For now, I’ll use the recessive black eyed yellow budgerigar as a model. This bird is genetically a dilute green, and its cousin the white is the dilute form of blue. ‘Dilute’, like pied is not actually a colour, it’s a modifier, this time recessive. The modifier genes may be represented by ‘C’ for non-modified or full colour and ‘c’ for the dilute or modified colour. Therefore yellow would be BBcc. BB denoting a green series bird and cc the dilute form of the green series. BBCC then is the normal green bird. You can see that the formula for the normal green bird changes depending on which mutation, relative to it, is under study, following on then, BBCc would be green/yellow, and going further still Bbcc would be yellow/blue. Do you can see in revision that the normal colour gene ‘C’ has been modified by the recessive dilute gene c which is a double dose become visible and when acting on a green series bird gives yellow. To further complicate things, when a yellow and a blue were mated all the young would be green, not because yellow and blue make green but because yellow is actually only a double factor dilute green, which, although it may be recessive to normal green it is still dominant to blue.

The parent bird can only pass on one gene at a time from each pair it passes. Thus, we can see in the square at left (see Graph 3 below) in a mating between two green/split blue birds i.e. Bb to Bb. Each bird can pass on either ‘B’ (green) or ‘b’ (blue) and depending on which gene from one pair combines with which gene from the other partner we may get either BB, Bb or bb.

Now, if we are going to speak of colours and modifiers at once, say when breeding a yellow with a blue (see Graph 4 below). Blue is bbcc, the double factor blue genes bb giving the blue colour, and the dominant non-dilute factor being shown by the capitals CC. one gene from each pair, which ever way they are passed on, end up in being available as ‘bC’. Now the yellow bird BBcc can likewise pass on as ‘Bc’ (or ‘B’c). if ‘bC’ and ‘Bc’ are applied as at left, all the offspring end up as BbCc. Bb = green/blue and Cc is full colour/dilute, so we have Green birds split for yellow and for blue. Blue is a mutation and dilute is a mutation. We need green plus the dilute mutation to get yellow. White is not a direct mutation, rather a combination of two other mutations, blue and dilute. You’ve probably already guessed that I’m leading up to breeding a white bird.

In colours we get 9 greens, 3 yellows, 3 blues, and 1 white (as Graph 5 below). With this mating the chances of breeding a white one are 1 in 16. This could be improved of course by mating the blue/whites, if you can pick them. It must be remembered that any colour can only be split for something recessive to it.

Blue can be split for white but not for yellow. Green is dominant to yellow which is dominant to blue which is dominant to white. Similarly white is recessive to blue, which is recessive to yellow, which is recessive to green. Nothing can be split for green and white can’t be split for anything. Well that will do for now on ‘recessive’. I hope you understand it.

‘Sex-linked’ is an often misused and misunderstood genetic term, and really it is quite simple. The sex of any bird is determined by its possession of a particular pair of similar sex chromosomes called ‘x’ chromosomes and the hen has one of these and an accompanying dissimilar one called the ‘Y’ chromosome. About 50% of the females egg cells contain x chromosomes and the other 50% Y chromosomes. All the male sperm contains x chromosomes. The sex of the offspring is determined at fertilization. When a sperm (always x) fertilizes an egg cell carrying an x chromosome the resulting bird will be a cock x/x. likewise when a sperm fertilizes an egg cell carrying a Y chromosome we get x/Y resulting in a hen. So the chances of make o female are 50/50 either way. Like genes, chromosomes are always in pairs, one half of the set coming from the cock’s sperm, and the other half from the female egg cells. These to halves combine and form the first living cell, which of course divides and multiplies until it grows into the finished product and hatches. It must be remembered that x and Y refer to chromosomes, not genes. The genes are carried on the chromosome.

When we refer to a ‘sex-linked’ bird, we mean one which carried its colour controlling genes on the sex chromosomes, everything else is the same. All known sex-linked genes are reclusive, so that the only difference between the genes (apart from the actual colour) causing a bird to be blue or lutino, is that the lutino ones are carried on the sex chromosomes and he blue genes are carried on some other pair of chromosomes. When I say ‘the only difference’ I refer to the difference in breeding behavior. Both birds still have the same number of pairs of chromosomes. (I believe budgies have 13pr) and of course the blue bird still has sex chromosomes. Just to wonder a moment, it’s interesting to note that only birds and butterflies have xx and female XY. I believe that all other things, including humans, have make XY and female xx. Anyway, the colour genes of the sex-linked bird are carried on the sex chromosomes, but only on x chromosomes. As far as I can tell this is the only exception to the rule concerning pairs of genes, because the hen has only one x chromosome and apparently the y chromosomes has no gene carrying function. Now for formulas. There are six established budgerigar mutations I know of which are sex-linked, but for now we’ll interest ourselves with only two: Albino and Lutino. However these are really the same mutation. The effect if the ‘Ino gene’ on a green bird produces Latino and on blue or grey gives Albino. It all follows the same sort of pattern as the ‘dilute gene’ which can produce yellow or white, depending on whether it acts on green series or blue series birds. Therefore with blue and Lutino we could breed Albino etc. We shall use I or i to denote the Ino gene and it’s opposite. So with the Ino gene carried only on the x chromosome, the hen has a recessive Ino gene carried only on the x, and none on the Y making her Xi/Y. an albino cock has recessive Ino genes on both so he is xi/xi. The normal cock has a dominant ‘non-Ino’ gene on each X making him XI/XI likewise when discussing the Ino mutations the normal hen is denoted by XI/Y. Now of course the only combination left is the cock with a dominant ‘non-ino’ (normal) XI providing a recessive xi which is a normal/split albino. You can see now that only two possible combinations exist with the hen, XI/Y which is normal, or Xi/Y which is albino. Therefore with no other possibilities, she can never be split for any mutation carried on sex chromosomes, i.e. a sex-linked mutation. We can use a fairly similar system for finding the expectations from any sex-linked mating, as for the recessives.

Above in Graph 6, I have mated a normal/albino cock to a normal hen. XI/Xi mated to XI/Y. Always put cock on top (easier to remember) and then the cock bird expectations will also be on top. With sex-linked expectations of course we are interested in sex as well as colour. So you can easily see from the above mating that we will get 25% normal cocks XI/XI, 25% split cocks XI/Xi, 25% normal hens XI/Y and 25% albino hens Xi/Y. there are a limited number of different sex-linked matings, five in all combinations and are correct for all sex-linked varieties.

The only other item which may be of interest at this time is the dark factor. Anyway the dark factor is semi-dominant, and as with most semi-dominants is a colour modifier rather than a colour. We haven’t really looked at genetic formula for dominants (or semi-dominants) yet, except to say to use the capital letters of the lower case letter used for the recessive mutation. We use a capital D for dark, but the small letter means non-dark (or light) d. Remembering that we will always have genes in pairs, DD will be the dark shade, Dd the medium shade, and dd the light shade. Being a modifier it is added to the original colour formula so that Bd/bd or Bbdd informs us that this bird is light green/blue. Just for practice BCD/Bdc is jade/yellow, taking for granted that jade is the mid colour. So now any mating which involves depth of colour as well as colour can include D and d. In the jade/yellow formula B/B tells us that it’s a dominant non-blue bird, so is green in colour. D/d gives us the depth of colour, middle, and C/c tells us the intensity, full colour split for dilute colour. You will notice that D/d is not dark/light. Nothing can be split for dominant or semi-dominant. Dilute was a recessive modifier, dark is a semi-dominant modifier. The breeding results for dark and light are similar to the semi-dominant pieds we discussed before. Dark x dark gives 100% dark, light x light gives 100% light. Dark x light gives 100% medium. Medium x medium gives 25% dark, 50% medium, and 25% light. Any others can be put across a square and worked out as shown previously.