Test Breedings I
Purpose: to get the genotype of an individual.
Sue Ann Bowling
Test breedings can be carried out for either of two distinct purposes: to determine the genotype of a specific individual, or to determine the fundamental genetics of a trait. Here we will discuss the first option, looking specifically at the determination of whether a dog carries a recessive gene. Note that as DNA studies advance and the carrier state becomes easier to distinguish via DNA testing, the type of test breeding described here should become less and less relevant.
The primary reason for doing test breedings historically has been to identify dogs carrying a trait that produces a health problem. For simplicity, we will use the black-brown dominant-recessive pair discussed earlier. The problem to be solved is to determine whether a black dog is carrying brown as a recessive. The analysis applies to any case in which a dog with two doses of the recessive is available for breeding, including a number of recessive health problems that are not actually lethal. Note that this type of test breeding is useful only after the mode of transmission (simple recessive) is firmly established.
We already know that if a BB black is bred to a bb brown all of the puppies will be black. If a Bb black is bred to a bb brown, each puppy has a 50% chance of being black and a 50% chance of being brown. So we breed the dog we want to test to a brown. If we get a brown puppy, the dog carries the brown recessive. We can never prove that the dog is pure for black, but we can calculate the probability that the observed number of black puppies would occur by chance if the dog were in fact carrying brown.
Remember each puppy has a 50% chance of being brown. The color of each puppy is independant, so the probility of a specific pair of puppies both being black is 50% x 50% or 25%. In fact, the probability that a Bb x bb mating with n puppies will produce all blacks is given by (.5)n. In tabular form, this is:
Black puppies in litter
Probability that a Bb parent could produce litter
The exact number of black puppies needed to "prove" that the black does not carry brown depends on how sure you want to be, but the probability that the parent is Bb even though it has produced a number of black puppies and no browns to a brown mate never quite goes to 0.
In our example, we assumed a fertile bb mate was available. What about the case in which the homozygous recessive is not viable or is infertile, such as gray-lethal in Collies? Suppose we imagine a locus called L, with alleles L for live and l for lethal. (We assume that the ll lethal can be distinguished at birth or shortly thereafer, not that it is an early embyonic lethal.) We want to determine whether a particular dog is LL or Ll in genetic constitution. We cannot test breed to an ll, because there are no living, fertile ll dogs. The best we can do is observe that any dog of the opposite sex that has produced an ll puppy must itself be Ll in genetic constitution. If we mate our test subject to a known Ll mate, the principle is the same, but this time it takes more puppies to reach the same level of certainty. If the test animal is LL, only LL and Ll puppies will be produced. If it is Ll, the chances of getting the given number of puppies, all healthy is:
Number of non-affected puppies
Probability that an Ll parent could produce litter
A test breeding utilizing a known carrier rather than an affected individual requires over twice as many offspring to get the same degree of certainty that an animal is not a carrier. For obvious reasons bitches were rarely test bred, especially in breeds with small litters - too much of her reproductive life would be lost in demonstrating that she was not a carrier. As the new DNA tests become available, this kind of test breeding will probably become very rare.
The use of test breeding to determine the mode of inheritance, however, may still be needed.
Genetics index page
last updated May 11, 2010