Test Breedings II

Purpose: to determine the genetic basis for a trait.

Sue Ann Bowling

Suppose we have a list of various types of a particular trait, and we want to know how they are inherited. The first step is to make a guess. It should be an informed guess - for instance, you may know that in other mammals a particular trait is inherited in a particular way, so as a first guess you assume that the inheritance is similar in the animal you are investigating. The point is, this first guess is just that - a guess. In order to elevate that guess to the level of a hypothesis, you need to work out what your guess predicts in terms of what parents can produce what, and then breed (or investigate breeding records) to see if that is really what happens.

Let's take a first guess we know is wrong. Labrador Retrievers come in black, brown and yellow, as explained earlier. Suppose we don't know the genetics of this. We have observed the three colors, and a reasonable initial assumption is that there a locus for color which has three alleles: black, brown and yellow. As we start to look at Stud Book data, we find that;

  1. Black to black can produce any color
  2. Yellow to yellow can produce only yellow
  3. Brown to brown usually produces browns, but can produce yellow
  4. Black to any other color can produce black.

This information adds to our initial guess. If black to black can produce any color then black must be the top dominant in the series. Likewise, if yellow to yellow can produce only yellow, then yellow must be the bottom recessive. Brown looks as if it is recessive to black but dominant to yellow. Our tentative hypothesis, then, is that we have a locus, J, with three alleles:

  • Jblk black
  • jbrn brown
  • jyel yellow.

Now we set up our Punnett squares and work out what each mating will produce. We find that

  1. JblkJblk x JblkJblk gives black to black producing all blacks
  2. JblkJblk x Jblkjbrn gives black to black producing all blacks
  3. Jblkjbrn x Jblkjbrn gives black to black producing black and brown
  4. JblkJblk x Jblkjyel gives black to black producing all black
  5. Jblkjyel x Jblkjyel gives black to black producing black and yellow.
  6. Jblkjbrn x Jblkjyel gives black to black producing black and brown
  7. Jblkjbrn x jbrnjbrn gives black to brown producing black and brown
  8. Jblkjbrn x jbrnjyel gives black to brown producing black and brown
  9. Jblkjbrn x jyeljyel gives black to yellow producing black and brown
  10. Jblkjyel x jbrnjbrn gives black to brown producing black and brown
  11. Jblkjyel x jbrnjyel gives black to brown producing black, brown and yellow
  12. jbrnjbrn x jbrnjbrn gives brown to brown producing all browns
  13. jbrnjbrn x jbrnjyel gives brown to brown producing all browns
  14. jbrnjyel x jbrnjyel gives brown to brown producing brown and yellow
  15. jbrnjyel x jyeljyel gives brown to yellow producing brown and yellow
  16. jyeljyel x jyeljyel gives yellow to yellow producing all yellows

The key point is that none of the black to black or black to yellow matings can, on this hypothesis, give us a litter with all three colors represented. Three colors is only possible if black carrying yellow is mated to an animal which is brown carrying yellow. Blacks always have the potential to produce some blacks, but if a brown is produced than the black must carry brown, and there simply isn't room for the yellow allele at the locus, which can hold only two alleles at once. While the individual matings seem to agree with with our incorrect hypothesis, the hypothesis falls down when it is applied to colors within a single litter.

The problem is that while it's fairly easy to go through a stud book and determine what parent color combinations can give a particular puppy color, it is much harder to pull out a whole litter. In the AKC Stud Books it is almost impossible, as the only dogs listed are those who have produced registered litters. The point is that without determining whether the observed distribution of phenotypes within a litter agrees with the hypothesis, the hypothesis is still little more than a guess.

There are two possible test breeding strategies to expose this problem. The first involves looking at as many litters as possible in which one parent is the top dominant (black) and the other is the bottom recessive (yellow). If such a litter includes both browns and yellows, then our one locus - three allele hypothesis cannot be true.

The second case is a variant - identify blacks with one parent yellow or chocolate, so you "know" that the black is Jblkjyel or Jblkjbrn, and examine litters to yellow and to brown mates. Again, the presence of all three colors in one litter disproves the hypothesis, but it will take fewer litters in total, as the initial selection of the blacks eliminates those that are pure for black.

Note that in most cases, this means a fair number of breedings. This again is a case where there is no way to prove the hypothesis correct. You may have nine litters with black to yellow producing only yellow or brown (with black in each case) but that doesn't prove the hypothesis is correct. Only a few black to yellow litters may even have the right parental genotypes, and especially if the number of puppies is small, one possible color may be missing by pure chance. As usual with scientific hypotheses, the hypothesis cannot be proven, but it can be disproven.

In this particular case, I knew the thypothesis was incorrect. I have friends who breed Labs, and one bred a black to a black and got all three colors in the litter. It's not considered unusual in Labs. I even used the litter in a genetics Science Forum article. There are, however, other loci in dogs where the assignment of one or more genes to the locus is questionable. Probably the most important are the A series and the E series.

Dominant black is a very unlikely top dominant of the A series. This series is known in a number of mammals, and more yellow is almost always dominant over less yellow. The key breeding here would need a breed with dominant black, sable and tan-point. Basenji breedings of this type (black to tan-point) have been reported to include all three colors. The only remaining doubt comes in whether the "reds" from these breedings are sable or ee reds. e is not known to occur in the breed, but without further test breeding of the red offspring, there remains some uncertainty. Still, I am inclined to treat As at this point as belonging to another locus entirely. This hypothesis agrees with recent DNA work, and in fact dominant black is now assigned to the K locus.

There was another possible problem in the A series, this one involving the recessive black seen in Shelties and German Shepherds. If the recessive black was in the A series, with sable dominant to tan-point which in turn is dominant to recessive black, then it shoud not be possible to get a litter with all three colors from a sable to sable or a sable to recessive black breeding - a sable could be black-factored or tanpoint factored but not both. There was some evidence from Shelties that such three-color litters do occur. This suggests that the presence or absence of tan points in the classic tan-point pattern may depend on a extra locus.

DNA work has now established that Sheltie black is indeed aa, so the three-color litters are now suspected to be (1) cases where the true parents are not those on the registration, (2) misidentification of the colors of one or more pups, or (3) rare mutations. However, the A (agouti) locus is still under study.

E is defined to include E, which allows the agouti series to show through, and e which in double dose makes the dog produce only phaeomelanin in the hair coat, effectively hiding what is present at the A locus. The two other proposed members of the E series, Ebr (brindle) and Ema (masked) are still at the hypothesis stage. Even Little, who is often quoted as the source for putting brindle and mask in this locus, prefaced almost everything he said with "if they are at the same locus." In particular, none of the test matings he carried out really clarified the relationship of e to Ema or to his proposed Ebr. Test breeding is definitely needed at this locus. Some work has been done in greyhounds that suggests that the brindle gene might be at the same locus (called "K" by the researcher) with dominant black, but this was preliminary at the time this website was first put together. The hypthesisthat both brindle and dominant black are at the K locus has since been shown to agree with DNA studies.

Last updated May 11, 2010