The A locus is responsible for a number of common coat patterns in the dog. Expression of all of them requires any combination of two Ky or Kbr alleles at the K locus, and at least one E or EM allele at the E locus. The gene involved is the Agouti gene, and variations in it are responsible for fawn and sable dogs (A^y), wild type (a^w), tan points (a^t), and recessive black(a).

analysis proves absence or presence of the mutation typically responsible for fawn or sable. In fawn/ sable dogs this test shows if other agouti alleles are present but hidden (only one copy of A^y). It also demonstrates how many copies of this allele are hidden in dogs, which cannot express agouti types (K^BK^B,  K^B  k^br, K^B k^y, at the k locus and/or ee at the E locus).

shows whether a black dog is black due to “recessive black,” or the more common black at the K locus. It also reveals whether a non-black animal carries “recessive black.” (e.g. German Shepherd Dog, Shetland Sheepdog)

a^t allele (tan points, tricolors) There is no direct test yet for this recessive allele (, but in some breeds, carriers may be deduced by use of the other two agouti tests. In breeds where only the A^y  and a^t alleles are present, the A^y test can be used to see if the fawn/sable dog is A^y/A^y (homozygous) or only has one A^y (heterozygous). If it only has one, the other allele must be a^t. (e.g. Afghans, Collies, Cardigan Welsh Corgi, Dachshund, Norwich Terrier, Staffordshire Terrier) In breeds where only A^y, a^t and “a” alleles are present, both the A^y test and the “a” test need to be performed. Any alleles unaccounted for by these two tests will be a^t. For example, if a dog is A^y/A^y both alleles are accounted for. If a fawn/sable dog only has a single A^y, then the other allele must either be an “a” or an a^t, and this can be determined by running the recessive black (“a”) test. Examples of breeds:< Shetland Sheepdog, Belgians (Malinois, Groenendael).