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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 possibilities 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 approximately 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% probability 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 over dominance, if the heterozygote is the desired state. I prefer incomplete dominance, recognizing 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 probability 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 non merle.

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?

Test Breedings II

Purpose: to determine the genetic basis for a trait.

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.

There is another possible problem in the A series, this one involving the recessive black seen in Shelties and German Shepherds. If the recessive black is 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 is 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 different locus.

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" bu the researcher) with dominant black, but this is preliminary at this time.

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 maa being the gene associated with the healthy gene, and mab 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 Hh being the normal, healthy gene and hd being the recessive gene which causes the problem. In the original test population, maa was always on the same chromosome with Hh, and mab was on the same chromosome with hd. In other words, chromosomes are either maaHh or mabhd, never maahd or mabHh. If a dog has maa on both chromosomes, it is also Hh on both chromosomes, a genetic clear. If it has maa on one chromosome and mab on the other, it also has one Hh gene and one hd gene, and is a carrier. If it has mab on both chromosomes, it has hd 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 maaHh or mabhd 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 maahd and/or mabHh. 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, (mabHh) 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 (maahd) 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.

 

 

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