Simple Recessive

With simple recessive a bird must have both of the mutant halves (alleles) making up the gene to show the mutation visually.

 Dutch Blue (Case 1)

 symbol = y for mutant allele, with male or female they must have two mutant alleles (yy) to display the muation. A Yy will only carry the mutation but not display it. We call this "split to" the mutation.

 Father 

normal green (YY)

 In each of these Dutch Blue mutation samples, it does not matter if the genetics of the father or mother are switched. The same offspring will occur if the Father is the Dutch Blue and the Mother is the normal green.

  Mother 

Dutch blue (yy)

 Offspring

 green/Dutch blue, 100% (Yy)

Dutch Blue (Case 2)

If you buy birds that you do not know the genetics of, a baby of a different color will be a surprise. In this way you can begin to figure out the genetics of the parents.

 Father

green/Dutch blue (Yy)

Remember! These are only the probabilities. Each clutch of babies may be different. For example: if the pair produces 100 babies, you can expect bout 25 of the babies to be Dutch Blue. Note the problem: you will not be able to tell the difference between the normal green offspring and the green offspring split to the mutation.

Mother 

green/Dutch blue (Yy)

Offspring

normal green, 25% (YY)

Offspring 

green/Dutch blue, 50% (Yy)

Offspring

Dutch blue, 25% (yy)

Dutch Blue (Case 3)

 Father

green/Dutch blue (Yy)

 Once again, it doesn't matter which parent is the Dutch blue and which is the green split to dutch blue. The outcome of the babies will be the same as outlined here.

   Mother 

Dutch blue (yy)

 Offspring 

green/Dutch blue, 50% (Yy)

 Note that the babies can be of either sex. You would expect equal numbers of each combination.

 Offspring

Dutch blue, 50% (yy)

Dutch Blue (Case 4)

Father

Dutch blue (yy)

When both parents are visually recessive, all the offspring will also be. If you get an offspring who does not look just like the parents, then you can assume that one of the parents is carrying an additional and different mutated recessive gene.

   Mother 

Dutch blue (yy)

Offspring

Dutch blue, 100% (yy)

 Dominant

 With a dominant mutation the bird only needs one mutant gene to visually display the mutation. The mutation will be displayed as fully and no more so with one gene that the complete pair.

Dominant Green Pied

symbol = P for mutant allele, either male or female need only have one mutant allele. Pp will display the mutation.

 Father

green pied (Pp)

 In the case of the most common form of green pied mutation in peach face love birds, the gene is dominant. Only one of the parents needs to have one mutant gene in the pair in order to display the trait visually and pass it on to the offspring. In this example the father has one mutant gene.

 Mother

normal green (pp)

Offspring

green pied, 50% (Pp)

 In the case of dominant genes, it doesn't matter which parent carries the gene. Also, both male and female offspring will inherit the mutated gene with the same probability.

 Offspring

normal green, 50% (pp)

 Sex Linked Recessive

With the sex linked mutation like the lutino it is extremely important to note which parent carries the trait. The female only needs one mutant allele but the male needs two. The gene is on a chromosome that does not have a pair. This is the opposite of sex linked genes in humans.

We will look at more cases with the sex linked recessive mutation. These possibilities are more complex than simple recessive or dominant.

 Lutino (Case 1)

symbol = Zi for mutant allele, males must be ZiZi, and females Zi to display the mutation.

 Father

normal green (ZIZI)

In this scenerio, the mother is visually the mutant as she cannot carry the gene without also displaying it. The father is a normal green bird with no mutation.

 Mother

lutino (--Zi)

Male Offspring

green/ino 50% (ZIZi)

All male offspring will be split to ino (in this case lutino). The females will be normal green birds and will not carry the mutant gene.

 Female Offspring

normal green 50% (--ZI)

 Lutino (Case 2)

  Father

lutino (ZiZi)

In this scenerio, the father is visually the mutant and therefore both alleles are mutated. The mother is a normal green bird with no mutation.

  Mother

normal green (--ZI)

 Male Offspring

green/ino 50% (ZIZi)

All of this pairs female offspring will be lutino. All of the male offspring will be green split to ino. We use the term ino because when other mutations are combined with the "ino" we can get creaminos and splitcinnaminos that are not lutino but it is the same gene creating the visual mutation.

Female Offspring 

lutino 50% (--Zi)

 Lutino (Case 3)

This is where surprises show up in the aviary. Neither of the parents display the mutation yet some of the offspring do.

Father

green/ino (ZIZi)

   Mother

normal green (--ZI)

 Male Offspring

normal green 25% (ZIZI)

 Female Offspring 

lutino 25% (--Zi)

  Male Offspring

green/ino 25% (ZIZi)

 Female Offspring

normal green 25% (--ZI)

 Co-Dominant

With one mutant allele the bird will show a partial color change toward the muation. With two mutant alleles the bird will display the full color change. It does not matter which parent has the mutated gene the probability of the offspring of either sex inheriting the mutation is equal.

 Orangeface (Case 1)

 symbol = L for mutant allele, symbol = l for normal allele.

 Father

Orangeface (LL)

In order to clearly see the difference between a normal green and a split orangeface, it is best to compare the birds side by side. They look more like the normal green than they do the orangeface parent.

Mother

normal green (ll) 

Offspring

Split Orangeface 100% (Ll)

 Orangeface (Case 2)

Father 

normal green (ll)

Mother

Split Orangeface (Ll) 

Offspring

normal green 50% (ll)

Both male and female offspring will inherit the mutated gene equally. These graphics do not depict any difference of male and female offspring.

Offspring 

Split Orangeface 50% (Ll)

 Orangeface (Case 3)

Father

Split Orangeface (Ll)

Mother

Split Orangeface (Ll)

Offspring

Orangeface 25% (LL)

Offspring

Split Orangeface 50% (Ll)

Offspring

normal green 25% (ll)

 Orangeface (Case 4)

Father 

Orangeface (LL)

With the Orangeface love bird this scenario is very straight forward. With some of the other co-dominant mutations it can be much more complicated.

Mother 

Orangeface (LL)

Offspring

Orangeface 100% (LL)

Multiple Factors

Father

Lutino/Dutch Blue (ppYyZiZi)

This is just one example of of a pairing of birds that have multiple mutations. The father has both the ino (lutino) and the Dutch Blue mutations. The mother has both the dominant pied and Dutch Blue mutations.

Mother 

Pied Dutch Blue (Ppyy--ZI)

 Male Offspring

green/ino/Dutch Blue (ppYyZIZi)

From this pairing, we have the potential to get female offspring that we cannot verify are pied or not. It is generally agreed that breeding the pied in with the ino is not a good idea.

Female Offspring 

Lutino/Dutch Blue (ppYy--Zi)

Male Offspring

green pied/ino/Dutch Blue (PpYyZIZi)

Male Offspring

Dutch Blue/ino (ppyyZIZi)

Female Offspring 

Creamino (ppyy--Zi)

Male Offspring 

Pied Dutch Blue/ino (PpyyZIZi)

 Female Offspring 

Pied Lutino/Dutch Blue (PpYy--Zi)

Female Offspring  

Pied Creamino (Ppyy--Zi)

The purpose of the page is to give a novice some help understanding the complexities of genetics in a visual manner. All these examples are very simplified. The explanations are only partially complete.

 *The Creamino name can be substituted with "Dutch Blue ino". It is one and the same. The "ino" is a sex linked mutation affecting a particular allele creating the Lutino (creamino when in the Dutch Blue) and Australian Cinnamon or even the splitcinnamino when the Lutino and Australian Cinnamon occur together in the male.