Short Cuts for Calculation

(1) The estimation of chances in F₂

From the above calculation, it was found that the chances of occurrence of the genotypes were similiar to the coefficients of terms in the expansion of formulae according to the binomial theory. For example,

In a 2 genes inheritance, the chances would be as follow.

Item	Number
The maximum number of genes in each individual	4
Number of individuals in a set of F₂	16
Number of individuals getting 4 dominant alleles is as AABB	4 !
	----------------	= 1
	4 ! X 0 !
Number of individuals getting 3 dominant alleles and 1 recessive allele is as AABb and AaBB	4 !
	----------------	= 4
	3 ! X 1 !
Number of individuals getting 2 dominant alleles and 2 recessive alleles is as AAbb, aaBB, AaBb	4 !
	----------------	= 6
	2 ! X 2 !
Number of individuals getting 1 dominant allele and 3 recessive alleles is as Aabb and aaBb	4 !
	----------------	= 4
	1 ! X 3 !
Number of individuals getting 0 dominant allele and 4 recessive alleles is as aabb	4 !
	----------------	= 1
	0 ! X 4 !

If X stands for dominant allele and Y, recessive, then, X⁴ stands for the individuals possessing 4 dominant alleles; X³Y, 3 dominant alleles and 1 recessive allele; X²Y², 2 dominant alleles and 2 recessive alleles; XY³, 1 dominant allele and 3 recessive alleles; and Y⁴, 0 dominant allele and 4 recessive alleles,

To expand (X + Y)⁴, and get X⁴ + 4 X³Y + 6 X²Y² + 4 XY³ + Y⁴.

The coefficients of the items, 1,4,6,4,1, just fit the result.

(2) Analysis of random combination

In multi-genes inheritance, if you want to make deductions, you can estimation the chance of each gene, then, put them together by multiplication and get the result. As

(Example 1) In the cross between AABbcc and aaBbCc, what is the genotypic frequency and phenotypic ration in the offsprings ?

For the first gene : AA crossed with aa, the genotype of offsprings must be Aa. (The chance was 1).

For the second gene : Bb crossed with Bb, the genotypic frequency in offsprings must be BB, 1/4; Bb, 1/2; bb, 1/4ĄC

For the third gene : cc crossed with Cc, the genotypic frequency in offsprings must be Cc, 1/2 and cc, 1/2.

So, if AABbcc was crossed with aaBbCc, their offsprings should have 1 X 3 X 2 = 6 types of genotypes.

The genotypic frequency is :	AaBBcc = 1 X 1/4 X 1/2 = 1/8
	AaBBCc = 1 X 1/4 X 1/2 = 1/8
	AaBbcc = 1 X 1/2 X 1/2 = 1/4
	AaBbCc = 1 X 1/2 X 1/2 = 1/4
	Aabbcc = 1 X 1/4 X 1/2 = 1/8
	AabbCc = 1 X 1/4 X 1/2 = 1/8

(Example 2) In the above question, if AABbcc was crossed with aaBbCc, what phenotypes would be produced and what would be the phenotypic ratio ?

For the first gene : if AA was crossed with aa, the phenotype of their offsprings must be A. (The chance is 1.)

For the second gene : if Bb was crossed with Bb, the phenotypic ratio in their offsprings must be : B, 3/4; b, 1/4 ĄC

For the third gene : if cc was crossed with Cc, the phenotypic ratio in their offsprings must be C, 1/2 and c, 1/2.

Then, in the cross between AABbcc and aaBbCc, their descendents should have 1 X 2 X 2 = 4 phenotypes.

Their ratio is :	A-B-C- = 1 X 3/4 X 1/2 = 3/8
	A-B-cc = 1 X 3/4 X 1/2 = 3/8
	A-bbC- = 1 X 1/4 X 1/2 = 1/8
	A-bbcc = 1 X 1/4 X 1/2 = 1/8

(16.09.2006)