Db), Melanotic (Ml) and Pattern (Pg) sit close together on the same chromosome, so they are usually inherited as a set. That linkage is a big part of why a pattern like mille fleur holds together, and it is built into how the Breeding Outcomes tool calculates a cross. For the wider system, see the Genetics Guide.
The three genes
Each of the three does a different job in building a pattern — for what they do, see the pattern genes page. This page is about how they inherit.
| Gene | What it contributes |
|---|---|
Dark Brown (Db) | Shifts the gold ground colour towards light orange or red-brown. |
Melanotic (Ml) | Adds and extends black through the plumage. |
Pattern (Pg) | Organises the black into structured markings — lacing, spangling and the like. |
Why linkage matters
Genes that sit close together on a chromosome do not assort independently — they travel together far more often than chance would allow. Db maps to the SOX10 region and Ml to the GJA5 region of chromosome 1, with Pg showing strong empirical linkage to both.1 Only an occasional crossover during meiosis reshuffles them:
| Pair | Estimated recombination |
|---|---|
Db ↔ Ml | ~10% |
Ml ↔ Pg | ~10% |
Db ↔ Pg | ~17-20% |
In practice that means roughly 81% of a bird's gametes carry the parental combination of these three intact, where three unlinked genes would keep the same combination only about a quarter of the time.2 The trio behaves almost like a single unit.
A worked example: two ways to make the same F1
Linkage does not only slow the reshuffling down. It also makes it matter which of the two chromosomes each allele arrived on — and that is settled by the grandparents, not by the bird in front of you. Here is a cross where that one fact moves the answer eighty-one-fold.
Two genes are enough to show it. Db and Ml sit about 10% apart, and Pg is held homozygous (Pg/Pg) all the way through, so it rides along without ever splitting and changes nothing.
Route A. A mille fleur cock (Db/Db · Ml/Ml · Pg/Pg) over a partridge hen (db+/db+ · ml+/ml+ · Pg/Pg). Every chick takes Db and Ml from its father, together on one chromosome, and both wild-type alleles from its mother on the other.
Route B. A cock carrying Db but not Ml (Db/Db · ml+/ml+ · Pg/Pg) over a hen carrying Ml but not Db (db+/db+ · Ml/Ml · Pg/Pg). Every chick takes Db from its father and Ml from its mother — on opposite chromosomes.
Both routes produce an F1 written exactly the same way: Db/db+ · Ml/ml+ · Pg/Pg. On paper the two birds are indistinguishable. On the chromosome they are not. Each chromosome below is drawn as a vertical bar, with Db near the top and Ml further down; the crossover falls in the gap between them.
Route A — coupling
Both dominants came from the father, on one chromosome
A gamete carrying both Db and Ml is a parental type: 45%
Route B — repulsion
One dominant from each parent, on opposite chromosomes
A gamete carrying both now needs a crossover: 5%
The circled gamete is the same one in both panels — the one carrying Db and Ml together, which is what you need to fix the pair. In Route A it is a parental type and comes out nine times in twenty. In Route B the identical gamete is a recombinant and comes out once in twenty. Now mate F1 to F1, where both parents have to produce it:
| Route A (coupling) | Route B (repulsion) | |
|---|---|---|
One parent gives Db and Ml together | 45% | 5% |
Both do, giving Db/Db · Ml/Ml | 0.45 × 0.45 = 20.25% | 0.05 × 0.05 = 0.25% |
Same written parents, same target, 81 times apart. (That 81 has nothing to do with the 81% figure earlier on this page — that one is 0.9 × 0.9 across the two intervals. The digits coincide; the quantities do not.) For scale: if Db and Ml assorted independently, the way most colour genes do, the answer would be 6.25% whichever way round they sat. So coupling makes this cross about three times easier than independence, while repulsion makes it twenty-five times harder. Linkage is not a penalty in itself — it locks in whatever arrangement the founders happened to have, and here Route A got the good one.
This is also why a written genotype is not the whole story. Db/db+ · Ml/ml+ describes both F1 birds above equally well, so anything reading a genotype off the page has to assume an arrangement. The Breeding Outcomes tool assumes coupling — it reads the order the alleles are written in, dominant first — which is the right assumption for Route A and the wrong one for Route B. Where you know the grandparents, you know the answer: whichever bird supplied both dominants supplied them stuck together.
In mille fleur
Mille fleur's genotype is eb/eb · Db/Db · Pg/Pg · Ml/Ml · mo/mo — and three of those five defining genes are exactly this linked group. Because Db, Pg and Ml ride together on chromosome 1, a mille fleur bird usually passes them on as a package rather than as three independent throws. That is a large part of why the pattern travels reliably through a line once it is fixed, and why related varieties — Tollbunt, porcelain — are built by swapping or adding a single gene around that same stable core.
How the Breeding tool handles it
The Breeding Outcomes tool marks Db, Ml and Pg with a small † and, when a cross involves more than one of them, shows a linked-gene note. The individual Db/Ml/Pg Punnett squares each show their own locus at the standard Mendelian ratios — but the combined F1 Genotype Frequencies table further down applies the actual linked recombination ratios by default.2 So the frequencies you read off already account for the linkage, instead of pretending the three genes assort freely.
eb/eb · Db/Db · Pg/Pg · Ml/Ml · ig/ig · mo/mo over eb/eb · Pg/Pg — and look for the † marking Db, Pg and Ml in the genotype table. Both parents are homozygous at every locus, so the F1 is a single row at 100%: Db/db+ · Pg/Pg · Ml/ml+, with Ig+/ig and Mo+/mo alongside. No F1 number can move here whether or not linkage is applied — but the arrangement is now known rather than assumed: Db and Ml both came in from the mille fleur father, on the same chromosome, so every F1 is a Route A bird. To make the numbers move, scroll to F2 Planning — there is nothing to pick, every gene has only one possible F1 outcome — and click ♂ F1 × ♀ F1 Intercross. Both parents are now heterozygous at Db and Ml at once, so the Mode: Linked ✓ button appears and changes the table under it. The F1-type row (Db/db+ · Ml/ml+) comes out at 10.25% under real linkage against 6.25% if the two genes assorted independently; the row tagged = Lemon Mille Fleur — Db/Db · Pg/Pg · Ml/Ml · ig/ig · mo/mo, the whole colour rebuilt — is 1.27% linked against 0.39% independent. That is the 20.25% against 6.25% worked out above, divided by sixteen for the ig/ig and mo/mo that have to land as well: coupling makes the mille fleur about three times easier to recover than free assortment would.
- Molecular mapping of the linked
Db/Mlpattern loci on chromosome 1 (SOX10 / GJA5 regions): Schwochow, D., et al. (2021) Pigment Cell & Melanoma Research; Sandve, S. R., et al. (2021) PNAS. See the pattern genes references. - Recombination-frequency estimates: Carefoot WC (1987) Test for linkage between the eumelanin restrictor (Db) and the eumelanin extension (Ml) genes in the domestic fowl. British Poultry Science 28:69–73. Db↔Ml (~10%) is that paper's own directly-measured backcross result; Ml↔Pg and Db↔Pg are cited within it from Moore & Smyth (1972a;b), tied to Pg via Carefoot's own 1985–1986 papers. The linked-ratio calculation used by the Breeding Outcomes tool is consistent with the chicken genetic map generally: International Chicken Genome Sequencing Consortium (2004) Nature 432:695–716.