RINGNECK GENETICS FOR THE BEGINNER
by John Fowler
(7/2001)

IV. PREDICTION METHOD

Every breeder wishes to know what the possible offspring are from a particular cross. In this article I want to share a simple method any breeder can use to determine possible results when the genotype of the parents is known. This article will utilize what we have learned in the previous articles. You may wish at any time to refer back to the previous articles by using the LINKS below.

The wild type or normal is the color standard of reference in the Ringneck Dove (Streptopelia risoria). This color is a dark gray with a violet head and breast. At the sex-linked locus there are three alleles -- Dark, blond, white. The normal color (wild type) is dark. Blond can be thought of as a dilute and white as an extreme dilute of the normal color. The gene symbols we will be using for the sex-linked locus are:

D⁺ = Dark or wild type or normal
d^B = Blond (recessive to D⁺)
d^w = White (recessive to D+ or d^B)
W = W chromosome (which signifies a female and lacks the "d" locus)

There are four autosomal recessive genes for color in the ringneck dove. These four include: albino (gene symbol = al), ivory (gene symbol = iv), pied (gene symbol = pi), and rosy (gene symbol = ry). Tangerine (gene symbol = Ta) is an autosomal co-dominant. Frosty (gene symbol = Fr) is likely a co-dominant also, but that is not yet conclusively proven.

The accepted method for symbolizing genes is small letters for recessive alleles and capitals for dominant alleles. Remember that the non-mutant allele at any locus is the wild type or normal allele. It is symbolized by a "+". A slash mark (/) will be used to symbolize the chromosome on which the allele is located.

In this example, we will pair a Blond Frosty male with a Dark Pied female. The male is homozygous for blond at the sex-linked locus and heterozygous for Frosty and Pied at the autosomal loci. The female is hemizygous for dark at the sex-linked locus and homozygous for wild type at the autosomal locus for Frosty and homozygous for Pied at the Pied locus. We will work a Punnet square for each locus and then apply the arithmetic method as we combine the gene pairs to determine offspring probability.

At the sex-linked locus the genotype for the male will be d^B//d^B and for the female will be D⁺//W. Because the male has two sex chromosomes the male gametes will segregate into d^B/ and d^B/. The female has only one sex chromosome and the gametes will segregate as D⁺/ and W/ (meaning that the sex chromosome is absent).

1/2 D⁺//d^B = Dark (heterozygous blond) male
1/2 d^B//W = Blond Female

Since autosomes occur in pairs (unlike the sex chromosomes), both males and females have equal numbers of genes on the autosomes. The genotype of the male at the Frosty locus is Fr//+ and the gametes will segregate as Fr/ and +/. The genotype of the female at the Frosty locus is +//+ and the gametes will segregate as +/ and +/. Because Frosty is a dominant or co-dominant gene it will express itself in the heterozygous condition.

Here's the Punnett Square for the Frosty locus.

1/2 Fr//+ = Frosty
1/2 +//+ = Wild Type

The genotype of the male at the pied locus is +//pi and the gametes will segregate as +/ and pi/. The genotype of the female at this locus is pi//pi and the gametes will segregate as pi/ and pi. Because Pied is a recessive gene it must be in the homozygous condition to be expressed

Here's the Punnett Square for the pied locus.

1/2 +//pi = Heterozygous Pied = Wild Type
1/2 pi//pi = Pied

Now we will use the arithmetic method of multiplying the fractions as we combine the gene pairs. This will give us an expected probability of offspring that will prove true over many offspring.

(1/2 D⁺//d^B)(1/2 Fr//+)(1/2 pi//pi) = 1/8 Dark Frosty Pied = Ice Male

(1/2 D⁺//d^B)(1/2 Fr//+)(1/2 +//+) = 1/8 Dark Frosty Male

(1/2 D⁺//d^B)(1/2 +//+)(1/2 pi//pi) = 1/8 Dark Pied Male

(1/2 D⁺//d^B)(1/2 +//+)(1/2 +//+) = 1/8 Dark Male

(1/2 d^B//W)(1/2 Fr//+)(1/2 pi//pi) = 1/8 Blond Frosty Pied = Ice Female

(1/2 d^B//W)(1/2 Fr//+)(1/2 +//+) = 1/8 Blond Frosty Female

(1/2 d^B//W)(1/2 +//+)(1/2 pi//pi) = 1/8 Blond Pied Female

(1/2 d^B//W)(1/2 +//+)(1/2 +//+) = 1/8 Blond Female

From this example, you can see that when you pair a female of a darker color to a male of a lighter color that the sons will be the color of the mother and the daughters the color of the father.

You can find an excellent table of genotypes and description of ringneck plumage colors by referring to Dr. Wilmer J. Miller's website (www.ringneckdove.com).