Quote Originally Posted by ChelseaV View Post
... I want to make sure I fully understand how recessive work. The same goes for any recessive really.
http://www.redtailboas.com/f115/no-f...s-guide-53782/

http://ghr.nlm.nih.gov/handbook

The links above may be helpful.

Genotype = the actual identity of the genes.
Phenotype = The creature's appearance and any other physical or behavioral manifestation. The phenotype is produced by all the genes and the environment working together.

How to find the probability of a given snake being het piebald (or any other het recessive).

1. In the simplest case, two genes (A and a) produce three possible gene pairs (AA, Aa, aa). Determine the genotypes of the parents and progeny. Remember that each parent gives one member of each gene pair to each baby.

2. Determine the phenotypes associated with each of the genotypes.
Example -- A = normal gene. a = piebald gene.
Gene pairs -- AA = Genotype is homozygous normal. Phenotype is normal.
Aa = Genotype is heterozygous (het) piebald. Phenotype is normal.
aa = Genotype is homozygous piebald. Phenotype is piebald.

Mating:
Aa x Aa --> 1/4 AA (normal), 2/4 Aa (normal), 1/4 aa (piebald)

3. Add the fractions of the like phenotypes together.
1/4 AA (normal) + 2/4 Aa (normal) = 3/4 normal phenotype

4. Divide the AA fraction by the total. Divide the Aa fraction by the total.
1/4 AA divided by 3/4 total = 1/3 = 33% (rounded down) probability
2/4 Aa divided by 3/4 (total) = 2/3 = 66% (rounded down) probability

Dominant genes are the mirror image of recessive genes. With codominant genes, each of the three gene pairs has its own phenotype so there is no need to calculate probabilities. By the way, 100% probability het ____ is a long way of writing het _____.

Clear as mud?

Here are six possible matings. Try working out the genotype probabilities. A = the normal gene, a = the corresponding albino gene. The a gene is recessive to the A gene and the A gene is dominant to the a gene.
AA x AA --> 1/1 AA
AA x Aa --> 1/2 AA, 1/2 Aa
AA x aa --> 1/1 Aa
Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Aa x aa --> 1/2 Aa, 1/2 aa
aa x aa --> 1/1 aa

Try working out the genotype probabilities of those 6 matings. A = the pinstripe gene, a = the corresponding normal gene. The a gene is recessive to the A gene and the A gene is dominant to the a gene.