Genetics of the zebra finch : Instructions for use
The raising and presentation of zebra finch has grown considerably over the past fifteen years. To improve the size of the new mutations, breeders have also resorted to the classic "carrier" birds.
Some manage to combine several mutations. All this made it essential to know a minimum of applied genetics. This is the minimum that I would like to introduce to new breeders.
It is not a complete course of genetics, but the simple presentation of the method I use preceded by some basic concepts.
2. The zebra finch and its mutations
A zebra finch has a number of visible characters (size, shape, pattern, color, sex) that make up its phenotype. It may have, in addition to other characters not expressed (it is said that it is a carrier). The whole of the characters, expressed or not, is called the genotype.
A young zebra finch comes from an egg cell, the result of the fusion of the nucleus of a spermatozoon of the father and the nucleus of the ovum of the female. The genetic program of the bird is already there: A sequence of cell divisions and coded information will trigger (or not) the appearance of the characters. The coded information is carried by genes located on long filaments contained in the nucleus: Chromosomes.
All chromosomes go in pairs: each chromosome has its counterpart.
There are two categories of chromosomes :
- The sex chromosomes:
• XX in the male
• XY in the female
- The chromosomes autosomes.
The gray zebra finch living in Australia is the origin of all our zebra-reared finch. It has a whole set of genes distributed in its chromosomes.
Whenever a new mutation has appeared, it is because there has been a modification of a gene of origin (and that it has proved to be hereditary). The gene of origin and the mutated gene are located at the same location called locus on each of the homologous chromosomes.
Both genes are alleles.
A bird is pure (homozygous) when all its alleles carry identical information.
A bird is heterozygous when at least one pair of alleles carries different information about the same trait.
There are currently about twenty different mutations of gray zebra finch.
We distinguish :
a) The dominant mutations
Pastel, crested, cheeks (gray, brown), black face (black-face), cheeks clear.
A mutation is dominant when it is expressed while the mutated gene exists in only one copy. This gene is located on an autosome chromosome.
There are therefore no zebra finch carrying these mutations.
Note: When the two homologous chromosomes each carry a dominant gene, the young is not viable. This is called a lethal factor.
b) Recessive mutations
White, variegated, saddled, white breast, black breast, orange breast, black cheeks, isabelle, agate, yellow beak, eumo.
A mutation is recessive when it is expressed only if the two autosome chromosomes each have the mutated gene.
If there is only one mutated gene, the character is not expressed. The bird is simply "carrier" of the mutation.
c) Gender mutations
Brown, pale back, masked old type, masked new type.
A mutation is linked to sex when the genes responsible for this mutation are located on the X chromosome (s) of the bird (the Y chromosome of the female being empty of genes).
The mutation is expressed in females since they receive from their father the mutated X chromosome. For it to be expressed in males, the mutated gene must be carried by each of the two X chromosomes. Otherwise, the male is only the carrier of the mutation; however, he can pass it on to half of his daughters.
Notes: The "Light Back" and "Masked" genes are alleles of the same non-mutated gene. A Gray male may carry Pale Back and Masked.
A pale-backed male can be a masked bearer, but not the other way around. In this case, even in a single copy, it is he who expresses himself.
The same factor (Pale Back) can be recessive compared to Gray, but dominant over Masked.
The gene "Brown" also located on an X chromosome does not have the same locus as the previous genes.
Anterior to the other two, it is on a different chromosome.
For these genes to be linked (Pale Brown Back, Brown Masked), it took the appearance of a phenomenon that is the subject of another article: Crossing-over.
d) Combined mutations
Many mutations as well as gray can be combined with each other. One can theoretically associate a lot but in practice, it is better to remain cautious: In addition to the many necessary crossings, it is necessary that the bird obtained remains typed and corresponds to the criteria of the standards.
The most famous are :
• Brown pastel
• Gray or Brown cheeks
• Isabelle Black Chest
Black Brown Black Breast or Brown White Brown Pastel combine, for example, a sex-linked mutation, a dominant free mutation and a free recessive mutation.
It is therefore necessary to know how to choose the best crossings to achieve this.
3. Crossing technique
a) Assign each mutation a symbol
We begin by assigning each mutation a symbol: By analogy with the atomic symbols, we can choose one or two letters of the name of the mutation.
The dominant mutations are in upper case, the others in lower case.
Personally, I use the following symbols (From French abbreviations of mutations) :
Fn: Black face
bj: Yellow beak
po: Orange Breast
pb: white chest
pn: black breast
J: Cheeks (Gray or Brown)
jn: Black cheeks
dp: Pale Back
ma: Masked Old type
mn: Hidden new type
Scientists have a + sign followed by the symbol of the unmutated gene.
Example: H (Huppe); H + (not Crested); pb (White breast); pb + (no white breast).
Personally, I find it more logical to write: H + (Huppé); H- (not Crested); pb + (White breast); pb- (no white breast).
In the end, the results will be the same.
b) Write the genetic formula of each bird
On either side of a fraction bar, symbols of the genes carried by each homologous chromosome are transferred, starting with the sex chromosomes.
c) Sex chromosomes
Gray male: XN / XN; Gray female XN / Y
In this case, N means Normal
d) Gender mutation
Brown male: Xbr + / Xbr +; Brown female: Xbr + / Y
Same formulas with dp +, my +, mn +.
e) dominant free mutation
Male pastel gray: XN / XN PL + / pl-; Gray pastel female: XN / Y pl- / PL +
Same formulas with H +, BF +, J +.
Non-mutated recessive factors are written in lower case.
f) Free recessive mutation
Black-chest male XN / XN pn + / pn +; Black breasted gray female XN / Y pn + / pn +
Same formulas with pb +, po +, jn +, pa +, se +, is +, and so on.
g) Combined mutations
Brown male black face black cheeks: XN br + / XN br + Fn + / fn- jn + / jn +
Male pastel pale yellow pastel: XN dp + / XN dp + Pl + / pl- bj + / bj +
Gray male / (/ means carrier) Pale back: XN dp + / XN dp-
Female Brown / Black cheeks: XN br + / Y jn + / jn-
Gray female black face / black breast: XN / Y Fn + / fn- pn + / pn-
Gray male black face / black breast: XN / Y Fn + / fn- pn + / pn-
Male pale back gray / Masked NT (new type): Xdp + / Xmn +. In this case, one could write DP +, since the Pale Back dominates its allele, the NT Mask.
4. Place these formulas in a cross table
We must first remember:
• That each parent transmits to his or her young only one of the two chromosomes of each pair.
• That the grouping in each gamete (spermatozoon or ovum) of these chromosomes is by chance: it is the genetic mixing.
The more the parent has mutated genes on different chromosomes, the more combinations will be possible. This is the only difficulty in this method, but it is inevitable.
Let's start with a simple crossover :
a) Brown male: (XN br + / XN br +) X Female gray XN br- / Y
|XN br+||XN br+|
|XN br-||XN br+/XN br-||XN br+/XN br-|
|Y||XN br+/Y||XN br+/Y|
Each chromosome of the male (in this case, the sex chromosomes) finds its homologous chromosome provided by the female. It only remains to translate each formula.
Results : XN br + / XN br- Male gray / brown (50%); XN br + / Y Brown female (50%)
b) Male gray / brown: (XN br + / XN br-) X Brown female: XNbr + / Y
|XN br+||XN br-|
|XN br+||XN br+/XN br+||XN br-/XN br+|
|Y||XN br+/Y||XN br+/Y|
Results : XN br + / XN br + Brown male (25%); XN br- / XN br + (25%); XN br + / Y Brown female (25%); XN br- / Y Gray female (25%).
Once the method is acquired, it is possible to find the result of any cross. It takes time, logic and patience (or a computer).
c) Male black-face (other name: Black face) brown: (XN br + / XN br + Fn + / fn- pl- / pl-) X Pastel gray: XN br- / Y fn- / fn- pl-: Pl +
|XN br+ Fn+ pl-||Genotype||XN br+ fn- pl-||Genotype|
|XN br- fn- Pl+||(XN br+ Fn+ pl-)/(XN br- fn- Pl+)||Male black pastel face gray / brown||(XN br+ fn- pl-)/(XN br- fn- Pl+)||Male pastel gray / brown|
|XN br- fn- pl-||(XN br+ Fn+ pl-)/(XN br- fn- pl-)||Male black face gray / brown||(XN br+ fn- pl-)/(XN br- fn- pl-)||Gray / brown male|
|Y fn- Pl+||(XN br+ Fn+ pl-)/(Y fn- Pl+)||Black brown pastel face female||(XN br+ fn- pl-)/(Y fn- Pl+)||Brown pastel female|
|Y fn- pl-||(XN br+ Fn+ pl-)/(Y fn- pl-)||Black-faced female||(XN br+ fn- pl-)/(Y fn- pl-)||Brown female|
This crossover involves a mutation linked to sex and two dominant free mutations. This leads to eight different phenotypes (12.5% each).
It seems important to me :
• Do not go down to lower percentages (6.25% or 3.125%). The chances of getting the desired bird are too low: One bird out of 16 or 32. Few couples provide 16 youngsters in the year!
• To be able to distinguish the birds carrying those, of the same phenotype, which would not be.
This method should allow any breeder :
• To predict the theoretical results of each couple that it forms.
• Analyze the results obtained and learn from them (some birds are found to carry a mutation that was unknown).
• To check for possible errors in the crossover results offered by books and magazines.
However, it is essential to know perfectly the genotype of each of its birds.
That supposes :
• A breeding in individual cages.
• A rigorously kept breeding book and an individual record (pedigree) for each bird.
• That the breeder who gives you a bird gives you his individual record.
Hoping to have been able to help all those who start or who are engaged in breeding combinations of mutations, I am at the complete disposal of breeders who would like clarification or additional information. The easiest way is to join me by mail ( firstname.lastname@example.org ). I will be happy to answer you.
René DRUAIS judge CNJF-OMJ Exotic beaks rights.