| MadSci Network: Genetics |
The ratios can be calculated with probabilities. In the simplest case, take a single gene with alleles A (dominant) and a (recessive).The assumptions are that you are performing heterozygous crosses (Aa X Aa, e.g.) and that your genes are independantly assorting, so it's a 'coin flip' as to whether a progeny will receive one gene or another - 50/50 probability. The last assumption you make is that A shows complete dominance over a.
Now do a coin toss --
From the first parent the probability is 50% (p=.5) that you'll get the A allele expressing the dominant phenotype. Whether the second parent contributes an A or a is irrelevant. Your genotype is AA or Aa = dominant phenotype.
If the first allele is a (p=.5), then there's a 50% chance that the second allele will be the dominant allele A; p = .5 X .5 = .25 that you'll get the genotype aA (phenotypically A). Thus the probabilitiy of having the A phenotype is .5 + .25 = .75.
Lastly, the probablility of the recessive phenotype is a on the first hit, and a on the second, or p=.5 X .5 = .25
Adding things up, for an independently assorting allele the probability is .75 of obtaining the dominant phenotype and .25 of obtaining the recessive phenotype. If your genes of interest show linkage to one another, the following method for calculating ratios doesn't hold since you can't assume the 'coin toss' of a 50/50 probability for receiving one allele or another. Otherwise these probabilities hold whether you are looking at one independently assorting gene, or 16 genes!
For a double heterozygote AaBb X AaBb, simply multiply the probabilities calculated for the individual phenotypes of the possible progeny -
The probability of two dominant phenotypes AB is .75 X .75 = .5625
Ab phenotype = .75 X .25 = .1875 remember, aB phenotype = .25 X .75 = .1875 **phenotype**, not genotype! ab phenotype = .25 X .25 = .0625As a check make certain all probabilities add to one, otherwise something is missing --
.5625 AB
.1875 Ab
.1875 aB
.0625 ab
---------
1.0000
These numbers represent the same ratios as 9:3:3:1, etc. If you want
integer numbers instead of the decimal notation, calculate the total
number of permutations (or possible results of your cross) you can have
with your genes and alleles. If you're dealing with just two alleles for
each gene, the number of permutations is 4^n where n = number of genes.
For the gene A (alleles A and a) = 4^1 = 4.
If this doesn't make sense, think of a Punnet Square:
Parents: Aa X Aa (single heterozygote cross)
Gametes: A a
-----------
A | AA | Aa | F1 progeny in the boxes
----------- (First filial generation)
a | aA | aa | - note the *4* boxes..
-----------
2 possible gametes from each parents => 4 possible 'genetic combinations' in
the offspring, but only two 'phenotypic possibilities' -
.75 X 4 = 3 offspring with the A phenotype, and .25 X 4 = 1 with the
a phenotype.
For genes A and B (Alleles Aa and Bb) = 4^2 = 16 possible results.
Taking the probabilities calculated from above:
.5625 AB X 16 = 9
.1875 Ab X 16 = 3
.1875 aB X 16 = 3
.0625 ab X 16 = 1
---------
1.0000
It's no coincidence that the ratio is 9:3:3:1, as expected!
For three gene heterozygous crosses (AaBbCc X AaBbCc; n = 3 genes), the possible genotypes of progeny = 4^3 = 64.
ABC = .75 X .75 X .75 = .421875 X 64 = 27
ABc = .75 X .75 X .25 = .140625 X 64 = 9
AbC = .75 X .25 X .75 = .140625 X 64 = 9
Abc = .75 X .25 X .25 = .046875 X 64 = 3
aBC = = 9
aBc = etc.. = 3
abC = = 3
abc = .25 X .25 X .25 = .015625 X 64 = 1
--------- --
1.000000 64
I leave it to you to calculate the probabilities for quadruple and
quintuple heterozygotes, etc. By now you can probably see that the number
of phenotypically different classes = 2^n (n = number of genes).
The number of genotypically different classes = 3^n.
- Lynn Bry, MD/PhD Student Molecular Microbiology
Medial School, St. Louis, MO.