American Pansy

American Pansy is a newly classified mutation. Originally thought to be basic Redhead or Sparkly, or some combination with fawn, it was sold in the United States as early as 2010 through James Marie Farms (1)(2), and later by Myshire Farms and other sources (3), likely imported from Germany. During this time, there was rumors of a separate Redhead that was compatible with EB, often called "Chocolate Pansy" (4). Recent test breeding by Jen Jordan and Rebecca Lynch in 2024 have found an isolated, separate mutation that was likely mixed with regular Redhead during the importation that causes the look of the fabled "Chocolate Pansy", and renamed it "American Pansy" for testing purposes (1)(2).

It acts similar to Sparkly and regular Redhead, in that it increases red and black pigment, creating larger "splotches" on the bird. In homozygous, isolated form, it is believed to cause a deep red color with white tips and black centers, almost mimicking the white laced red variety in chickens. (1)(2)

Though it presents with a very faint heterozygous phenotype on pharaoh - see the slight narrowing of the striations on the back, and some slight pattern changes through the breast - it is almost imperceptible and thus labelled as recessive here. (1)

It is believed to be on the E locus, meaning the bird can only have one copy of Redhead and American Pansy at the same time. (1) While it looks similar to typical Redhead, it causes much more prominent red feathering and a white tipping around the edges of each feather.

The proposed order of dominance for this mutation on the E locus is Extended Brown > Sparkly > Pharaoh > American Pansy > Redhead, with Redhead being the least dominant. The gene symbol here is proposed to be E^a.

Breeding with American Pansy:

AP x AP = 100% AP

(aa x aa = 100% aa)

AP x Wildtype = 100% Het. AP

(aa x a+a+ = 100% a a+)

AP x Het. AP = 50% Het. AP, 50% AP

(aa x aa+ = 50% aa, 50% aa+)

Het. AP x Het. AP = 25% AP, 50% Het. AP, 25% Wildtype

(aa+ x aa+ = 25% aa, 50% aa+, 25% a+a+)

Het. AP x Wildtype = 50% Het. AP, 50% Wildtype

(aa+ x a+a+ = 50% aa+, 50% a+a+)