As you can see in the wonderful world of colorbred canaries with all it’s various types let’s get into understanding what is happening when you cross your canaries. It’s time to get acquainted to words and laws that will help you get a better understanding.
Genotype: A set of genes that individually possesses and determines expression of certain mutations. A bird that carries another mutative gene. (Heterozygous)
Example: A Intensive Yellow carrier of Recessive White.
Phenotype: Mutations that can be seen externally in an individual. Phenotype is an expression of mutation in the Genotype. It is also influenced by environmental and hormonal factors. A bird that carries the same expressive gene. (Homozygous)
Example: A Non Intensive Red Ivory Factor
Recessive: Carrying a character than will be intervened by another character.
Example: Red Agate Mosaic x Recessive White Agate Opal= Red Agate Mosaic carriers of Opal and Recessive White.
Linked to Sex: Males can only be carriers. Doesn’t follow Mendel’s 3rd Law of Inheritance of Character, genes normally tend to stay together unless cross linking occurs. Cross Linking is when a linked to sex mutation male breeds to a classic hen the results would be all hens will be that linked to sex mutation and all males will be carriers.
Example: Yellow Brown Pastel Mosaic Male x Yellow Brown Mosaic Hen= Yellow Brown Mosaic Males carrier of Pastel and Yellow Brown Pastel Mosaic Females.
Mendel’s First Law
Law of Uniformity
By crossing two different individual homozygous for certain characters all offspring are heterozygous.
Example: Parents are a Yellow Male x Recessive White Female
The male’s genotype is (cb+/cb+) meanwhile the female’s genotype is (cb/cb)
To check possibilities you must use the first expressed gene in the males equation to both of the hens. (cb+/cb) (cb+/cb). By taking the male first gamete cb+ I paired it with the first of the hens gamete cb thus giving us (cb+/cb). Using the same first gamete I paired it again with the hen’s second gamete which is again cb, giving us the same results.
Male Gene (cb+/cb+) Hen Gene (cb/cb) Male Gene (cb+/cb+) Hen Gene (cb/cb)
| | | |
cb+ cb cb+ cb
(cb+/cb) (cb+/cb)
Male Gene (cb+/cb+) Hen Gene (cb/cb) Male Gene (cb+/cb+) Hen Gene (cb/cb)
| | | |
cb+ cb cb+ cb
(cb+/cb) (cb+/cb)
Full Answer: (cb+/cb) (cb+/cb) (cb+/cb) (cb+/cb)
The gametes for the male are cb+ and the gametes from the hen are cb.
This now shows that all offsprings will be all Yellow carriers of Recessive White. (cb+/cb)
Mendel’s 2nd Law
Law of Separation
Crossing the two offsprings from our first example above, now that we have both male and female being Yellow carriers of Recessive White you will now see the how phenotype will have a chance to be expressed.
Example: Parents are Yellow Carrier of Recessive White Male x Yellow Carrier of Recessive White Female.
The male and female genotype is (cb+/cb) (cb+/cb)
Male Gene (cb+/cb) Hen Gene (cb+/cb) Male Gene (cb+/cb) Hen Gene (cb+/cb)
| | | |
cb+ cb+ cb+ cb
(cb+/cb+) (cb+/cb)
Male Gene (cb+/cb) Hen Gene (cb+/cb) Male Gene (cb+/cb) Hen Gene (cb+/cb)
| | | |
cb cb+ cb cb
(cb/cb+) (cb/cb)
As you can see this now has 3 different outcomes and also shows the percentages since each option.
The full answer is (cb+/cb+) (cb+/cb) (cb+/cb) (cb/cb).
25% Chance of being Yellow (cb+/cb+)
50% Chance of being Yellow Carrier of Recessive White (cb+/cb)
25% Chance of being Recessive White (cb/cb)
Mendel’s 3rd Law
Law of Inheritance
In regards to coloration of eyes may come into play it adding ino+ or ino will show if the outcome of the eyes.
Example: Parents are a Yellow Male and a Recessive Albino Hen
Male genotype is (cb+/cb+) (ino+/ino+)
Female genotype is (cb/cb) (ino/ino)
Male Gene (ino+/ino) Hen Gene (ino/ino) Male Gene (ino+/ino+) Hen Gene (ino/ino)
| | | |
Ino+ ino ino+ ino
(ino+/ino) (ino+/ino)
All babies will have black eyes carrier of having red eyed offsprings.
The whole answer would be (cb+/cb) (ino+/ino) Yellow Carrier of Recessive White and Ino.
Now let’s combine all 3 laws together here in this new example to show you how many multiple possibilities may occur. From using the 3rd law’s offspring answer what would happen if you would breed a Yellow Carrier of Recessive Albino to a Yellow Carrier of Recessive Albino.
Parents: Yellow Carrier of Recessive Albino Male x
Yellow Carrier of Recessive Albino Hen
The parents genotype is (cb+/cb) (ino+/ino)
Male Gene (cb+/cb) Hen Gene (cb+/cb) Male Gene (cb+/cb) Hen Gene (cb+/cb)
| | | |
cb+ cb+ cb+ cb
(cb+/cb+) (cb+/cb)
Male Gene (cb+/cb) Hen Gene (cb+/cb) Male Gene (cb+/cb) Hen Gene (cb+/cb)
| | | | cb cb+ cb cb
(cb+/cb) (cb/cb)
Male Gene (ino+/ino) Hen Gene (ino+/ino) Male Gene (ino+/ino) Hen Gene (ino+/ino)
| | ` | |
Ino+ ino+ ino+ ino
(ino+/ino+) (ino+/ino)
Male Gene (ino+/ino) Hen Gene (ino+/ino) Male Gene (ino+/ino) Hen Gene (ino+/ino)
| | | |
Ino ino+ ino ino
(ino+/ino) (ino/ino)
The answer is: (cb+/cb+) (cb+/cb) (cb+/cb) (cb/cb) (ino+/ino+) (ino+/ino) (ino+/ino) (ino/ino)
You can also do it another way combining both genotypes together.
Male Gene (cb+/cb) (ino+/ino) Hen Gene (cb+/cb) (ino+/ino)
| |
cb+ ino+ (cb+/ino+)
Male Gene (cb+/cb) (ino+/ino) Hen Gene (cb+/cb) (ino+/ino)
| |
cb+ ino (cb+/ino)
Male Gene (cb+/cb) (ino+/ino) Hen Gene (cb+/cb) (ino+/ino)
| |
cb ino+ (cb/ino+)
Male Gene (cb+/cb) (ino+/ino) Hen Gene (cb+/cb) (ino+/ino)
| |
cb ino (cb/ino)
There are 16 different possibilities shown at the current moment. Now I think it’s time to implement the punnet square by using the gametes.
cb+/ino+ cb+/ino cb/ino+ cb/ino
cb+/ino+ cb+/cb+ cb+/cb+ cb+/cb cb+/cb
ino+/ino+ ino+/ino ino+/ino+ ino+/ino
cb+/ino cb+/cb+ cb+/cb+ cb+/cb cb+/cb
ino+/ino+ ino/ino ino+/ino ino/ino
cb/ino+ cb+/cb cb+/cb cb/cb cb/cb ino+/ino+ ino+/ino+ ino+/ino+ ino+/ino
cb/ino cb+/cb cb+/cb cb/cb cb/cb
ino+/ino ino/ino ino+/ino ino/ino
Out of the 16 possibilities displayed above: 9 out of 16 are Yellow with black eyes, 3 out of 16 are Recessive White with black eyes, 3 out of 16 are Lutinos and 1 out of 16 are Albino. Typically this is the 9:3:3:1 portion of Mendel’s 3rd law.
Linked to Sex Mutations
Crossing Over Genetics
To what some may already know the Isabel mutation was created by the cross of an agate male and a brown hen. The phenomenon that occurred was due to the crossing over the original 4 gametes with the 0 gamete in the hen. Example: Parents are Agate Male and Brown Hen.
Male Gene z+rb/z+rb Hen Gene zrb+/0 Male Gene z+rb/z+rb Hen Gene zrb+/0
| / | /
| / | /
| / | /
z+rb/ zrb+ z+rb/ 0
z+rb+ |
zrb ------ zrb/ 0
Isabel
The crossover occurred when we realized that by combining both eumelanin gene with the dilution gene are together changing the dominant gene which is the black gene z+rb+ to crossover losing the black eumelanin and became diluted to zrb thus the first Isabel to be created.
Mutations that are linked to sex are: Brown, Agate, Isabel, Pastel, Ivory, Satine, Greywing and Jaspe SD/DD.
Recessive Mutations
With these specific mutations following under the recessive category, the answer is pretty simple. If you breed a recessive to a recessive all you will get is recessive however unlike the linked to sex gene if you breed a recessive to a dominant all babies will be carriers. As well as if you breed a carrier to a recessive you will get 50% carriers and 50% recessive. The outcome is very predictable, the quality depends on the breeder.
Example: Male Yellow Black Topaz x Female Dominant White Black
The male genotype is (G+/G) (z+rb+/z+rb+) (to/to)
The female genotype is (CB+/CB+) (z+rb+/z+rb+) (to+/to+)
(G+/G) (z+rb+/z+rb+) (to+/to) Yellow Black Carrier of Topaz
(CB+/CB+) (z+rb+/z+rb+) (to+/to) Dominant White Black Carrier of Topaz
All Offsprings are 100% Carriers
Example 2: Male Yellow Black Topaz x Female Dominant White Black Carrier of Topaz
The male genotype is (G+/G) (z+rb+/z+rb+) (to/to)
The female genotype is (CB+/CB+) (z+rb+/z+rb+) (to+/to)
(G+/G) (z+rb+/z+rb+) (to/to) Yellow Black Topaz
(CB+/CB+) (z+rb+/z+rb+) (to/to) Dominant White Black Topaz
(G+/G) (z+rb+/z+rb+) (to+/to) Yellow Black Carrier of Topaz
(CB+/CB+) (z+rb+/z+rb+) (to+/to) Dominant White Black Carrier of Topaz
All Offsprings are 50% Carrier and 50% Recessive
Example 3: Male Yellow Black Carrier of Topaz x Female Dominant White Black Carrier of Topaz
The male genotype is (G+/G) (z+rb+/z+rb+) (to+/to)
The female genotype is (CB+/CB+) (z+rb+/z+rb+) (to+/to)
(G+/G) (z+rb+/z+rb+) Yellow Black
(CB+/CB+) (z+rb+/z+rb+) Dominant White Black
(G+/G) (z+rb+/z+rb+) (to+/to) Yellow Black Carrier of Topaz
(CB+/CB+) (z+rb+/z+rb+) (to+/to) Dominant White Black Carrier of Topaz
(G+/G) (z+rb+/z+rb+) (to/to) Yellow Black Topaz
(CB+/CB+) (z+rb+/z+rb+) (to/to) Dominant White Black Topaz
All Offsprings have a 25% chance of being a pure classic mutation losing topaz completely, 50% chance they carry topaz still and 25% chance of obtaining topaz.
Mutations that are recessive are: Recessive White, Ino, Urucum, Opal, Topaz, Eumo, Cobalt, Onyx, Mogno and Phaeo
