Example 3: Two genes

This is an example with two genes I've already used in the previous examples: the spangle gene Sp and the blue gene bl. This gives 9 possible genotypes:

Sp⁺ bl⁺ ------ : lightgreen Sp⁺ bl⁺ Sp⁺ bl⁺ ------ : spangle(1F) Sp bl⁺ Sp bl⁺ ------ : spangle(2F) Sp bl⁺

Sp⁺ bl⁺ ------ : lightgreen/blue Sp⁺ bl Sp⁺ bl⁺ ------ : spangle(1F)/blue Sp bl Sp bl⁺ ------ : spangle(2F)/blue Sp bl

Sp⁺ bl ------ : skyblue Sp⁺ bl Sp⁺ bl ------ : blue spangle(1F) Sp bl Sp bl ------ : blue spangle(2F) Sp bl

spangle(1F) = single factor spangle
spangle(2F) = double factor spangle

These 9 gentypes produce 6 different phenotypes (indicated with different backgrounds). Notice that the horizontal bar between the two homologues genes is interupted: this indicates that the two genes (Sp and bl) lie on different chromosomes.

From these 9 genotypes there are 45 possible parings. One of the simplest is:

Pairing 3.1: double factor spangle cock x skyblue hen

Genetic formula:

Cock Sp /Sp bl⁺/bl⁺

Hen Sp⁺/Sp⁺ bl /bl

The gametes produced by the cock are:

For the bl gene: bl⁺

For the Sp gene: Sp

Combining these possibilities:

bl⁺ Sp

The gametes produced by the hen are:

For the bl gene: bl

For the Sp gene: Sp⁺

Combining these possibilities:

bl
Sp⁺

Punnet square:

bl⁺ Sp

bl
Sp⁺ bl⁺/bl
Sp/Sp⁺

Solution:

100% spangle(1F)/blue

As before you can interchange the cock and the hen.

Suppose you want to breed "white" birds: blue spangles(2F). This means combining both the blue and the spangle gene in a homozygote state into a bird. How can you accomplish this? Since a spangle(1F)/blue carries both the blue and spangle gene, the most obvious way is by pairing two spangle(1F)/blue birds:

Pairing 3.2: spangle(1F)/blue cock x spangle(1F)/blue hen

Genetic formula:

Cock Sp⁺/Sp bl⁺/bl

Hen Sp⁺/Sp bl⁺/bl

The gametes produced by the cock are:

For the bl gene: bl⁺ bl

For the Sp gene: Sp⁺ Sp

Combining these possibilities:

bl⁺ Sp⁺ bl⁺ Sp

bl Sp⁺ bl Sp

The gametes produced by the hen are:

For the bl gene: bl⁺ bl

For the Sp gene: Sp⁺ Sp

Combining these possibilities:

bl⁺
Sp⁺ bl⁺
Sp bl
Sp⁺ bl
Sp

Punnet square:

bl⁺ Sp⁺ bl⁺ Sp bl Sp⁺ bl Sp

bl⁺
Sp⁺ bl⁺/bl⁺
Sp⁺/Sp⁺ bl⁺/bl⁺
Sp⁺/Sp bl⁺/bl
Sp⁺/Sp⁺ bl⁺/bl
Sp⁺/Sp

bl⁺
Sp bl⁺/bl⁺
Sp/Sp⁺ bl⁺/bl⁺
Sp/Sp bl⁺/bl
Sp/Sp⁺ bl⁺/bl
Sp/Sp

bl
Sp⁺ bl/bl⁺
Sp⁺/Sp⁺ bl/bl⁺
Sp⁺/Sp bl/bl
Sp⁺/Sp⁺ bl/bl
Sp⁺/Sp

bl
Sp bl/bl⁺
Sp/Sp⁺ bl/bl⁺
Sp/Sp bl/bl
Sp/Sp⁺ bl/bl
Sp/Sp

Solution:

6.25% lightgreen

12.50% lightgreen/blue

6.25% skyblue

12.50% spangle(1F)

25.00% spangle(1F)/blue

12.50% blue spangle(1F)

6.25% spangle(2F)

12.50% spangle(2F)/blue

6.25% blue spangle(2F)

We become 6.25% blue spangle(2F). This means that a young from such a pair will have a chance of 1 in 16 to be blue spangle(2F). We have to breed a lot of offspring (or use multiple pairs) to have a fair chance of breeding a blue spangle(2F). The following table illustrates this:

Number of offspring	Chance of breeding no blue spangle(2F)'s
1	93.75%
2	87.89%
10	52.45%
16	35.61%
25	19.92%
36	9.79%
46	5.14%
71	1.02%

In order to have a 50% chance to breed at least one blue spangle(2F), you need to breed more than 10 young.

There are other ways of breeding a blue spangle(2F). They are better in some respect, but will take more time. One of the best methods starts with the pairing:

Pairing 3.3: spangle(1F)/blue cock x skyblue hen

Genetic formula:

Cock Sp⁺/Sp bl⁺/bl

Hen Sp⁺/Sp⁺ bl /bl

Remember: Different chromosomepairs are not linked during meiosis!
Hence, the two genepairs split independently when the gametes are formed.

The gametes produced by the cock are:

For the bl gene: bl⁺ bl

For the Sp gene: Sp⁺ Sp

Combining these possibilities:

bl⁺ Sp⁺ bl⁺ Sp

bl Sp⁺ bl Sp

The gametes produced by the hen are:

For the bl gene: bl

For the Sp gene: Sp⁺

Combining these possibilities:

bl
Sp⁺

Punnet square:

bl⁺ Sp⁺ bl⁺ Sp bl Sp⁺ bl Sp

bl
Sp⁺ bl/bl⁺
Sp⁺/Sp⁺ bl/bl⁺
Sp⁺/Sp bl/bl
Sp⁺/Sp⁺ bl/bl
Sp⁺/Sp

Solution:

25% lightgreen/blue

25% skyblue

25% spangle(1F)/blue

25% blue spangle(1F)

For the next pairing blue spangle(1F) birds, and to a lesser extend the spangle(1F)/blue birds, can be used.

Pairing 3.4: blue spangle(1F) x blue spangle(1F).

Genetic formula:

Cock Sp⁺/Sp bl/bl

Hen Sp⁺/Sp bl/bl

I will solve this using a schorcut:

Since the mutated allele of the blue gene (bl) is the only allele of the blue gene that is involed, we know that all of the offspring will only have this allele.
In other words:
the offspring will always be bl/bl ('blue').
Hence we only have to concider the spangle gene: spangle(1F) x spangle(1F).
This we already did in pairing 1.1:

25% normal

50% spangle(1F)

25% spangle(2F)

Adding the fact that all the offspring will be bl/bl, the solution becomes:

25% skyblue

50% blue spangle(1F)

25% blue spangle(2F)

We become 25% blue spangle(2F). If your are not sure about my shortcut, you can try writing it out.

There are many more combinations possible. It is not my intention of giving all the solution, for all possible combinations. In stead I want to explain how to calculate a pairing.

Finding a solution for multiple genes is a natural extention of the two gene case:

For mutiple genes involved in the pairing when those genes are located on different chromoses:

Find the gametes for each gene seperatly as if this gene is the only one involved.

Make all the possible combinations, one allele for each gene, to become the gametes.

Note: Each gamete is equally likely to occure.

If at this point there was something you didn't understand, re-read the previous parts. The next part will depend heavily on the previous parts. If you still have problems or you think I made a mistake somewhere, please let us know.