Barebones genetics guide

[paulh]

A while back I tried to write a genetics guide with the minimum of jargon and the maximum information in the fewest words. What material was left out that should have been included?

GENETICS GUIDE

Definitions:
1. Genotype = the actual identity of the genes.
2. 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.

3. Normal = wild type = 1. The most common phenotype in the members of a given species in the wild. 2. The most common gene found in a given gene pair in the members of a given species in the wild.
4. Mutant = 1. NOT the most common phenotype in the members of a given species in the wild. 2. NOT the most common gene found in a given gene pair in the members of a given species in the wild.

5. Homozygous = the two genes in a gene pair are the same. A homozygous snake may have either a normal or mutant phenotype. It depends on the genes in the gene pair.
6. Heterozygous (slang, het) = the two genes in a gene pair are NOT the same. A heterozygous snake may have either a normal or mutant phenotype. It depends on the genes in the gene pair.

Every vertebrate has thousands of gene pairs. For simplicity, we ignore the homozygous normal gene pairs. This is like a repairman. He concentrates on the malfunctioning machine or machines in an assembly line and ignores all the machines that are working as expected.

Each sperm or egg gets one member of each gene pair. If the gene pair is homozygous, then every sperm or egg has the same gene. If the gene pair is heterozygous, then half the sperm or eggs get one gene and the other half of the sperm or eggs get the other gene. When a sperm fertilizes an egg, the gene pairs are reestablished.

Simplest case in genetics: All but one of the gene pairs are homozygous normal, to the best of our knowledge and belief. There is one gene pair of concern. Each gene in the pair could be either gene A or gene a. That makes three possible gene pairs: AA, Aa, aa. These gene pairs make nine possible matings (male x female) with half of the babies with each genotype expected to be males and half females:
AA x AA --> 1/1 AA
AA x Aa --> 1/2 AA, 1/2 Aa
AA x aa --> 1/1 Aa
Aa x AA --> 1/2 AA, 1/2 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
aa x Aa --> 1/2 Aa, 1/2 aa
aa x aa --> 1/1 aa
The fractions are the probability of a given genotype. Each baby's genotype is a new roll of the genetic dice.

When sex is not an issue, the nine possible matings can be reduced to six:
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

The phenotype of the heterozygous (genotype = Aa) snake determines whether the mutant gene is dominant, codominant, or recessive to the normal gene.

If A is the mutant gene and a is the normal gene and A is dominant to a,
AA --> homozygous mutant phenotype
Aa --> phenotype is the same as the AA phenotype
aa --> homozygous normal phenotype

If A is the mutant gene and a is the normal gene and A is codominant to a,
AA --> homozygous mutant phenotype
Aa --> heterozygous mutant phenotype (can be distinguished from both AA and aa phenotypes)
aa --> homozygous normal phenotype

If a is the mutant gene and A is the normal gene and a is recessive to A,
AA --> homozygous normal phenotype
Aa --> phenotype is the same as the AA phenotype
aa --> homozygous mutant phenotype

Lets say there is a different gene pair of interest, the B gene pair. There are two possible genes: B and b. These make three possible gene pairs: BB, Bb, bb. The six possible B gene pair matings are the same as the six possible A gene matings -- simply put in B for A and b for a.

Often a mating has two or more gene pairs of interest. In such cases, determine the parental genotypes for each gene pair. Find the appropriate result for each single gene pair. Then match all the results of the second pair with each result from the first pair and multiply the fractions.

Example: Parental genotypes are Aa Bb mated to Aa Bb.
Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Bb x Bb --> 1/4 BB, 2/4 Bb, 1/4 bb

Final result:
1/4 AA - 1/4 BB = 1/16 AA BB
1/4 AA - 2/4 Bb = 2/16 AA Bb
1/4 AA - 1/4 bb = 1/16 AA bb

2/4 Aa - 1/4 BB = 2/16 Aa BB
2/4 Aa - 2/4 Bb = 4/16 Aa Bb
2/4 Aa - 1/4 bb = 2/16 Aa bb

1/4 aa - 1/4 BB = 1/16 aa BB
1/4 aa - 2/4 Bb = 2/16 aa Bb
1/4 aa - 1/4 bb = 1/16 aa bb
(The fractions are the odds of a given outcome for each baby, not for a whole litter.)

For a third pair of interest, match all of the results of the third pair of genes with each result from the first two pairs of genes. And so on.

When the final genotypes are determined, convert them to the appropriate phenotypes.

From here on, it is mostly learning shortcuts and practice, practice, practice.

Here is one shortcut. Often people are primarily interested in the probability of getting one genotype from among the possible ones.

Example 1. The corn snake's albino mutant gene and the anerythristic mutant gene are in different gene pairs. Both mutant genes are recessive to the corresponding normal genes. What is the probability of getting an albino anerythristic (AKA snow) baby corn snake from a het albino, het anerythristic (AKA het snow) corn snake mated to another het albino, het anerythristic (AKA het snow)?

Symbols:
A = normal, a = albino
B = normal, b = anerythristic

Parental genotypes: Aa Bb x Aa Bb
Snow genotype = aa bb

Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Bb x Bb --> 1/4 BB, 2/4 Bb, 1/4 bb

Combine the probability of aa with the probability of bb:
1/4 aa - 1/4 bb = 1/16 aa bb (snow)

Example 2. The ball python's pinstripe mutant gene and pastel mutant gene are in different gene pairs. The pinstripe mutant gene is dominant to the corresponding normal gene. The pastel mutant gene is codominant to the corresponding normal gene. What is the probability of getting a pinstripe pastel baby ball python from a pinstripe pastel ball python mated to a pinstripe? (The pinstripe pastel ball python came from a pinstripe x pastel mating. The pinstripe ball python came from a pinstripe x normal mating.)

Symbols:
A = pinstripe, a = normal
B = pastel, b = normal

Parental genotypes: Aa Bb (pinstripe pastel) x Aa bb (pinstripe)
pinstripe pastel genotype = either AA Bb or Aa Bb = A_ Bb

Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa = 3/4 A_, 1/4 aa
Bb x bb --> 2/4 Bb, 2/4 bb

Combine the probability of A_ with the probability of Bb:
3/4 A_ - 2/4 Bb = 6/16 A_ Bb = 3/8 A_ Bb (pinstripe pastel)

[dr del]

I'm not sure - I skimmed so are all the allelic combo's covered?

The tend to be the ones that cause the most confusion.

[paulh]

Multiple alleles did not make it in. The file was intended to be short, so a lot of stuff was left out, like meiosis, chromosomes, Punnett square, the difference between codominance and incomplete dominance (because at the breeder's level it is not significant), etc.