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, ay is sable and at is tan-point (black and tan = referred to as tricolor if a white spotting gene is present.) An ayat dog is sable, but generally somewhat darker than an ayay dog. The difference is generally of the same order as the difference within ayay or within ayat, so it is not possible to be absolutely sure whether at 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 atat dogs and sable merle on ayat and ayay.
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 sisw, so sw, color headed white, tends to be maintained in the breed as well. (I am keeping it simple by ignoring sp, piebald, which may also occur in Shelties.) swsw dogs are predominantly white with color on the head and perhaps a few body spots. While healthy, they cannot be shown. Of course all of this assumes that Little's attribution of white markings to different alleles at the same locus was correct.
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 ayatMmsisw. 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 apparant 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 MMswsw 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 sp, piebald.) An SS dog is solid color, an Ssp dog may have minor white marking (and is often unshowable) and an spsp 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. spsp and probably Ssp 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. I will assume it is at a separate locus K, with KB being dominant black, epistatic to anything at the A series, while kyky 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 Tervuran (AyAyEEkyky) and a Golden Retriever (??eeKBKB). Note that ee is epistatic to the A series, so we do not know what the normal A allele is in the Golden. The gametes are AyEky for the Terv and ?eKB for the Golden. Every puppy inherits Ay?EeKBky 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 KBky or KBKB, 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 Ay gene in at least one dose, and with dominant black moved to the K locus Ay 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!
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.
Genetics index page
sbowling@mosquitonet.comLast updated April 7, 2010
