and the conventional testing disagree (e.g, affected dog tests clear or clear dog tests affected) consider the possibility of a crossover, and notify the organization doing the test.

Basic Genetics IV:
the relationship of genes to traits (single locus)

With the exception of the few DNA tests available, we cannot know the genetic makeup of our dogs, only the physical makeup, or phenotype. We tend to break that phenotype up into traits, some breed specific, some more general. For instance, we might know that a Sheltie is 15" tall, a black-nosed sable merle with full white collar, feet and Teletype and a narrow face blaze, OFA good, is missing one premolar, has natural ears, and had double rear decals. All of these "traits" are defined by human beings. Very few of them actually refer to single genes that might be inherited as dominant, recessive, incompletely dominant or co-dominant.

In some cases we can break down a trait into a specific combination of genes. In the case of color, for instance, we know of a considerable number of genes that affect color through specific processes. In some cases, this knowledge has fed back on what we consider to be traits. Thus in the case given, the dog is:

• Sable a^y- (as opposed to black with or without tan-point markings).
• Black (as opposed to brown) B-
• Merle Mm
• Irish-marked sⁱsⁱ or sⁱs^w
• Possibly a face-marking gene

In addition, the dog's color can be affected by minor genes (such as the modifier genes determining how much of the dog is white) by random factors (which probably influence the exact pattern of both white spotting and the location of the dark patches in the merling) and by environmental factors (such as uterine environment, nutrition or excessive exposure to the sun.) The point is that very few of the traits that humans have chosen are in fact due solely to the effect of a single pair of alleles at a single locus. We have looked at some such simple traits as regards color.

However, the height of the dog, the ears, the hip rating, the missing premolar, and the double rear decals are probably not single-gene traits, but rely on the interaction of several pairs of genes, with perhaps some influence from the environment.

In general I am using dominant, recessive, co-dominant or intermediate to refer to genes at the same location on a single pair of chromosomes, i.e., alleles at the same locus. There are cases where genes at one locus can "hide" genes at another locus. An example in dogs is recessive yellow, ee, in which recessive yellow, although a recessive at its own locus, can hide whatever the dog carries at the A locus and the proposed K (dominant black) locus. This type of relationship among different loci is called epistatic. The locus that is hidden is referred to as hypostatic. In some cases (e.g., E at the E locus) an epistatic locus has an allele that allows the hypostatic locus to show its effects.

We will consider a number of types of inheritance. The first group actually refer to single-gene traits. Any of these types of inheritance may also be involved in the inheritance of multiple-gene traits.

Single-locus inheritance

• Dominant-recessive
• Intermediate
• Co-dominant
• Sex-limited
• Sex-linked

More complex inheritance will be covered on the next page, and includes

• Modifier genes
• Polygenic additive
• Threshold traits
• Variable expression
• Incomplete penetrance
• Polygenic recessive or dominant
• Mixed polygenic

Dominant-recessive inheritance

Black and brown provide a clear example of a dominant-recessive relationship among alleles. Every dog has two genes at the black/brown locus. If both genes are for black, or if one is for black and one is for brown, the dog is black, most readily identified by nose color. If both genes are for brown, the dog is brown, again most readily identified by nose color. BB cannot be distinguished from Bb without genetic tests or breeding tests.

Many genetic diseases, especially those that can be traced to an inactive or wrongly active form of a particular protein, are inherited in a simple recessive fashion. van Willebrand's disease (vWD) for instance, is inherited as a simple recessive within the Shetland Sheepdog breed.

Intermediate inheritance

Warning! Although this type of inheritance is common, it has a variety of names (incomplete dominance and overdominance are two common ones) some of which are also used for other things entirely. Here I will use it to refer to the type of inheritance in which the animal carrying two identical alleles shows one phenotype, the animal carrying two different identical alleles shows a different phenotype, and the animal carrying one copy of each of the alleles shows a third phenotype, usually intermediate between the two extremes but clearly distinguishable from either.

In dogs, merle color is a good example of this type of inheritance. If we define M as merle and m as non-merle, we find we have three genotypes:

• mm non-merle, with normal intense color
• Mm merle, with normal color diluted in a rather patchy fashion
• MM homozygous merle, extreme dilution, dog mostly white if a white-spotting gene is also present, and often with anomalies in hearing, vision and/or fertility.

Note that there is really a continuum between dominant-recessive and intermediate inheritance. In Shetland Sheepdogs, for instance, sables carrying one gene for tan-point have on average more dark shading than dogs with two sable genes. However, the darkest shading on dogs pure for sable is probably darker than the lightest shading on dogs carrying a gene for tan-point. In practice, intermediate inheritance is often treated as if it were a special case of dominant-recessive inheritance, as can be seen by the symbols used for merle and non-merle - usually the capital letter refers to a dominant gene and the lower-case letter refers to a recessive gene. I think a separate name is justified because it could be equally well argued that homozygous merle is an undesirable recessive for which the merle color is a marker that the dog carries the merle gene.

Many of the standard color genes normally treated as dominant-recessive do in fact have intermediate inheritance, the heterozygote generally much more similar to one homozygote than the other, between at least some alleles in the series. Coat color gene loci with at least some allele pairs leaning toward intermediate inheritance include A (agouti, patterning of black and tan), C (color, intensity of color), and S (white spotting). I suspect the same is true for T (ticking), G (graying) and even D (dilution) if another diluting gene, such as merle, is present. This may be much more generally true than is recognized.

Co-dominant inheritance

The dividing line between intermediate inheritance and co-dominant inheritance is fuzzy. Co-dominance is more likely to be used when biochemistry is concerned, as in blood types. Co-dominance means that both alleles at a locus are expressed. Co-domininance in X-linked genes is a special case that will be treated under sex-linked inheritance.

Sex-limited autosomal inheritance

Please, don't confuse sex-limited inheritance with sex-linked inheritance. They are two totally different things. Sex-linked inheritance is discussed below. I do include sex-influenced traits under the sex-limited heading, though some genetics texts separate sex-influenced and sex-limited traits.

A classic example of a sex-limited trait in dogs is unilateral or bilateral cryptorchidism, in which one or both testicles cannot be found in their usual position in the scrotum. Since a bitch has no testicles, she cannot be a cryptorchid - but she can carry the gene(s) for cryptorchidism, and pass them to her sons. Likewise, genes affecting milk production are not normally expressed in a male. The main problem with sex-limited inheritance is that it is impossible to know even the phenotypes of the unaffected sex in a pedigree, which makes it difficult to determine the mode of inheritance.

In sex-influenced inheritance, the genes behave differently in the two sexes, probably because the sex hormones provide different cellular environments in males and females. A classic example in people is male early-onset pattern baldness. The gene for baldness behaves as a dominant in males but as a recessive in females. Heterozygous males are bald and will pass the gene to about 50% of their offspring of either sex. However, only the males will normally be bald unless the mother also carries the pattern baldness gene without showing it (female heterozygote.) If the mother is affected with baldness (homozygous) but the father is not, all of the sons will be affected and all of the daughters will be non-affected carriers. A bald man may get pattern baldness from either parent; a bald woman must have received the gene from both parents.

Sex-linked inheritance

In order to understand sex-linked traits, we must first understand the genetic determination of sex. Every mammal has a number of paired chromosomes, that are similar in appearance and line up with each other during gamete production (sperm and eggs). In addition, each mammal has two chromosomes that determine sex. These are generally called X and Y in mammals. Normal pairing of chromosomes during the production of gametes will put one or the other in each sperm or ovum.

In mammals, XY develops testicles which secrete male sex hormones and the fetus develops into a male. An XX fetus develops into a female. Thus sperm can be either X or Y; ova are always X. Sex linked inheritance involves genes located on either the X or the Y chromosome. Females can be homozygous or heterozygous for genes carried on the X chromosome; males can only be hemizygous.

X-linked recessive:

The most common type of sex-linked inheritance involves genes on the X chromosome which behave more or less as recessives. Females, having two X chromosomes, have a good chance of having the normal gene on one of the two. Males, however, have only one copy of the X chromosome - and the Y chromosome does not carry many of the same genes as the X, so there is no normal gene to counter the defective X.

An example of this type of inheritance is color blindness in human beings. Using lower case letters for affecteds, we have

• Affected male: xY Color blind
• Non-affected males XY Normal color vision
• Affected female xx Color blind
• Carrier female xX Normal color vision
• Clear female XX. Normal color vision

Now the possible matings:

xY to xx (both parents affected) xx females and xY males, all offspring affected.

xY to Xx (affected father, carrier mother) half the females will be xX and carriers, half will be xx and affected. Half the males will be XY and clear, half will be xY and affected.

xY to XX (affected father, clear mother) all male offspring XY clear, all daughters Xx carriers.

Note that the daughters of an affected male are obligate carriers or affected. The unaffected sons of an affected male cannot carry the problem.

XY to xx (father clear, mother affected) xY males (affected) and xX daughters (carriers.)

XY to Xx (father clear, mother carrier) half the males affected (xY) and half clear (XY); half females clear (XX) and half carriers (Xx)

XY to XX (father and mother both genetic clears) all offspring clear.

Note that all female offspring of affected males are obligate carriers (if not affected.) Likewise, any female who has an affected son is a carrier. Non-affected sons of affected fathers are genetically clear.

This type of inheritance may be complicated by the sublethal effect of some X-linked genes. Hemophilia A in many mammals (including dogs and people) is a severe bleeding disorder inherited just like the color-blindness above. Many affected individuals will die before breeding, but for those who are kept alive and bred for other outstanding traits, non-affected sons will not have or produce the disease. All daughters, however, will be carriers.

X-linked dominant:

Here I will use X⁺ for the dominant gene on the X chromosome, and X for the gene on the normal X chromosome. The actual possibilities are similar to those for an X-linked recessive, except that X⁺X females are now affected. In X-linked dominant inheritance, more females than males will show the trait. Possible matings are:

Affected to homozygous affected (X⁺Y to X⁺X⁺): All offspring affected.

Affected to heterozygous affected (X⁺Y to X⁺X): All daughters affected; half of sons affected.

Affected to homozygous normal (unaffected female): (X⁺Y to XX): All daughters affected, all sons normal.

Normal to homozygous affected (XY to X⁺X⁺): all offspring affected, but daughters are heterozygous affected.

Normal to heterozygous affected: (XY to X⁺X): Half of offspring affected, regardless of sex. Affected daughters are heterozygous.

Normal to normal (XY to XX) all offspring clear.

X-linked co-dominant:

Mammalian cells, even in females, get along fine with just one X chromosome. In fact, more than one X chromosome within a cell seems to be a problem if both are active. So in female cells, one or the other X chromosome must be inactivated. This occurs more or less at random, so any female mammal has patches of cells with one X chromosome inactivated, and patches with the other not active. If the gene being discussed codes for an enzyme that is spread throughout the body, it may not be obvious that the different patches of cells are behaving differently, and we will get what looks like dominant, recessive, or intermediate inheritance.

However, if the gene is expressed directly within the cell, the mosaic nature of the female may become obvious. The tortoiseshell cat provides an excellent example of this.

In cats, the orange color is on the X chromosome. It is designated as O, and the "wild-type" gene that allows black (eumelanin) to appear in the coat is designated +. Note that a cat homozygous or hemizygous (male) for + may be solid or tabby with the eumelanin pigment showing only in the tabby stripes, ticks and blotches (in extreme cases only on the tips of the hairs) and the "black" may just as well be chocolate or blue. A cat with only O genes will be some shade from cream to deep red., with no black/blue/chocolate pigment in the coat, but usually with tabby markings.

However, a cat with the gene for orange on one X chromosome and the gene for non-orange on the other is neither orange nor non-orange, but has patches of both colors. This color is known as tortoiseshell, and I am going to use the broad definition, including blue/cream or chocolate/yellow tortoiseshells. Most of the time cats with two X chromosomes are female, and since two X-chromosomes are required for tortoiseshell, most tortoiseshell cats are female.

Now and then a cell does not divide properly when it is making a germ cell, and you might, for instance, get an XY sperm cell. This would produce an XXY male, which would look male (he has a Y chromosome) but also have two versions of X and thus could be a tortoiseshell. However, the XXY makeup, corresponding to Klinefelter's syndrome in human beings, is believed to produce sterility. A similar syndrome involving females with only one X chromosome but no Y is called Turner's syndrome in human women, and again appears to produce sterility. We will therefore consider only matings between animals with two sex chromosomes.

Non-orange male to non-orange female (+ to ++): all non-orange offspring.

Non-orange male to tortoiseshell female (+ to +O): Males 50% orange and 50% non-orange; females 50% non-orange and 50% tortoiseshell.

Non-orange male to orange female (+ to OO): all males orange; all females tortoiseshell.

Orange male to non-orange female (O to ++): All males non-orange; all females tortoiseshell.

Orange male to tortoiseshell female (O to O+): males 50% orange and 50% non-orange; females 50% orange and 50% tortoiseshell.

Orange male to orange female (O to OO): All offspring orange.

Y-linked inheritance:

The Y chromosome in most species is very short with very few genes other than those that determine maleness. Y-linked inheritance would show sons the same as their fathers, with no effect from the mother or in daughters. In humans, hairy ears appear to be inherited through the Y chromosome. Padgett does not list any known problem in dogs as being Y-linked.

 

Canine Color Genetics

Dogs have a wide variety of genes that influence color. Further, the same genes may give a very different effect on different types and lengths of coats. While this site is primarily concerned with Shetland Sheepdog colors and a long, working-type (double) coat, I will use comparisons from other breeds and even other species whenever it seems useful. References, including other mammalian color genetics, are on a separate page.

One of the biggest problems people have with genetics is the assumption that a defined trait - size, ear type, color, yappiness - is due to a single gene. In fact, genes code for two types of things. One, which is relatively well understood, is the structure of a particular protein. The normal equivalent of the albino gene, for instance, codes for tyrosinase, an enzyme which breaks up the amino acid tyrosine as a first step in producing melanin, the major pigment in mammalian skin and hair. In an albino, this enzyme cannot be produced, and as a result melanin cannot be produced. A second type of gene controls when and where other genes are turned on or off. These genes are the subject of vigorous ongoing study, and probably have a major impact on such things on the number of vertebrae in the spine or the age at which growth is complete. I've included a page which defines some of the terms used in genetics, as well as explaining dominant, recessive and incompletely dominant genes. Right now, let's look at some of the gene series (loci) known to influence canine color, and try to get a feel for what they do.

Before starting our list, we need to know that mammals have two forms of melanin in their coats. One, eumelanin, is dark, though it can vary somewhat in color due to variations in the protein that forms the framework of the pigment granule. The base form of melanin is black. Melanin can also appear brown (often called liver in dogs) or blue-gray. The second pigment, which varies from pale cream through shades of yellow, tan and red to mahogany (as in the Irish Setter), is called phaeomelanin. There are at least two and possibly as many as four gene series that determine where, on the dog and along the length of the hair, eumelanin and phaeomelanin appear.

The generally recognised color series (loci) in dogs are called A (agouti), B (brown), C (albino series), D (blue dilution) E (extension), G (graying), M (merle), R (roaning), S (white spotting) and T (ticking.) There may be more, unrecognised gene series, and in a given breed modifying factors may drastically affect the actual appearance. Thus one school of thought holds that the round spots on a Dalmation are due to the same gene that produces the roaned areas on a German Shorthair Pointer, but with vastly different modifiers.

A, the agouti series. The standard assumption, based on Little's research, is that this series contains four alleles (different forms of the gene). A fifth allele may exist in Shetland Sheepdogs, and a sixth in certain "saddle-tan" breeds.

• A^s produces black without any tan on the dog. White markings are due to a different gene, and there are other genes that can modify the black to liver (chocolate Lab) or blue dilute (blue Great Dane.) If A^s is present, in most cases the dog will be able to produce only eumelanin pigment (but see the E series). Note that the agouti series is known in a number of mammals, and dominant black is almost always found in a different series, so there is a strong possibility that dominant black is not really in the agouti series.
• a^y in the absense of A^s produces a dog which is predominantly tan (phaeomelanin) sometimes with black tipped hairs or interspersed black hairs. The usual term for this color is "sable." In examining dogs from a^y breeds, I have generally found that even if there is no other black on the coat, the whiskers (the course, stiff vibrissae, not the "beard" seen with some terrier coats) are black if they originate in a pigmented area. Examples of a^y dogs include Collies, fawn Boxers and Great Danes, and some reds (Basenji red is thought to be a^y, for instance.) a^y is recessive to A^s, but incompletely dominant to a^t. That is, an a^ya^t dog is on average darker (more black hairs) than an a^ya^y dog, but the difference is generally within the range of color for a^ya^y within the breed.
• a^t, present in double dose, produces a dog which is predominantly black, with tan markings on the muzzle, over the eyes, on the chest, legs, and under the tail. A Dobermann or Rottweiler is a good example of the classic black and tan pattern. The Bernese Mountain Dog shows the effect of black and tan combined with white markings, often called tricolor.
• a^w is the fourth allele considered by Little. This is the wild "wolf-color" seen in Norwegian Elkhounds and possibly in some salt-and pepper breeds. It differs from sable in two ways. First, the tan is replaced by a pale cream to pale gray color. Second, the hairs are normally banded - not just the scattering of black-tipped hairs sometimes seen in a sable, but several bands of alternating light and black pigment along the length of the hair. Little was unable to determine the dominance relationship of this gene, or even to say with certainty that the banding and the reduction of tan pigment were due to the same gene.

Although Little did not make any distinction between the Dobermann black and tan and the "saddle tan" seen in many terrier breeds (black "saddle" but extensive tan on legs and head), it seems likely that a fifth gene exists in the a series. For the moment I'll call it "saddle tan," a^sa. It seems recessive to a^y sable, but other dominance relationships in the series need more investigation.

Finally, at least two breeds (Shetland Sheepdog and German Shepherd) have a fully recessive black. Since black is the bottom recessive of the A series in many other mammals, it seems logical to assign this color to recessive black, a, and state that recessive black is caused by aa at the agouti locus. There is an alternative theory in Shelties which suggests the existence of a recessive gene that removes tan points from a genetic black and tan or a dominant, widespread gene that forms tan points on all colors but dominant black.

Little's assignment of dominant black in dogs to the A locus (A^s) is totally against experience with this locus in other species, where more yellow is generally dominant to more black. There may be a third locus controlling dominant black, in which case A^y would be the top dominant in the A series.

B, the brown series. This series is relatively simple. B, in single or double dose, allows the production of black pigment. A bb dog produces brown pigment wherever the dog would otherwise have produced black. The gene apparently codes for one of the proteins that makes up the eumelanin pigment granule, so the bb granules are smaller and rounder in shape as well as appearing a lighter color than those of a dog carrying B. This gene is responsible for a number of liver and chocolate colors, especially in the sporting breeds. The same gene produces some "reds" (in Australian Shepherds, Border Collies, and Dobermanns, for example), and probably the bronze Newfoundland. It has some effect on the iris of the eye and on the skin color, including the eye rims and the nose leather. Phaeomelanin (tan) is very little affected, so the color of the tan points on a red Dobermann (a^ta^tbb), for instance, is little affected. I have seen little discussion of the effect of brown on a sable dog, but I would expect a brown nose leather and eye rims, with the coat shaded brown rather than black. Probably the dog would closely resemble a sable, perhaps with an orangey cast and a light nose. Note that some shades of liver, though a eumelanin pigment, overlap some shades of tan, a phaeomelanin pigment. In particular the deadgrass color (bbc^chc^ch) can overlap recessive yellow (ee)

C, the albino series. This again is a fairly complex locus, especially in other mammals. The top dominant, C, allows full color to develop, and is probably the structural gene for tyrosinase. The bottom recessive, c, does not appear to occur in dogs, but in other mammals it completely prevents the formation of any melenin in the coat or the irises of the eyes, giving a pink-eyed or red-eyed white. It is worth pointing out that human albinos from dark-skinned parents often show some yellowish or reddish hair and even skin color, but it seems this is not due to granular melenin. c, therefore, is a form of tyrosinase which cannot act as it is intended to in the formation of melanin. Since c is simply a non-working form, there may be more than one form of c gene (lots of ways to get something not to work), and there is some evidence that when two different forms are mated, colored offspring may result.

There are a number of intermediate genes where the mutation apparently produces a partly active form of tyrosinase. Some C alleles known in other mammals are:

• C full color, allows full expression of whatever pigment is prescribed by other genes. Most dogs are CC.
• c^ch, chinchilla or silver, when present in double dose removes most or all of the phaeomelanin pigment with only a slight effect on black pigment. This is named after a small fur-bearing South American rodent called the chinchilla. Black and silver replacing black and tan, or a wolf-like color without the extra banding (see a^w, above) may also be due to a c^chc^ch genotype. Dogs with very light tan probably are c^chc^ch or something similar. Liver dogs show lightening even of eumelanin pigment, and the "deadgrass" color of the Chesapeake Bay Retriever is thought to be due to a bbc^chc^ch genetic makeup. The possibility of other, rufous modifiers affecting the shade of phaeomelanin pigment needs to be kept in mind, as does the possibility of more than one form of chinchilla in the dog - rabbits are thought to have three.
• c^e, extreme dilution, has also been proposed for the dog. This gene may be part of the makeup of some "white" dog breeds where the white color is due to extreme dilution of tan. The West Highland White Terrier may be c^ec^eee. A cross to a black and tan breed would be interesting from the point of view of color genetics. Eyes may be lightened in some species, but this is doubtful in dogs.
• c^h, Himalyan, is not known to occur in the dog. In homozygous form, it makes the formation of eumelanin dependant on the temperature of the skin. Thus a genetically solid black animal will have reduced black on the extremities (seal brown) and an almost white color on the body. The effect on tan/orange pigment is confusing - the tan in agouti hairs is removed, but that resulting from the orange gene in cats (not in dogs) remains intense on the extremities. There is reason to suspect that this gene, as well as some forms of chinchilla, also affects the organization of the brain, particularly in the neural pathways from the eyes to the brain. There may be a reason for Siamese cats to be cross-eyed. Eyes are normally blue or pink.
• c^p, platinum, is optically similar to albino but retains very slight tysonase activity and in the mouse is described as retaining some luster in the coat as opposed to the pure white seen in albino. Although there is a total absense of proof one way or the other, I would hypothesize that the white Doberman, with pale blue eyes and pink nose, is due to a homologous gene.
• c, albino, is not known to occur in the dog as a regular part of any breed color, though possible candidates for mutations to c have been recorded. As mentioned above, the c gene cannot produce working tyrosinase, and a cc individual cannot produce melanin pigment.

As seen from the above, C is known to have a number of different forms and effects. The usual assumption is that dogs have at least one mutant allele, c^ch which when homozygous lightens phaeomelanin (yellow) pigment to cream and more weakly affects liver and longhaired black. A second proposed allele, c^e may be responsible for further reduction of cream to white in some breeds, or modifying alleles may be responsible for the further lightening in these cases. While some forms of C modify eye pigment (e.g., blue eyes in Siamese cats) there is little evidence for this in dogs unless "white" Dobermans are indeed due to a C-locus mutation. Although C appears to be fully dominant over any of the other alleles, the dominance relationship between the others generally goes in the direction of more color incompletely dominant over less color, the heterozygote generally resembling but not necessarily identical to the homozygote with more pigment

D, the dilution series. This, again, is a relatively simple series, containing D (dominant, full pigmentation) and d (recessive, dilute pigment). In contrast to C, which has its strongest effect on phaeomelanin, or B, which effects only eumelanin, D affects both eumelanin and phaeomelanin pigment. It is thought to act by causing the clumping of pigment granules in the hair. Like B, it often affects skin and eye color, and in some breeds dd has been associated with skin problems. "Maltese blue" is a term often used to describe dd blacks. If a solid liver dog also is dd, the result is the silvery color seen in Weimararners and known as "fawn" in Dobermans. (In most breeds, fawn refers to a^y yellows.)

While dd acting on black or liver is a part of the genotype of several breeds, dd acting on sable is relatively rare. For one thing, the action of dd on phaeomelanin has been described as a flattening or dulling of color. The cinnamon color in Chows is probably due to an a^ya^ydd genotype, but otherwise the combination of dd with phaeomelanin coat color seems limited to breeds in which color is of little importance (e.g., blue brindle in Whippets.)

Although D is usually described as completely dominant to d, I have seen one blue merle Sheltie bitch who suggested that this may not always be the case. The black merling patches in this bitch were actually an extremely dark blue-gray. Other than this she was an excellently colored blue merle. The owner insisted that she was not a maltese blue, but that she had relatives who were. I suspect that this bitch may have been Dd, with the additional diluting effect of the merle gene allowing the normally hidden effect of a single dose of d to show through.

E, the extension series. This series is probably the least satisfactory of those generally assumed to exist in the dog. In most mammals, the E series includes E^d (dominant black), E (normal extension) and e (recessive red or yellow, and sometimes some intermediate alleles called Japanese brindles. In dogs, this is clearly not the case; breeding experiments have conclusively proven that dominant black and recessive red are not in the same series. This has led to dominant black being thrust into the A series, which as already mentioned conflicts with results in other mammals.

In this summary, I will give the genes as postulated by Little, followed by a brief discussion of other possible explanations and a suggestion for matings that might clarify the situation. Note that the question is not in whether the genes occur, but whether they are in fact alleles in the same gene series. With regard to e and E, recent sequencing of the e and E genes in dogs show definite homology with those in other species.

• E^m, mask factor. This gene replaces phaeomelanin (tan) with eumelanin (black) over part of the dog. There is considerable variation in the area of replacement, probably affected by modifiers but possibly involving more than one form of E^m. At its weakest the mask factor may produce black hair fringing the mouth, or a slightly smutty muzzle. At its strongest (Belgian Tervuren) most of the head is black, and there is considerable blackening of chest and legs. The effect of E^m shows to its fullest extent on clear sable dogs (a^ya^y), but is visible on the tan points of black and tan dogs (a^ta^t) as well. In its strongest version, it can change a black and tan to a pseudo-black, with tan so restricted in its distribution that it may not be immediately apparent that the dog is not black. The occasional "black" puppy produced by two Tervuren parents is probably this type of black, with two a^ya^tE^mE^m parents producing an a^ta^tE^mE^m puppy. A similar but not quite as strong blackening of the head of a genetic black and tan occurs in German Shepherds.
• E^br, brindle. This gene probably got into the E series by mistaken homology with Japanese brindle, which behaves quite differently from brindle in the dog. In Japanese brindle, the patchy color is believed to be due to two alleles of the E series side by side on the same chromosome. Only one can be expressed, and different parts of the animal will show the expression of different genes. The result is a coat made up of random small patches of tan and black pigment, rather like a tortoiseshell cat. If a Japanese brindle animal also has the genes for extensive white spotting, the tan and black pigmented areas tend to become larger and more compact, similar to what one sees in a calico cat (genetically, a tortoiseshell with white markings.) There is a canid which might be Japanese brindle with white spotting, the Cape hunting dog, Lycaon pictus. This animal has a coat which is a rather random patchwork of black, yellow and white. The color has very little similarity to brindle in the dog.
  Brindle in dogs consists of black, vertical stripes on a sable/fawn background, usually rather soft-edged, but much more regular that a typical Japanese brindle, and showing no tendency for the tan and black patches to become more distinct in the presense of white spotting genes. Genes that affect eumelanin will affect the dark stripes, so a bb brindle, for instance, will have brown rather than black stripes. Brindle on a black and tan will show only in the tan areas, while brindle on a black cannot be distinguished at all. If in fact recessive red (ee) is in the same series with brindle, it is not possible for brindle (or mask) to occur on an ee dog as one of the E genes would have to be E^br (or E^m), leaving no room for ee. Little implies that brindle and mask were co-dominant, with masked brindles being E^brE^m, in which case masked brindle could not breed true.
• E, normal extension of black, allows the A-series alleles to show through with no masking or brindling. It is apparently recessive to both E^m and E^br.
• e, recessive red, overrides whatever gene is present at the A locus to produce a dog which shows only phaeomelanin pigment in the coat. Skin and eye color show apparently normal eumelanin, although some ee dogs appear to show reduced pigment on the nose, especially in winter (snow nose.) A number of breeds show recessive red as a normal or even breed-wide characteristic - Irish Setters, Golden Retrievers, yellow Labradors. In a few breeds such as the Cocker Spaniel "reds" may be either a^ya^y or ee, and crossing the two can produce unexpected blacks. I believe there may be a key in the color of the whiskers, which on my observations seem to be black in a^ya^y breeds and straw to cream (dilute red) in ee breeds, always assuming the whisker base sprouts from a pigmented area. Little hypothesized that dogs with both forms of red (a^y-ee) were not viable and would be lost before birth.

The dominance relationships in the Little proposal are not simple. He assumes that E^m and E^br are co-dominant. In an a^ya^y dog, then, brindle without a mask could be E^brE^br, E^brE, or E^bre. A masked dog without brindling would be E^mE^m, E^mE or E^me. A masked brindle would have to have the genotype E^mE^br. This assumption makes some predictions which should be readily testable:

1. Two masked brindles, mated together, should produce appoximately a 1:2:1 ratio of masked fawn to masked brindle to brindle without masking. In other words, masked brindle should not breed true.
2. A masked brindle could not carry E or e. Thus a masked brindle, bred to sable a^ya^yE- would pass either mask or brindle. The expectation would be a litter of brindles without masks and masked sables (fawns) without brindling, but no sables without either mask or brindle and no masked brindles.
3. If a masked brindle is bred to an ee red, the results would depend on the A series genes in the ee red, but there would be neither ee nor a^ya^y reds with neither masking nor brindling. Some blacks might occur, but if the puppy had areas of tan pigment, the tan would be either masked or brindled, but never both and never tan without either mask or brindle.

My impression in talking to breeders of masked brindles is that these predictions are not fulfilled. Possible revisions of the E series include:

1. Remove E^br from the E series, instead recognising that in many ways it is closer to tabby (Ta) in the cat family. This is the gene series responsible for the various stripes, ticking, spots and rosettes seen in both wild and domestic cats. Granted, the pattern is not the same (striped cats normally have stripes ringing the legs), but brindle is also a black striping gene which is visible primarily on an a^y background. This would leave E^m, E and e in the E series, giving a prediction that E^m- bred to ee could produce either 100% masks if the mask is E^mE^m, half masks and half sables without masks if the mask is E^mE, or half masks and half recessive reds if the mask is E^me. The one outcome that would be missing is that a masked to recessive red breeding could produce unmasked sables and unmasked recessive reds in the same litter. Given the difficulty in distinguishing sables from recessive reds, this might prove difficult.
2. Remove E^br from the E series, possibly putting it in the same series with dominant black (currently in the A series.) The new series (here called K - the last letter of black - for convenience) would have three genes, K^d dominant black, K^br producing eumelanin stripes on any phaeomelanin (tan) pigment on the dog. The assumption is that K^d is dominant over K^br which in turn is dominant over k (more black dominant over less black.) The prediction would be that a dominant black (K^d-) bred to a clear sable would produce either all dominant blacks if the black is K^dK^d, a fifty fifty mix of dominant black and brindle if the black is K^dK^br, or a fifty fifty mix of dominant black and clear unmasked sable if the black is K^dk, but never a litter with all three colors. Unpublished studies on racing greyhound litters agree with this prediction.
3. E^m might still be in the E series, but this should be tested. The test breeding would be difficult, because of the difficulty in being sure whether a "red" dog is ee or a^ya^y, but the test is whether a masked dog, bred to another mask or to a recessive red ee, produces both ee red and fully expressed, unmasked tan-point or sable in the same litter. Probably some cross breeding would be required to be sure of the genotypes of parents and offspring.
4. If both removals hold up, this would leave the E series with just two alleles, normal expression of the A series (E -dominant) and recessive red (e - recessive.) It has now been reported in the scientific literature (Newton et al, 2000) that the genetic sequence of canine e/E correponds to the E-locus (specifically recessive red) in several other species (fox, cow, human and mouse.)

G, the graying series. Although only two genes were recognised in this series by Little, this may be a more complex locus, or genes that affect graying may reside at more than one locus. The effect of G, in single or double dose, is the replacement of colored by uncolored hairs as the animal ages, very much like premature graying in human beings. This gene should be suspected in any breed where a dark puppy pales and washes out with age, and the paling is due to interspersed white hairs. The gene is almost certainly present in some Poodles, Old English Sheepdogs, and terriers. The fading may start immediately after birth or after a period of weeks to months has elapsed, and may go as far as it is going to by the first adult coat or may continue through the animal's lifetime. G may or may not be the gene involved in the graying of muzzle and over the eyes in aged dogs, or in the lightening of black to steel blue without interspersed white hairs. This is a series that definitely needs more work.

M, merle. This is another dilution gene, but instead of diluting the whole coat it causes a patchy dilution, with a black coat becoming gray patched with black. Liver becomes dilute red patched with liver, while sable merles can be distinguished from sables with varying amounts of difficulty. The merling is reportedly clearly visible at birth, but may fade to little more than a possible slight mottling of ear tips as an adult. Merling on the tan points of a merled black and tan is not immediately obvious, either, though it does show if mask factor is present, and may be discernable under a microscope. Eyes of an Mm dog are sometimes blue or merled (brown and blue segments in the eye.)

Although merle is generally treated as a dominant gene, it is in fact an incomplete dominant or a gene with intermediate expression. An mm dog is normal color (no merling). A Mm dog is merled. But an MM dog has much more white than is normal for the breed (almost all white in Shelties) and may have hearing loss, vision problems including small or missing eyes, and possible infertility (Little). The health effects seem worse if a gene for white markings is also present. Thus the dachsund, which is normally lacking white markings, has dapples (Mm) and double dapples (MM) the latter often having considerable white, but according to Little other effects are limited to smaller than normal eyes. In Shelties, Collies, Border Collies, and Australian Shepherds, all of which normally have fairly extensive white markings, the MM white has a strong probability of being deaf or blind. The same is probably true with double merle Foxhounds and double merles from Harlequin Great Danes with the desired white chest. A few double merles of good quality have been kept and bred from, as a MM double merle to mm black breeding is the only one that will produce 100% merles.

It is possible that merle is a "fragile" gene, with M having a relatively high probability of mutating back to m. The observed pattern would then be the result of some clones of melanocytes having suffered such a back mutaion to mm while they are migrating to their final site in the skin, producing the black patches, while others remained Mm. This hypothesis also explains why a double merle to black breeding occasionally produces a black puppy, the proposed back mutation in this case occurring in a germ cell. On the other hand, the observed blacks from this ype of breeding may actually be cryptic merles - genetically Mm, but with the random black patches covering virtually all of the coat.

Merle is a part of the pattern of ragged black spots seen in the harlequin Great Dane. There appears to be an additional gene which removes the dilute pigment, leaving the "blue" area clear white. The fact that harlequins continue to produce merles argues that animals pure for this proposed extra factor may not exist, and one possibility is that a homozygote for this whitening factor is an embryonic lethal. Interestingly, there are recent reports of Shelties born with a harlequin pattern, but in this case the "blue" area actually develops color with time, winding up a light silvery blue. These dogs appear to have larger than normal black areas, at the extreme being so-called cryptic merles, that is, no blue is visible without an extensive search. Other shelties born harlequin or "domino" retain the white body color.

Although Danes are usually solid color, the harlequin color description includes a preference for a white neck and front. Since the black patching is as apt to be on neck and front as anywhere else, this requires incorporation of a gene for white spotting (probably irish spotting, sⁱ sⁱ). Given that SS double merles seem to fare better than their sⁱ sⁱ counterparts, I would expect that double merles from harlequin Danes with patched fronts and necks might be healthier than from those that fit the standard better. The harlequin description also faults black hairs in the white area. The harlequin - silver blue pattern in Shelties could be an extreme case of black hairs in the white area. Both harlequins and the silver-blue merle Shelties have occasional patches of gray (merle?) as well as black, though this is not considered desirable.

R, roan. This may or may not be a true series. Both Little and Searle suggest that roan may simply be a very fine ticking, with dark hairs growing in an initially white area of the coat. A second type of roan, in which white hairs develop in an initially dark coat, could be due to gray or could be a type of roaning different from the progressive development of dark hair in a light area. In any event, roan (R) appears to be dominant to non-roan (rr). It is not clear whether this is full dominance or incomplete dominance. I will here treat roan as being at the ticking locus.

S, white spotting. This is another somewhat unsatisfactory series, and one in which modifying genes appear to have a very large effect. Certainly there are genes for solid color, for a more regular white spotting, and for basically white with some colored markings. But the variability within each type makes it unclear how many alleles actually occur at this locus. In general dominance is incomplete, with more color being dominant over less color. Heterozygotes commonly resemble the more-pigmented homozygote, but with somewhat more white.

• S, solid color. This is the normal gene in breeds without white markings. An SS dog can completely lack white, but it can also express very minor white markings - white toes, white tail tip, or a star or streak on the chest. SS breeds generally fault these markings.
• sⁱ, irish spotting. Irish spotting is generally confined to the neck, the chest, the underbody, the legs and the tail tip. White does not cross the back between the withers and the tail, though it may appear on the back of the neck. Breeds with "Collie markings" which breed true for the markings are generally sⁱ sⁱ.
• s^p, piebald. This is a more difficult gene to identify. Certainly some breeds, such as parti-color Cockers, seem to breed true for piebald. Crosses of parti-color and solid in Cockers, however, often have minor white marking. Piebald and irish spotting seem to overlap in phenotype in one direction, while piebald and extreme white overlap in the other. In general, it seems a piebald has more than 50% white, white often crosses the back, and the pattern gives the impression of fairly large colored spots on a white ground.
• s^w, extreme white piebald. Extreme white piebalds range from the color-headed whites (Collies, Shelties) which may also have a few colored spots on the body, especially near the tail, through dogs with color confined to the area around the ear or eye (Sealyham, White Bull Terrier, Great Pynenees) to some pure whites (Dalmation ideal). There is some anecdotal evidence that s^ws^w dogs without color on or near the ear have a higher probability of deafness than dogs with color on the ears, but this varies with breed and it is not known whether a separate allele of S might be involved. In Boxers, some whites are produced from show-marked parents. Little believed that the Boxer lacked the gene for sⁱ, the irish-type spotting desired in the show ring being produced by heterozygosity for S and s^w. Since the Boxer club is adamantly opposed to any breeding of whites, even test breeding, this has not been independantly confirmed.

All of the spotting genes are assumed to be affected by the action of modifiers, with + (plus) modifiers being generally understood to increase the amount of pigment (decrease white) while - (minus) modifiers being assumed to decrease the amount of pigment (increase white.) Merle appears to act as a minus modifier, in addition to its effects on coat color.

It is not clear to what extent the S series affects head pigment. Color-headed white shelties, for instance (s^ws^w), can have completely colored heads - not even a forehead star or white nose. On the other hand, relatively conservatively marked dogs can appear with half white or all white heads. There is probably at least one other gene series that affects head markings. It is at least possible that the plus and minus modifiers affect head and body markings simultaneously.

T, ticking. Some dogs develop flecks of color in areas left white by genes in the S series. The clearest and most obvious ticking is seen in Dalmations, where additional modifier genes have enlarged and rounded the ticks. A large number of irish, piebald and extreme white breeds also have variable ticking, though not often as obvious as the Dalmation. The color of the ticking seems to be the color the coat would be in that area if the white spotting genes were not present. Thus a genetically black and tan Dalmation (a fault) will have tan spots where a black and tan would have tan markings. A ticked sable, a^ya^yTT or a^ya^yTt, may not have obvious ticking, becasue there is not much contrast between the tan and the white. Careful examination, however, will often show tan flecks on the legs. Ticking on a long-haired dog is also difficult to discern. The Border Collie on the front page of my site is ticked and probably sⁱs^w, as well as having the gene(?) for half white head. The tick marks in her ruff are not visible in the photo, but they are present (if difficult to find) on the living dog.

The usual dominance relationship given is that T (ticking) is dominant over t (lack of ticking.) Some breed-specific sources suggest that ticking acts as a recessive. I am inclined to suspect incomplete dominance of T. In Border Collies, for instance, a color called blue mottle is in fact a very heavily ticked piebald. The dam of the Border Collie mentioned above was such a blue mottle, presumably TT, while Dot is apparently Tt.

Ticking is also very much affected by genes which modify the size, shape and density of tick marks. In fact roan, which can develop by the gradual growth of pigmented hair in white areas of the coat, may simply be a form of ticking.

 

Size Genetics

The purpose of the following article is to give an explanation about how breeders are able to breed a dog for a certain size.  This article uses the Sheltie breed as an example. This theory is also true for the APBT.

Size as an example of additive inheritance

Sheltie breeders, as a group, tend to be hung up on size. We have reason to be. The breed was produced by mixing large and small breeds, few if any of which were in the size range we accept as correct today (13" to 16", with preference for 15" and up). Size, however, is not a simple genetic trait. At a minimum, it depends on the genetic codes for growth hormones, the genes that dictate when and how much growth hormone is produced, probably genes that code for receptor proteins that respond to growth hormones, genes that control bone shape and angulation between bones, and other genes affecting various metabolic processes. We don't even know the whole list, or how to determine what makes a particular dog large or small. In some cases the gene for larger size would be dominant, in some cases recessive, in some cases the dog heterozygous at a particular locus would be intermediate is size. It is, however, possible to use a very simple model to explain some of the oddities of Sheltie size. Remember this is a greatly simplified model! The real situation is almost certainly more complicated. We may even have a few genes for correct size hidden in there somewhere!

Suppose we assume we have four loci affecting size. Assume also that each locus has two alleles, one derived from the Collie part of our breed's ancestry, and the other from the original small island Sheltie (remember that at one time 12" was pushed as the maximum height) and toy breeds crossed in late in the 19th Century. We'll call these genes f, i, j, and k. The alleles for large size will be f⁺, i⁺, j⁺ and k⁺; those for small size will be f^-,i^-, j^-, and k^-. The size of the dog will be a base of 15" plus 3/4 times the sum of the "+" genes minus 3/4 times the sum of the "-" genes. A dog with all "+" genes, for instance, would be 21" tall, while a dog with all "-" genes would by 9" tall.

Suppose a breeder, breeding fairly close within her own line and related dogs, winds up with a consistent size genotype of f⁺f⁺ i⁺i⁺ j^-j^- k^-k^-. All gametes will be f⁺i⁺j^-k^-. All puppies will have the same size genotype as their parents, and will be 15" tall. But she's had to inbreed quite a bit, and she is looking for an outcross that will give her back what she's lost without sacrificing her predictable size.

She finds another breeder, again breeding fairly closely within her own line, who also gets all 15" Shelties, and whose line is strong for exactly the traits she needs. Breeder A breeds her best bitch to breeder B's best stud dog, and breeder B, who is missing a couple of things A has managed to fix, breeds her own bitch to a stud from breeder A's lines. The puppies arrive, grow up, and all are 15".

Then two of these 15" pups, from litters level in size stemming from strains level in size, are bred to each other. The result could easily be puppies all over the map in size. What happened?

Breeder B had a consistent size genotype of f^-f^- i^-i^- j⁺j⁺ k⁺k⁺, and her dogs consistently produced f^-i^-j⁺k⁺ gametes. The uniformly in -size puppies from the strain cross, then, all had the genotype f⁺f^-i⁺i^-j⁺j^-k⁺k^- and could produce any of 16 types of gamete, ranging from f⁺i⁺j⁺k⁺ to f^-i^-j^-k^-. I'm not going to try to draw a 16 x 16 Punnett square, but the expected size distribution in 256 puppies is:

1- 9 inch (all -)
8- 10 1/2 inch (7 -, 1 +)
28- 12 inch (6-, 2+)
56- 13 1/2" (5-, 3+)
70- 15" (4+, 4-)
56- 16 1/2" (5+, 3-)
28- 18" (6+, 2-)
8- 19 1/2" (7+. 1-)
1- 21" (all +)

If we assume some minor size genes as well, so the various categories are smeared out somewhat, the results don't look too unfamiliar. Note that if you breed the 16 1/2" but do not breed the 13 1/2", the result will be a gradual loss of - genes, and an overall upward creep in height. Also, there is no way to look at a 15" dog and determine whether it is fully heterozygous ( f⁺f^-i⁺i^-j⁺j^-k⁺k^-) or homozygous ( f^-f^- i^-i^- j⁺j⁺ k⁺k⁺, for instance) It's not as simple as breeding only from in-size dogs with in-size littermates!

How about genes for correct size? Could we have an additional allele, f, i, j, k at each locus, with ffiijjkk dogs being uniformly 15", and dogs with 7 normal genes and one + gene being 15 3/4"? It would certainly be nice, as then we'd just have to eliminate the + and - genes to have a breed that breeds true for size. It would be a slow process, if only because dogs with the alleles for correct size would so easily be confused with dogs with a balance of + and - genes. Given the background of our breed, though, the source of such genes for correct size is an open question. Most of our breed's ancestors were larger or smaller than 13" to 16". The standard advice on breeding for size, though, is to breed to correct size, which is based on the unstated assumption that the alleles f, i j and k exist.

Other complications undoubtedly occur. The hypothetical small and large genes may differ in their effect - f⁺ might contribute more to oversize than j^- does to undersize, for instance. There may be additional loci that have a dominant-recessive effect - NN or Nn might add 2" to the height while nn would allow the fjkl loci to control size. On the other hand, QQ or Qq might allow the fjkl loci to control size while qq would be 2" less than the fjkl size. The important thing to remember is that size is based on more than one gene pair, and as a result can do some very strange things.