Saud's Aviary

This blog will detail hand rearing of two African grey chicks. You may read about their parents and the clutch  here . The two chicks hatched on Jan 6 and Jan 9, 2012. They were pulled on Jan 22 with one aged 13 days and the other 16 days. The chicks were close banded on Jan 22. The band id of the chicks are SAUD-202 and SAUD-203 respectively. Here is their first photo and their data. SAUD-202 (L) SAUD-203 (R) 13 and 19 days Old ID: SAUD-202 Hatch Date: Jan 6, 2012 Pulled on: Jan 22, 2012 (16 days) Weight: 129g ID: SAUD-203 Hatch Date: Jan 9, 2012 Pulled on: Jan 22, 2012 (13 days) Weight: 117g The chicks are housed inside a homemade brooder made out of an old fish tank. The bottom is covered with wood shavings and lined on top with tissue paper. The brooder temperature is maintained at 90'F. The chicks hatched 3 days apart but their size does not differ significantly. The older chick had its eyes open at the time of pulling while

Pair #5 This is a Red-Factor pair. The female has several red feathers on her chest and legs and the male has a splash of red feathers on its back. They have laid their first egg today on Jan 19, 2012 . I am very eager to see if their chicks will show any red feathers. However, I am not getting my hopes up as the males is a young bird. Possibly only 4 years old. It is not very likely that the eggs are fertile. I will know for sure in about 2 weeks.  Jan 21, 2012 There are now two eggs. The hen has started incubation. The pair spends the night inside the box. The nest box does not have provision for a camera. It is a new box that I purchased a few months back but go lazy about providing a hole for a camera. This means physical nest box inspections throughout the incubation period. Jan 24, 2012 Still two eggs. I think this will be the clutch size. The eggs are not showing fertile yet. I will check them again in a few days. Jan 29.2012 Both eggs are still showing cle

In this final chapter of our tutorial we will list secondary mutations that are a combination of two or more primary mutations. Knowing which primary mutations combine together to create a particular secondary mutation is important to calculate genetic outcome using both the Punnett squares method and the genetic calculator. For example, albino is a combination of Blue and Ino, when both mutations are visually present together we get a visual albino. Following is a list of common names for ringnecks and the primary mutations that are involved in each. Please note that most of these names and combinations are common to budgerigars, lovebirds and some other species.

In the last chapter we used the Punnett squares method to predict outcome in cases where birds were carrying multiple mutations. We saw that the Punnett squares method increased in complexity as the number of mutations in the parents increased. If the parents have a single mutant gene the Punnet Squares method has 4 cells. With two mutant genes involved, it can have up to 16 cells and with three the number could go up to 64. We therefore need a simpler method to work out the outcomes in such cases. Fortunately, there are genetic calculators that can be used in such cases. Here we will discuss the use of Gencalc ( http://www.gencalc.com/ ). If you have understood what we discussed in earlier chapters, then this should be very simple for you. Even if you got lost somewhere along the way it should still be possible for you to learn to use the genetic calculator.  On the main page, select the species that you want to work with. This will open up the calculator page for that species.

In Pied birds, the melanin (blue pigment) is not produced uniformly in the feathers. In some areas the pigment is produced normally while in other areas the melanin is not produced at all. In a green bird the areas where no melanin is produced appear yellow. These areas are usually random in shape and size.  There are two different types of pied mutations. The dominant pied and the recessive pied. Some species of birds have dominant pieds, some have recessive pieds and some have both. The budgerigars, ringnecks and lovebirds have both types of pieds; the cockatiels have recessive pieds only. The dominant pied gene (pi) is incomplete dominant to the normal. If you pair a dominant pied (SF) to a normal you will get 50% dominant pieds and 50% normals. Double factor dominant pieds show a heavier degree of piedness (yellow areas) than a single factor dominant pied  The recessive pied gene (s) is recessive to the normal. Pairing a recessive pied to a normal produces 100% normals split to re

In Chapter 4.2 Alleles, we defined alleles as two or more variants (versions) of the same gene that occupy the same position (or locus) on the DNA. We discussed blue and yellow-face blue genes as an example of alleles. Sex-linked genes can also have alleles. For example, in ringnecks ino and pallid are alleles of the same locus and in budgerigars ino and Texas clear-body are also alleles. Let us look at a diagram to understand the relationship between the ino and the Pallid genes. Fig 4.7 shows the X chromosome on which the ino and the pallid genes are located.  We see that the ino and the pallid gene share the same position on the X chromosome. Therefore one X chromosome can either have the pallid gene OR the ino gene but not both. Keeping this in mind all the possible combinations of the ino and pallid genes are shown in this figure. If a male has two ino genes it is a visual ino, if it has two pallid genes it is a visual pallid, if there is one pallid gene and one ino gene the

The violet gene (V) is another incomplete dominant gene that affects the birds in a manner similar to the dark factor gene. We had discussed that a dark gene affects the appearance of a bird as if you were looking at the bird through dark sunglasses. The effect of the violet factor can be thought of as looking at the bird through violet sunglasses. A green bird with a single violet gene V+/V will show a violet sheen on the green body. Two violet genes (DF) will cause a bird to appear darker than a Violet green Single Factor bird. A blue bird with a single violet gene bl/bl V/V will show a light violet blue color, Two copies of the violet gene V/V will show a rich violet color on a blue bird. The violet gene has the most striking effect when it is present along with a copy of the dark gene on a blue bird. This bird is called a cobalt violet or commonly a  visual violet . DF violet blues and DF Violet cobalts are also known as visual violets. The following table shows the various c

The dark factor gene is an incomplete dominant gene that causes the darkening of the melanin (blue/black pigment) in birds. Since the gene is incomplete dominant, two dark genes (DF) have a greater effect on the bird than a single dark gene (SF). To understand the effect of the dark gene on a bird, imagine that you put on dark sun glasses and look at a normal green bird. The green appears darker and the bird is called dark green. With a double dark factor imagine that you are looking through two pairs of dark glasses. The bird will appear darker still and this colour is called Olive. Similarly if the bird is a visual blue, a single dark gene causes the blue colour to appear darker and the bird is called a cobalt. Two dark genes further affect the blue colour and the bird is called a mauve. The following table shows the effect of the dark genes on green and blue birds.   In our markets a grey-green is often wrongly called an olive. Please note that an olive bird is genetically very d

We have talked about the grey gene and its effect on a green bird. A single grey gene gives a green bird a grayish tone and changes the color to grey-green. Two grey genes have the same effect and the bird shows the same grey-green color as with a single grey gene. If a blue bird (two copies of the blue gene) also carries one or two copies of the grey gene it shows a grey color instead of blue and is called a visual grey. A visual grey is therefore not a primary mutation but actually a combination of the grey and blue mutations. We will now look at how we may express this in standard genetic notation and use this notation to predict outcomes of birds carrying one or two copies of these genes. When dealing with multiple mutations in a single bird we must express all relevant genes in the standard notation. Following are some examples A visual single factor (SF) Grey bl/bl, G+/G A visual double factor (DF) Grey bl/bl, G/G A grey green (SF) split to blue bl+/bl, G+/G A grey green (DF) b

The Grey Gene We will now talk about the grey gene. The grey gene effects the appearance of melanin (blue/black pigment) in the feathers and gives the bird a grey appearance. This gene is dominant to the normal and therefore only a single gene is enough to cause a change in the visual appearance. We will use the symbol G to denote the grey gene. G+/G means that the bird has one grey gene and one Normal (G+) gene. Such a bird would appear grey green. A bird with two copies of the grey gene will be denoted by G/G. This bird also appears grey green and looks exactly the same as the bird with only one grey gene. When only one copy of the gene is sufficient to affect the appearance of the bird and a second copy has no further effect it is called complete dominance. A bird with one copy of the dominant gene is called a single factor (SF) and a bird with two copies of the dominant gene is called a double factor (DF) bird. Let us go back to the Punnett squares method of calculating genetic ou

Alleles Sometimes nature produces two three or even more variants (versions) of the same gene. These variants are called alleles. I will try to explain by a diagram.. We already know that in the DNA two copies of genes are present which may be the same or different. We therefore have two places where the blue or one of its variants may be present. We discussed the possibilities in the budgerigars of the two genes being bl/bl, bl+/bl or bl+/bl+. Things are a little more complicated as more than one type of the blue gene exists. These are called the Blue white face (same as simple blue), blue Yellow face type I, blue yellow face type II and the blue Golden face. All these genes are called alleles of the blue locus. The important thing to understand in this figure is that all these genes can only fit on the DNA in the place represented by the circle. Since there are only two circles on the DNA diagram, only two genes from the list of alleles given on the right can be present in

Chapter 4: Multiple Mutations So far we have discussed outcomes of pairs carrying only one mutation. Ino or blue or spangle. In this chapter we will learn how we can predict outcomes of pairings where one or both birds are carrying more than one mutation.  4.1 Genetic Notation We have used very simple notation to represent dominant and recessive genes. We used G to represent a green (normal) gene and B for a blue gene. This is not standard genetic notation and presents a very simplified view. We will now look at the placement of genes on the DNA to introduce the standard genetic notation. We have defined the gene as a section of the DNA which deals with a specific feature of the bird. A change in any gene which makes it different from the normal is called a mutation. Imagine that the DNA looks something like figure 4.1    The DNA has a long chain like structure with two copies of each gene. The different shapes shown represent different genes. Let us say that the circle represents

We have learned that a male bird has two copies of the X chromosome. A female bird has only one copy of the X chromosome and a shorter Y chromosome. A single copy of a sex linked gene is sufficient to express the mutation in the female where as two copies of the sex-linked gene are required for the male to show the same mutation. We will now learn how we can use this knowledge to predict outcomes of birds carrying sex-linked genes. We will denote a normal X chromosome with the letter X. An X chromosome with a ino gene will be denoted by XL as shown in the figure below.   Let us consider the pairing of a Lutino male to a Normal female The Lutino male is represented by XL XL as we know that for a male to appear Lutino it should have two X chromosomes with the ino gene. The normal female is represented by X Y. So the pairing becomes XL XL with X Y Putting these combinations in a Punnett square we get   To interpret the results above we must remember that the birds with two X chromosome