How
to predict the outcome of a breeding event when two pairs of genes
are involved
This is relevant in poultry genetics because there
are traits that depend on two or more gene pairs. The comb type
is an example. Shank and foot color are traits that depend on three
pairs of genes. What we do in a case such as comb type that is determined
by two pairs of genes, is to determine the combinations of the genes
of each parent for both sets of genes. We then realize that the
combinations of the first genes can occur with any combination of
the second genes. So, we have to consider all the possible combinations
of the genes of the first set with the genes of the second set.
An example illustrates this. Suppose a trait is determined by two
sets of gene pairs on different chromosomes. A male with a genes
(A, B) and (W, X) for this trait is crossed with a female having
genes (a, b) and (w, x) for the same trait. The possible combinations
from the first gene pair are: (A, a), (A, b), (B, a) and (B, b).
The possible combinations of the second set of genes are: (W, w),
(W, x), (X, w) and (X, x). We’re not finished because each
combination of the first set of genes can occur with any combination
of the second set of genes. To determine these ‘double’
combinations it is helpful to make a drawing:
The arrows in the figure above indicate the combinations of the
(A, a) gene pair with the four combinations of the second genes:
(A, a) with (W, w); (A, a) with (W, x); (A, a) with (X, w); and
(A, a) with (X, x). Next we do the same thing except using the (A,
b) gene pair instead of the (A, a) one. Then again with the (B,
a) gene pair and lastly with the (B, b) gene pair. This gives 16
‘double’ combinations.
Punnett diagram for the inheritance of two traits
The inheritance of two sets of gene pairs can be determined by
the use of a Punnett diagram. In these examples, I start with single
traits and work with Punnett diagrams. The strategy is to use Punnett
diagrams to determine the combinations for the single traits separately,
then use those Punnett square combinations to make the larger Punnett
square for the inheritance of both traits together. The Araucana
black large fowl is homozygous for both dominant black, E, and pea
comb, P. The Rhode Island Red is homozygous for recessive wheaten,
ey and single comb, p+ (which is really the lack of the pea comb
gene). A mating between an Araucana black large fowl and a Rhode
Island Red gives sons and daughters that belong to the F1 generation
and are all heterozygous for dominant black E, recessive wheaten,
ey, P (pea comb) and p+ (lack of pea comb). A mating between two
of these F1 chickens is what I consider here. So, this is what you
would get in the F2 generation.
The F1 sons and daughters are heterozygous for both traits. The
F1 fowls are (E, ey) and (P, p+). Since we know this already, we
don't need to use a Punnett square to determine the distribution
of traits in the F1 generation. We do, however, need to use a Punnett
square to determine how the genes combine with each other in the
F2 generation (F1 x F1). Then we use a Punnett square again to determine
how the two separate traits appear with each other. So the Punnett
square for first trait has E and ey on both edges. The other one
has P and p+ on both edges. These two Punnett squares determine
how the genes for these traits are inherited separately in the F2
generation from F1 matings. The Punnett square for the inheritance
of the individual traits are:
Now we make another Punnett square to determine how the two traits
appear with each other in the F2 generation. The combinations from
each of these single-trait Punnett squares above is used along an
edge of a new and larger Punnett square to determine how the genes
will appear together in the F2 chickens that come from mating F1
chickens to each other:
The gene combinations that give the same phenotype (appearance)
have the same color in the Punnett square above. So, the F2 generation
from a mating of Araucana and Rhode Island Red will have one fourth
that are double heterozygotes (heterozygous for both traits). These
are the four center squares above. This gene combination occurs
four times out of 16, so the percent is 25%. These will probably
be black with a 'poor' pea comb. It may be difficult or impossible
to distinguish them from the EE Pp+ birds. Only 1/16 will be black
with a good, homozygous pea comb (the EE PP combination in the corner).
Because sex-linked traits are popular with fanciers, this next example
considers the inheritance of two sex-linked traits. Consider the
cross between a male that is heterozygous for silver and barring
and a female that is red, non-barred. His genotype is (S, s+) with
respect to silver and (B, b+) for barring. Her genotype is (s+,
- ) with respect to silver and (b+ - ) with respect to barring.
First, I make a Punnett square for the inheritance of the silver
trait and another for the barring trait. I then make another Punnett
square with the results of these Punnett squares.
For the silver trait, the Punnett square is:
The Punnett square for the barring trait is:
The results of these two Punnett diagrams are used to make a larger
one. In the Punnet square below, I have written the results of the
Punnett square for the silver trait across the top and the results
of the Punnet square for the barring trait down the left-hand side.
The interior squares of the Punnett diagram are filled the same
way they are for the smaller Punnett diagrams.
Since there are 16 interior squares in this Punnett Diagram, there
are 16 possible combinations of the four genes. Because these genes
are linked, the appearance of each combination is not equally probable.
See the discussion of linkage below. If the two traits, silver/gold
and barring/non-barring had not been on the same chromosome, then
we could consider each combination in the Punnett diagram above
to be equally probable with the probability being the number of
times a specific combination appears in the Punnett diagram divided
by the number of squares inside the Punnett diagram.
Adding a third gene pair, as one would have to do to consider shank/foot
color (assuming those genes are on different chromosome), is essentially
the same procedure just expanded by a third gene pair. This gives
64 total combinations.
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