Cockatiel color inheritance
Genes have different relationships towards each other. This relationship defines whether the feature of the gene comes visible or stays invisible, but still carried inside the genotype. Some alleles dominate other, causing the “weaker” recessive allele to “step aside”. Sometimes also the sex of the bird may define what colors it can carry or show visually.
Below there are the most typical color inheritance types and geteical example diagrams showing how the genes are transmitted. When marking the charts it is typical that the dominant factor is written with as capital letter and the recessive allele with small letter symbol. Sometimes one can also use ready defined genetical symbols. In such cases, the alleles of the same family are usually marked with the same symbol but there’s a + next to the wild-type. In the diagram below each individual has two spots, each symboling one chromosome’s locus. One spot is the other allele of the gene pair.
Dominant inheritance means that the gene dominates over the “weaker” recessive gene. You only need one allele from one of the parents to get the color visible. Dominant colors can never be as splits. If your cockatiel doesn’t show visually some dominant color it doesn’t carry it at all. …Unless some other allele is even more dominant!
Below there’s an example of a dominant inheritance. In this chart the “n” means the wild-type allele without any mutation. “N” means the dominant allele that is a mutation of the “n” gene. So, the n is recessive and N is dominant. In this chart we have a female that has genepair nn and a male that has the gene pair Nn. This means that the female doesn’t carry the mutation but the male has one allele of the gene pair carrying the dominant mutation. So the male shows this color also visually. The genetic chart counts every combination that the alleles carried by the parents could form. This is how we get to count every possible genotype, whether the mutation is visible or not – and we also get to count the likelihood of one genotype appearing. So the large spots symbolize the gene pairs of the parents and from each gene there goes a line to two chicks, meaning that this gene was transmitted to the baby. So, the little spots on the row below symbolize the gene pairs of the chicks.
As we can see, we get two nn gene pairs and two Nn. Two fourth is the same as 50%. So the likelihood of each genotype’s appearance can be predicted as following:
Mother = nn
Father = nN
50% nn meaning that one half doesn’t carry the mutation at all.
50% nN meaning that the other half gets one mutant allele and therefore will show the color visually.
You can for example imagine that the female is grey and the male is dominant yellowcheek (aka yellowface). So the female has two wild-type genes in her tangerine locus and the male has one wild-type and one tangerine allele, causing the dominant yellowcheek coloration. Is the chart would describe such combination, half of the chicks would be dominant yellowcheeks and half of them would be normal greys.
Sometimes a cockatiel can embody features from both recessive and domianant allele at the same time. In such case it is said that the alleles are co-dominant. If the cockatiel gets a dominant gene from the other parent it is called a single factor. If it gets the gene from both parents, having two similar alleles in the gene pair, it’s called a double factor. Double factor affects usually about twice as strongly as single factor. Think about a cup of coffee. If you pour a splash of milk, the coffee gets lighter. If you pour two splashes it gets twice as light. So the double factor is sort of like the gene would be in its pure form – and single factor is affected partially by the wild type.
A subcategory under co-dominance is the incomplete dominance. The incomplete dominance, also known as the semi-dominance, means that the heterozygous (single factor) bird appears to be visually closer to the wild type’s semblance than the mutation allele’s semblance. Dominant silver belongs to this cathegory, as the single factor can appear very much alike to the wild type normal grey.
Chart below shows a case in which both of the parents have one dominant silver (or dominant edged) allele, “D”, and one wild-type allele, “d”, that is recessive in relation to the “D”. So the both parents are heterozygous dominant silvers – single factors. The gender doesn’t play any role in this case.
Female = Dd
Male = Dd
50 % Dd Single factor dominant silver
25% DD Double factor dominant silver
25% dd Doesn’t carry the dominant silver at all
Then again, if the other parent is single factor and the other one is double factor the chart could look like this:
Female = Dd
Male = DD
50 % Dd Single factor dominant silver
50% DD Double factor dominant silver
This means that every chick will inherit the dominant silver coloration but half of the chicks would get is as single factor, half as double factor. In practice this means that half of the chicks will be about half lighter than the other half.
Recessively inherited color needs two genes in the gene pair to become visible. If there’s only one copy the gene stands back as the other gene type dominates it. So there needs to be two genes in the gene pair so that there wouldn’t be anything to dominate the recessive allele. Among cockatiels, recessive colors are for example pied, whiteface (and the other blue family) and dilutions. Also some albinistic mutations like fallow are recessive. Recessive gene often is signed with small letter and the dominant with capital. So whereas in the previous case the mutant was written with large letter, in this case it’s written in small letter and the wild-type is the capital letter.
In the example below, the other parent has two wild-types, “NN”, and the other parent has two mutant alleles, “nn”. If the chart was about pied the male would be normal grey and the female a visinle pied. Just keep in mind that also in this case the gender doesn’t matter. We couls switch the parents’ genders and the results would still be same.
Male = NN
100% Nn Color is inherited as a split but won’t be visible
All the babies have inherited the allele combination Nn, meaning that they are heterozygous pieds, splits. Because the pied allele is recessive, the wild-type allele dominates over it and the mutation won’t be visible. Even though the parents have alltogether two mutant genes, it is not enough to produce visual pieds because the parents can still give only one allele for one chick.
In the chart below happens otherwise. Now both of the parents are visually wild-type but they carry the recessive allele as a split. So their genotype is Nn.
Female = Nn
Male = Nn
50% Nn Color stays as split but still exists in the genotype
25% nn Color becomes visible
25% NN Color isn’t inherited. It’s not visible, neither a split. This bird just doesn’t carry it.
Even though neither of the parents are visually pieds, their chick could now receive a mutant allele from both of the parents. Therefore one fourth of the offspring can inherit the pied color to their fenotype. Half of the babies get the color as split but there’s also a 25% chance that a baby gets a wild-type allele from both of the parents so that it just doesn’t carry the mutant allele at all.
Probabilities to get visual offspring will grow if the other one is visible pied.
Male = Nn
50% Nn Color stays as split but still exists in the genotype
50% nn Color becomes visible
And naturally: If both of the parents are homozygous in relation to the mutation, all the chicks will inherit the mutant color visually.
Sex-linked recessive inheritance
Sex-linked inheritance means that the gene causing a certain color is located in a sex chromosome Z. These sex-linked colors are actually recessive but because of the exceptional location of the locus the gene inheritance acts a bit differently.
As mentioned in the section about the basics of cockatiel genetics, males have two Z chromosomes and females have one Z and one W. Because the Z chromosome is responsible for carrying the sex-linked locuses, the female can have only one copy of the gene. As you might remember, normally there’s a place for two copies of the gene. Because of the possibility for only one allele in females, there’s never anything that could dominate the gene. This means that a female can never carry a sex-linked color as a split. If the female doesn’t look like a visual sex-linked coloration, it doesn’t have it in its genotype at all. This acceptional situation means also that the female needs only one gene to show the color visually. When it comes to males, the sex-linked genes behave quite much as recessive genes would. Males can be split to sex-linked colors and they do need two copies of the gene for the color to become visible.
When it comes to breeding, if only the mother carries a sex-linked color, male offspring can receive this gene only as split but never visible. Female offspring wouldn’t get the gene at all (because they get the W chromosome from their mother and W doesn’t carry the gene). If only the male carries a sex.linked color visually, the color is transmitted to every female chick but each male would get it only as split. This is because they would get the other Z (the one that carries the mutation) from their father but the mother would give them the Z chromosome with the dominant wild-type gene. For the males to receive the color visually they would need the gene also from the mother. This means that if the mother doesn’t carry the sex-linked color, you can not get any male offspring showing the color visually. But, the father doesn’t have to carry the color visually to transmit the gene for his sons. It is enough that the father carries the sex-linked color as split and the mother shows it visually. The most famous sex-linked recessive colors are for example lutino and cinnamon.
Note that because the female has only one Z chromosome it can receive only the other one of its father’s Z chromosomes – and the colors within. If the male carries for example pearl, cinnamon and sex-linked yellowcheek in its chromosome Z1 and only pearl in chromosome Z2 (meaning that the male is pearl split to cinnamon and yellowcheek), a daughter could inherit colors only from one chromosome. If the daughter receives Z1 it becomes sex-linked yellowcheek cinnamon pearl. If it receives Z2 it becomes a pearl.
Below there’s an example of how a sex-linked color could be inherited. Because one of the most common questions is “Why didn’t my lutino/pearl/cinnamon hen didn’t give her color to the offspring with grey male?” we’ll use such case as an example. So: We have a female carrying a sex.linked color. Let’s choose lutino to make it more concrete and let’s mark the wild-type as ino+ and the mutant allele as ino. The W chromosome is marked as grey spot without any genetical symbols since the chromosome won’t carry these colors. Each baby that has this kind of grey spot is female.
Female = ino –
Male = ino+ino+
100% ino+inoNormal split to lutino: the color is hidden but exists in the genotype.
100% ino+ – Color won’t be inherited.
Intermediate form of inheritance
Intermediate means that the color of the cockatiel is something between two separate colors. This means that if two alleles from the same gene family are placed in the gene pair’s locus, meaning that there is one of each allele of a same gene, together they will produce a coloration that is something between of what these colors would be if they were homozygous. An example of this is the platinum color of cockatiels. Platinum is actually parino – partial ino – meaning that it’s a sibling gene of ino that causes lutino coloration. If the platinum is homozygous, meaning that there are two parino alleles, the bird looks fully like platinum (warm smokey color). But if there’s one parino and one ino the result is something between these two colors: the cockatiel would be lighter than platinum but darker than lutino.
The blue family
“Blue” is widely known gene among parrot species. It is a recessive mutation that gets its name because it causes the blue color in many birds. Or actually, it removes all the yellow and red pigmentation (psittacine). If there are blue tones (for example under green), after removed yellow there is only blue left. But because cockatiel don’t have the spongy zone they can’t show blue or violet colors. So when it comes to cockatiels, don’t get confused even the blue gene won’t cause the cockatiel to be blue. Within cockatiels there are five known alleles, listed below from the lightest to heaviest psittacine reduction.
Wild-type, the gene that the normal grey would have in the blue loci
Goldcheek (mutation, parblue)
Pastelface (mutation, parblue)
Creamface (mutation, parblue)
Whiteface (mutation, blue)
These alleles are situated in the same locus. In practise it means that even though all the mutants are recessive compared to the wild-type, they form exceptional inheritance relationships towards each other. Pastelface and creamface both dominate over whiteface but the creamface is recessive in relation to pastelface. Normally if the cockatiel has only one pastelface gene the color won’t be visible. But if the bird has one pastelface and one whiteface allele, pastelface pops out visible.