MadSci Network: Genetics |

Posted By:

Date:

Message ID:

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

As a check make certain all probabilities add to one, otherwise something is missing --Abphenotype = .75 X .25 = .1875 remember,aBphenotype = .25 X .75 = .1875 **phenotype**,notgenotype!abphenotype = .25 X .25 = .0625

.5625 AB .1875 Ab .1875 aB .0625 ab --------- 1.0000These 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 --------- 1.0000It'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 64I leave it to you to calculate the probabilities for quadruple and quintuple heterozygotes, etc. By now you can probably see that the number of

- Lynn Bry, MD/PhD Student Molecular Microbiology

Medial School, St. Louis, MO.

MadSci Network

webadmin@www.madsci.org