For any given couple, it is always possible to know the probable phenotypes of the offspring and the phenotypes that are impossible to breed. This will help the breeder when trying to breed a specific phenotype. Of course the calculus give only probable results or expectations and these may not always exactly fit to the practical results obtained from just one or two breeding seasons. This does not mean that the calculations are wrong. Instead it means that the practical results deviated from the expected average.
To predict the percentages of possible different colours obtainable from a couple there are three steps:
Step 1 – Knowledge of the parent's genotypes.
Knowledge of the parent's genotypes is essential to the process of predicting the offspring's possible phenotypes. If the genotype for one or both parents is unknown, it can usually be inferred by comparison with a few close relatives. On the following examples I will use a dark green cock and a light green hen.
Genotype

Phenotype

Locus for melanin production

Locus for yellow pigment production

Locus colour depth


dil^{+}
dil (Heterozygous)

bl^{+ }bl^{1} (Heterozygous)

D^{+} D (Heterozygous)

Dark green cock (mutant I)

dil^{+}
dil (Heterozygous)

bl^{1 }bl^{1(Homozygous)}

D^{+} D^{+} (Homozygous)

Light blue hen

Step 2 – Application of Mendel's chart.
[ Example A will show the inheritance of the genes
carried on the locus for melanin production. Example B will show
the simultaneous inheritance of the loci for yellow pigment production
and colour shade. These two loci should always be studied together
because of the linkage between them.
[ Determine all types of sperms and ova the
couple can produce. For each situation draw a chart where the number of columns
equals the total number of sperms and the number of lines equals the total
number of ova.
[ Write all types of sperms on the top
horizontal squares (red) and all the types of ova on the left vertical squares
(red).
[ Fill each empty square (blue) with the sperm
on the top square and the ova on the left square.
Step 3 – Interpretation of Mendel's chart: determination of the offspring's expected phenotypes.
[ The number of squares where the cock and
hen's genes are combined (blue) corresponds to 100% of the possible offspring.
[ Each blue square represents a possible
genotype for the offspring. The same genotype may appear more than once and
different genotypes may express the same phenotype.
[ Because the total number of blue squares
is 4 (on both examples) than each blue square corresponds to 25% chances to
obtain a specific phenotype. However, example B concerns two linked loci.
When this happens, the production of normal sperms (bl^{+} D^{+}
and bl^{1} D) is greater then the production of "crossover" sperms (bl^{+}
D and bl^{1} D^{+}). Although we cannot be entirely sure of the
exact proportions of normal and "crossover" sperms, we can safely say that
more than 50% of normal sperms are produced (or more than 25% of each type of
normal sperm). On the other hand, less than 50% of "crossover" sperms are also
produced (or less than 25% of each type of "crossover" sperm).
[ Multiplying the number of squares where
genotypes express the same phenotype by the correspondent percentage value, you
get the probability of obtaining that specific phenotype.
[ Let's say, from example A you want to
calculate the probability of obtaining chicks with normal melanin distribution.
Three genotypes correspond to birds with normal melanin distribution; 3 x 25%
equals 75%. This means that 3 out of 4 birds (75%) from this couple are
expected to show normal melanin distribution.
[ From example B, you want to
calculate the probability of obtaining cobalt chicks. There is only one
genotype that produces cobalts. Because cobalts originate from one type of
normal sperms there are more than 25% chances to obtain this phenotype.
[ Combining both expectations the result is:
75% of all cobalt chicks are expected to show normal melanin distribution; or,
more than 25% of birds with normal melanin distribution are expected to be
cobalts.
