The diagram we use in genetic pairing always has the male on the left and the female on the right:

When figuring up possible autosomal (non-sex-linked) combinations, we use a diagram that looks like this, and then plug the results into the pairing diagram above:

 Back to the Top

Males, however, may carry one or more of the sex linked genes on only ONE of their two X chromosomes. If they do, the DOMINANT NORMAL will mask that trait, and the male will be considered SPLIT to that mutation. For example, if a male carries Pearl on only one of his X chromosomes, which would be genetically coded X XP, he is split to pearl. If he carries Cinnamon on the other X chromosome (XC XP), he would be split to Cinnamon AND Pearl since he only has one gene for each color. If, however, he had Cinnamon and Pearl on the SAME X chromosome (XCP X) he would be split to cinnamon-pearl. NOTE: It makes no difference whether the sex linked gene is carried on the first X chromosome or the second in the genetic code.

A male cockatiel will always donate one of his two X chromosomes to the offspring, but a female can donate either her X or her Y. If the mother passes along her one X chromosome, the offspring is MALE (X X). If she gives her Y chromosome instead, the result is a FEMALE (X Y). In sex linked mutations, it is good to remember that males will always throw daughters of the same mutation, and mothers will always have sons that are at least split to any mutation they carry.

Back to the Top

Pied, Fallow, Silver, and Whiteface are all mutations that are also recessive to the dominant Normal Grey. What makes them different from Sex Linked mutations? They are autosomal, or not attached to the X chromosome. This is important because both males AND females can be split to simple recessive mutations. The pied gene is represented with a small "p"; silver - "s"; fallow - "f", and whiteface - "w."

In order for one of these colors to be visual, a cockatiel must have TWO of the genes present in its genetic makeup, or Genotype. If only ONE recessive gene is present, the bird is considered SPLIT to that mutation, regardless of its sex.

Both parents must carry at least ONE of the genes for their offspring to be visually that mutation. Crossing a bird that doesn't carry a particular recessive mutation with one that is visually that mutation will produce offspring that appear normal but are split for that mutation. For example, a male not carrying pied mated to a female pied will produce all split pied babies.

When you combine the male's recessive genes with the female's, if one does not carry any genes for that mutation it is coded as NN (denoting non-mutated or normal). Don't confuse this with the NORMAL of Normal Grey! If the bird only carries ONE gene for a particular simple recessive mutation, we use one small letter depicting the mutation followed by one large "N." For example, for split to whiteface, we would code the recessive combination "wN." Here are some examples using the simple recessive mutation Pied:

Note: It doesn't matter whether the mutated gene is coded first or last in the pair, as both will show split.

 Back to the Top

As I said previously, when a male cockatiel's sperm is created, it contains half the chromosomes which determine that male's hereditary traits. The ovum in the female also contains half the chromosomes for HER hereditary traits. This is the reason fertilization of the ovum is necessary...without the two halves joining, there would be no paired chromosomes and therefore no cell division, etc., that eventually forms a baby. Even though a male has TWO X chromosomes, only ONE goes into the creation of his offspring. The other chromosome, either X or Y, comes from the female.

The way we show this in genetics is by using the algebraic FOIL method .... F(irst), O(uter), I(nner), L(ast). Let's make up a mating using numbers instead of colors ...

 

The FIRST chromosome from the male and the FIRST chromosome from the female are combined for the first possibility....

The OUTER two chromosomes are then combined (that is, the first of the male's and the last of the female's) and because there is a Y at the end, it is a female, so it is placed on the right side of the grid....

Next, the INNER two chromosomes are combined (the last of the male's and the first of the female's)...and because there are two X chromosomes, it goes on the left of the grid....

Finally, the LAST two chromosomes are combined (the last of the male's and the last of the female's)...and again, since it is a female, it goes on the right of the grid...

Looks simple, doesn't it? Let's work the pairing now with colors instead of numbers attached to those X chromosomes....

FIRST:

OUTER:

INNER:

LAST:

Now we'll interpret those genetic codes using percentages. We take the entire possibilities of this pairing's offspring and look at that as 100%. Since each of these genetic codes has an equal chance of happening based upon 100 babies produced, we say that:

Matings start to get complicated when you start adding in the recessive possibilities to the above grid. Let's say that the female in the above example is not only a Pearl, but a Pearl-Pied. Well, that would be simple! As I stated before, if one bird that visually carries a recessive mutation is mated to another that doesn't carry the mutation at all, all the babies will be split for that mutation. To show that using genetic calculation, we would make use of our recessive grid....

Using the FOIL method and bringing one gene per parent down, you get a possible four combinations for EACH recessive mutation. In this situation, all four of the combinations are the same, so 100% of the babies would be split to pied.

Let's suppose instead that both the male and the female were SPLIT to pied. The grid would then look like this....

As I stated before, it doesn't matter whether the recessive gene is on the right or left side of the combination (i.e., Np or pN), the baby is still split for that mutation, and we code it as pN when we "plug it into" the sex linked grid. So, now our mating is not so simple! We have a 25% chance of our babies being full pieds, a 50% chance of them being split to pied and a 25% chance of a baby that isn't split at all. Okay...let's combine the outcome of our sex linked grid and our recessive grid by charting out the possibilities of mating a Normal Grey male split to Cinnamon, Lutino and Pied with a Pearl hen split to Pied....

Remember, one of THESE:

Can occur with each one of THESE:

Which would give you THIS:

For now, we will leave in the duplicates (pN and Np), because it will be easier to see the percentages, based on 100% of the babies.

· 25% would be Normal split to Cinnamon, Pearl, and Pied;

· 12.5% would be Normal split to Cinnamon and Pearl;

· 12.5% would be Pied split to Lutino and Pearl;

·  25% would be Normal split to Lutino, Pearl, and Pied; and

· 12.5% would be Normal split to Lutino and Pearl.

· 12.5% would be Cinnamon-Pied;

· 25% would be Cinnamon split to pied;

· 12.5% would be Cinnamon;

· 12.5% would be Lutino-Pied;

· 25% would be Lutino split to pied; and

· 12.5% would be Lutino.

Here's a way you can figure out the number of Genotypes you will get from any combination of sex-linked and recessive genes.
 
• Separate the sex-linked chromosomes. Without the recessives, this pair would produce FOUR possible sex-linked combinations: XCL XP, XCL Y, XCP XP, and XCP Y.
• Calculate the number of combinations by multiplying the number of possibilities:
• Figure the number of combinations for EACH recessive pairing. In this case there is only ONE possible combination of the recessive genes: fN. There are THREE possible combinations of whiteface: ww, wN, and NN (which we leave blank). There are also THREE possible combinations of pied: pp, pN, and NN.

Now let's determine the PERCENTAGE OF OCCURRENCE. When you mate two birds that are SPLIT to a recessive mutation, 25% of the resulting offspring are FULL VISUAL mutations, 50% are split for that mutation, and 25% carry no genetic trait for that mutation. If, for example, you mate a split pied to a split pied, the chart looks like this:

Regardless of whether the recessive gene appears first or last in the genetic code (pN or Np), the outcome is still the same: SPLIT. Out of the four, one is pp, or fully visual (25%); two are split (50%); and one has no pied genes (25%). If, however, you mate birds that are split for more than one gene, the percentages change:

If you are figuring percentages, you divide 100 by the number of possibilities, and you come up with each combination occurring 6.25% of the time! You will notice, however, that certain combinations are coded differently but mean the same thing...for example, pp sN and pp Ns. There is only one pp ss and only one NN NN...so these each have a 6.25% chance of occurring. There are, however, two "visual pied split to silver," so that genetic code has a 6.25% x 2 (or 12.5%) chance of occurring. There are 4 "split to silver and pied"....6.25% x 4 = 25% chance of occurrence. There is only one visual pied not carrying silver, so that chance of occurring remains 6.25%. And so on... This is, of course, based on 100 birds produced by one pair, so if in one clutch you get four visual silver-pieds, you are very lucky!

6.25% visual pied and silver
12.50% visual pied split to silver
6.25% visual pied (not carrying silver)
12.50% visual silver split to pied
25.00% split to silver and pied
12.50% split to pied (not carrying silver)
25.00% visual silver (not carrying pied)
12.50% split to silver (not carrying pied)
6.25% not carrying pied or silver
100.00%

Now, let's compute the percentage of the previous pair with the addition of sex-linked mutations.

Here we have a male pearl split to pied and silver mated to a female normal split to pied and silver. Looking at just the sex-linked combinations, all the males will be normals split to pearl, and all the females will be pearl....so this will have no effect on the percentages, with the exception of splitting up the males and the females. Instead of dividing the percentages among the entire offspring, we look at the percentages separately for sons and daughters.
 

You will have to adjust the percentages when you add more sex-linked possibilities.

You will have to adjust the percentages when you add more sex-linked possibilities.

Let's change that male to being split to Cinnamon and Pearl and see what happens:

To determine the percentages, just take the ones you figured up before, divide them in half, and substitute CINNAMON for PEARL in the other half!

Enough of this! When the percentages get this small, they mean very little to you. What is important, usually, is knowing the possibilities that CAN happen, and, if you can, the sex of the babies.

Crossover is the "jumping" of one sex-linked gene (the flea) from one X chromosome (dog) to another! Since you need two X chromosomes for crossover to occur, it is obvious that it only happens with males. There are only a few rules regarding crossover:

1. The gene cannot occur on BOTH X chromosomes. For example, this means that a male with a genetic description XC XC cannot experience a crossover.

This would be the expected outcome of the above mating...50% of the males will be split to Pearl, 50% of the males will be split to Cinnamon; 50% of the females will be pearl and 50% will be cinnamon. But how do you think we ever got the beautiful double mutation Cinnamon-Pearl? It was the result of crossover.

 This makes the male�s genetic coding X XCP, and the mating chart like this:

Suddenly, he has produced Normal and Cinnamon-Pearl female offspring. Without crossover, he could only produce Pearl and Cinnamon hens, with all Normal males.

There is also such a thing as REVERSE crossover, where a male that contains a double mutation on one X chromosome passes on only a single sex-linked mutation!

The Dominant Silver mutation is the first which is, as its name indicates, equally as dominant as the normal Grey. When a normal cockatiel has one gene for a simple recessive mutation, we call him "split" to that mutation. If he is, however, carrying only one gene for Dominant Silver, he is considered a Single Factor (SF) Silver. A SF Dominant Silver is also known as a dilute Dominant Silver, meaning that the silver color is diluted, or washed out, by the normal Grey. Predictably, a cockatiel with both Dominant Silver genes is called a Double Factor (DF) Silver.

In the SF Dominant Silver mutation, the color of the normal cockatiel becomes a silvery shade of Grey. The male matures to carry the yellow face mask and orange cheek patches; both males and females have a slightly darker head and neck, as well as Grey legs and black eyes. If you recall, the recessive silver mutation has RED eyes; the Dominant Silver retains the dark eyes of the normal.

The DF Dominant Silver is an even further dilution of silver grey, giving an almost yellowish-white coloration to the wings and tail feathers. The eyes continue to be black, and the legs are, again, grey.

Inheritance of Dominant Silver is similar to that of the simple recessive mutations. The genes exist apart from the sex-linked chromosomes; therefore, both males and females can have the single factor. A normal Grey SF Dominant Silver cock would be genetically coded X X SN, while a normal Grey DF Dominant Silver cock would be X X SS. Of course, all the mutations can be combined with the Dominant Silver mutations.

 First, lets explore the inheritance of the Dominant Silver gene. A male Dominant Silver mated to a female normal Grey would be coded:

All the babies from this pair would be Single Factor Dominant Silver cockatiels, regardless of whether they are males or females. Similarly, a female Double Factor Dominant Silver mated to a normal Grey cock would produce all SF Dominant Silver babies as well.

 
Think back to our discussion of Simple Recessive mutations. The Dominant Silver mutation is inherited just like the pied, whiteface, fallow, and silver recessives. The only difference is, we code the Dominant Silver with a capital "S" to denote its equality with the normal Grey. So, when genetically coding the inheritance of Dominant Silver, treat it just like recessive silver (but in the upper case)!

Now, to combine the Dominant Silver mutation with other mutations, you add one of the combinations above to your genetic charts! For example, a SF Dominant Silver Pearl cock mated to a DF Dominant Silver hen would be charted like this:

1. Separate the sex-linked mutations out, and chart them - -

2. Then chart out any simple recessive or Dominant Silver mutations. (There are no simple recessives in this mating.)

3. Now, combine the two. Remember, each of the Dominant Silver combinations can occur with each of the sex-linked combinations!

50% of your male offspring are DF Dominant Silvers split to Pearl, 50% of your males are SF Dominant Silvers split to Pearl, 50% of your female babies are DF Dominant Silver-Pearls, and 50% of your female offspring are SF Dominant Silver-Pearls.

  Let's add a simple recessive into the picture. Hmmm...how about if the male is also visually whiteface, and the female is split to whiteface?

Adding this to the above chart, we get:

Now, 25% of your male offspring are DF Dominant Silver Whitefaces split to Pearl, 25% of your males are SF Dominant Silver Whitefaces split to Pearl, 25% of your male offspring are DF Dominant Silvers split to Pearl and Whiteface, 25% of your males are SF Dominant Silvers split to Pearl and Whiteface, 25% of your female babies are DF Dominant Silver-Pearl-Whitefaces, 25% of your female offspring are SF Dominant Silver-Pearl-Whitefaces, 25% of your female babies are DF Dominant Silver-Pearls split to Whiteface, and 25% of your female offspring are SF Dominant Silver-Pearls split to Whiteface.

All the male offspring produced by this mating would be normals split to Lutino and Whiteface. (By the way, dont ever make the mistake of saying "split to albino," or I�ll have to send the genetically-correct police after you!) The female babies would all be Lutino split to whiteface. Notice that NONE of the babies are "albino" in appearance! If there were a TRUE albino mutation, it would exist as one gene (probably sex-linked), and the female babies would all look like the father. But, since the "albino" mutation is really Lutino-Whiteface, the female babies take on all the fathers sex-linked genes, and have the appearance of Lutino while also inheriting the one gene for Whiteface.

Now that I have exhausted that subject, I will finally discuss how to produce a Lutino-Whiteface baby. There are several ways to do this. The easiest way, of course, is to go out and buy a Lutino-Whiteface pair! But that�s the easy way out, and I personally don�t find that too challenging. The better way is to develop a good Lutino line and a good Whiteface line, and then to use the best of both for your breeding of Lutino-Whitefaces.

 A good Lutino male (let�s try to pick one without a bald spot, please) mated to a nice whiteface hen (just a plain whiteface will do), will produce this:

The male offspring will all be split to both Lutino and Whiteface, and all hens will be Lutino split to whiteface. Now, to any of you who said, "Oh, good, now I can breed the brother to the sister to get albino babies!"....a slap on the wrist and a mighty TSK TSK. Don�t you know that inbreeding is the way we developed those nasty bald spots to begin with?

 Let�s take the biggest male produced by the above couple and mate him back to one of his female whiteface cousins on his mother�s side. That charting will look like this:
 

As you can see, 25% of your males will be normal split to Lutino and Whiteface, 25% will be Whiteface split to Lutino, 25% will be normal split to whiteface, and 25% will be whiteface; 25% of your females will be Lutino split to whiteface, 25% will be Lutino-Whiteface, 25% will be normal split to whiteface, and 25% will be whiteface. Of course, the only babies above that will be useful in producing further Lutino-Whiteface offspring will be those carrying both Lutino and Whiteface! Since you can�t tell which males are split to Lutino, you are better off keeping just the female Lutinos and Lutino-Whitefaces for future breeding.

 Back to the Top

 

The pastel face cockatiel is very similar to the Yellowcheek in coloration, with a yellow cheek patch instead of the orange of the Normal Grey. The mode of inheritance, however, is completely different than that of the Yellowcheek.

The pastel gene is autosomal rather than sex-linked, and is recessive to all mutations except the Whiteface. When bred to the Normal Grey (or any mutation other than Whiteface), both the male and female offspring are split to Pastel Face, as in the example below:
 

If the Male in this mating was, instead, a Pearl-Pied-Pastel, and the female was a Pied, the genetic charting would look like this:

All those p�s and P�s can get confusing!

When paired with the Whiteface, however, the equation changes radically. It is believed that Pastel and Whiteface reside on the same pair of autosomal chromosomes. When even one Pastel gene exists with a matching Whiteface gene, the Pastel becomes VISUAL! For Pastel to be visual in cockatiels WITHOUT Whiteface, however, there must be TWO Pastel Face genes present. Elsie Burgin, one of NCS�s most highly respected breeders, gave me some good information about possible expectations from breeding Pastels. I�ve charted them genetically for you below. Don�t be surprised if you have to read this three or four times before you understand it completely!

 Back to the Top

Cynthia Kiesewetter has been owned by cockatiels, as of this writing, for approximately 15 years. It was her father, James Dodson (one of the founders of the Connecticut Budgerigar Society), who instigated her interest in aviculture. His knowledge of budgerigar genetics was instrumental in spurring her to learn more about cockatiel genetics.

 
Cynthia was active in her local club, Connecticut Association for Aviculture, for 8 years as Secretary, Bulletin Editor, Show Advisory Committee Chairperson, and Membership Co-chairperson with her husband, Michael. They have also been members of the American Federation of Aviculture, Society of Parrot Breeders and Exhibitors, American Cockatiel Society, National Cockatiel Society, and the North American Cockatiel Society.

 
In the National Cockatiel Society, Cynthia (better known as Cindy to her friends)  served as Connecticut State Coordinator, Nomination Chairperson, and, from early 1993 to late 1997, Band and Membership Secretary and NCS Magazine Genetics Consultant. She was awarded the N.C.S. Presidential Award in 1993, 1994. 1995, and 1996 for outstanding contributions.  Cynthia and Michael also created and maintained an Internet site for the National Cockatiel Society for two years.  Cynthia resigned her NCS duties in late 1997 to form the North American Cockatiel Society and dedicate herself more to her birds and her newly expanding family.   

Copyright 2000 by Cynthia Kiesewetter

Copyright Notice & Disclaimer Statement