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title: "7.6 Testing parents for monogenic traits when no genetic markers are available"
canonical: "https://wiki.groenkennisnet.nl/space/TAB/296681668/7.6%20Testing%20parents%20for%20monogenic%20traits%20when%20no%20genetic%20markers%20are%20available"
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In molecular genetics many loci for qualitative (monogenic) traits are detected and alleles are characterized. The latter has resulted in a lot of genetic markers available for testing the presence of alleles for traits from animals that are considered for breeding. But often a genetic marker is not available. Then you have to perform “test” matings.

Suppose you want to mate your brown dog, a bitch, with a black male and you want to know what will be the chance on black puppies. You know that the black allele is dominant over the brown allele. It means that you want to know the chance that a random chosen black male is heterozygous. In the breed the allele frequency of the brown allele is 0.1 and from the black allele 0.9. This means that (assuming that the population is in Hardy Weinberg equilibrium) the fraction of homozygous black animals is 0.9<sup>2</sup> = 0.81 and the fraction of heterozygous animals is 2*0.9*0.1 = 0.18. So the chance for a black male to be heterozygous is 0.18 / (0.81+0.18) = 0.18. Roughly 1 out of 5 black males is heterozygous and will give 50 % black and 50 % brown puppies when mated to your brown bitch.

In the Dutch Groninger Blaarkop cattle breed the frequency of the dominant blaze allele is 0.95. The recessive homozygous allele does result in an unwanted spotted animal. How can we know with an accuracy of 95 % that a sire is homozygous for the blaze allele? Thus with an uncertainty of 5 % you want to get the answer. The best way is to cross the sire with Friesian spotted cows. A homozygous sire will get 100 % offspring with the blaze pattern in the cross with spotted animals. How many test matings have to be performed? Each calf born of a testcross has a chance of 0.5 to be spotted when the sire is heterozygous. With two calves the chance is 0.5*0.5 = 0.25. With 5 calves the chance is 0.5 <sup>5</sup> = 0.0325 and lower than 0.05. With less than 0.05 uncertainty you have to perform 5 successful test matings. Thus the number of offspring needed depends on the allele frequencies in the population and the required accuracy or the uncertainty of the test.