** Breeding for Two Traits**

**Genetics Mini-Series Article #8**

We discussed basic autosomal dominance back in article 2 of the genetics mini-series in the context of blue eggs. Using Punnett squares, we determined the probability that a particular offspring would inherit a certain trait, such as laying blue or white eggs.

But suppose we want to calculate the probability that offspring will inherit* two* desirable traits, such as laying blue eggs *and* having blue feathers – a blue blue egger. How could we figure this out?

As is explained on the Chicken Genetics for Exhibition Breeders website (which, if you’ve been following along, you are totally ready to dive into at this point), the probability of inheriting two traits together is equal to the probability of inheriting one trait times the probability of inheriting the other. [Probability(trait a, trait b) = probability(trait a) *x* probability(trait b)] In other words, the probability of inheriting blue feathering times the probability of inheriting blue egg laying equals the probability of inheriting both. **Be sure to convert the percentages into decimal values. **(100% = 1.0; 50% = 0.5; 25% = 0.25, etc.)

Suppose we had a hen that laid blue eggs, but only had one gene for it. She would be heterozygous for that trait or *Oo*. Suppose also that she is black, homozygous recessive for blue or *bl/bl*. Our theoretical rooster carries the genes for white eggs and is *oo*. His feathering, however, is blue, the result of a heterozygous *Bl/bl*. (Remember, *Bl/Bl* is splash.)

If we cross these two birds, their offspring have a 50% chance of inheriting the blue egg gene from their mom. They have a 50% chance of inheriting one blue feathering gene from their dad. Fifty percent of 50% is 25%. Thus, they have a 25% chance of inheriting both of these genes. Of course, they also have a 50% chance of being male! If I want a blue-egg laying, blue-feathered, *female* from this cross, there is a (0.5 x 0.5 x 0.5 = 0.125) 12.5% chance of getting one. (Also: ½ x ½ x ½ = 1/8.)

Here’s another twist. What if I want a blue-*barred* blue egger pullet? If I use a homozygous barred male (*BB*) with a blue female (*Bl/bl*), then I have a 50% chance for offspring which are both blue and barred. If I use the same male but with a splash female (*Bl/Bl*), then all of the offspring will be both blue and barred. They will be heterozygous for both traits (*B/- Bl/bl* females; *B/b Bl/bl* males), and both of these traits will show. If my female was also a homozygous blue egg layer (*OO*, such as a Splash Ameraucana), then all would inherit one blue egg gene, also dominant (*Oo*).

If my splash female is heterozygous (*Oo*), perhaps an Easter Egger, then half of the offspring would inherit the blue egg gene. Half of those would be female. So, in this cross, the sire is *BB blbl oo ZZ* and the dam is *bb BlBl Oo ZW*. Thus: 100% barred (1.0) x 100% blue (1.0) x 50% blue egg (0.5) x 50% female (0.5) = 25% chance for a blue barred blue-egg-laying female (*B/- Bl/bl O/o ZW*).

Take a look at combinations of genes on the Poultry Mutations page of the Genetics for Exhibition Breeders website. Remember that capitalized gene symbols are dominant and lower case symbols are recessive. Consider how you might create certain feather patterns by combining others.

One Earth Farms has created their Maiden Rock Bantams. Read the description on their website. What did they have to do to create blue-egg laying bantams with three different feathering patterns in silver or gold? It’s an inspiring project that shows what is possible.

WOW! Another great post in your mini-series. You have a knack for explaining difficult things in an understandable way.

Thank you so much, Max! I really appreciate that. 🙂