FI = S(1/2)p+m+1(1+FAi)
The mean coefficient of consanguinity is equal to the mean of the various individual coefficients weighted by the frequencies of the various types of crosses between related individuals.
To evaluate this, an inventory is compiled for the individuals of the different types of crossings between related individuals, and they are classified on the basis of the value of Fx.
EQUATION a= Fifi where fi is the frequency of the subjects with consanguinity Fi.
This means that each genotype is fertilized exclusively by itself (a situation that is possible in maize (corn), but not in Drosophila, or in Man).
In a population of plants, underHW in Go, which is then put in a situation of self-fertilization:
What is the frequency Hn of the heterozygotes in generation n?
Hn = 1/2 Hn-1 --> Hn = (1/2)nHo; tends towards zero.
Dn = Dn-1 + 1/4 Hn-1
Rn = Rn-1 + 1/4 Hn-1
--> at the equilibrium of self-fertilization: Deq = Do + 1/2 Ho; Heq = 0; Req = Ro + 1/2 Ho
Being under HW in Go, Do = p2; Ho = 2pq; Ro = q2
--> Deq = p2 + 1/2 2pq = p2 + pq = p (p +q) = p; similarly for Req -->
Genotype frequencies at equilibrium:
Thus, any population (and this includes a consanguin population) will behave as though:
The genotype frequencies at equilibrium will be:
F(AA)eq = p2(1 - F) + pF = p2 - p2F + pF = p2 + Fp (1 - p) = p2 + Fpq; similarly for F(aa); thus:
F(AA)eq = p2 + Fpq
F(Aa)eq = 2pq(1 - F)
F(aa)eq = q2 + Fpq
The risk that a consanguin subject will be homozygotic for the allele a is: F(aa) = q2 + Fpq
It is usual within the human population for there to be several common ancestors (e.g. below: AM and AF male and female ancestors, parents of GP 1 and 2). In practice, the equation is simplified to:
CcI = S(1/2)p+m+1
For a deleterious mutant recessive autosomal allele (rare by definition) with a frequency q, the risk that a consanguin child will be homozygotic for this allele is: q x Cc whereas it is q2 for the children of non-consanguin parents.
NOTE: the exact equation q2+ pqCc is replaced by the approximation: q x Cc. This is applicable to human genetics (genetic counselling) if/because q is very small.
In other words, the increase is: (q2 + Fpq)/q2 = 1 + Fp/q
In the human species, the percentage of heterozygotic loci, calculated from enzymatic polymorphism, has a value H = 0.067. We can take it that there are 30000 structural genes, and in consequence 2010 genes in the heterozygotic state in the human genome (30000 x 0.067 = 2010).
If an individual results from an uncle-niece cross:
Consequences: If regular consanguin crosses are made (for example brother/sister crosses in mice), at each generation:
--> Fi tends towards a value of 1,
--> the individuals will become totally homozygotic.
Within each familly, all the individuals will be identical in the genetic sense of the word.
exactly the same genome
exactly the same genes.
This leads to the concept of the isogenetic line.
The genotype frequencies at equilibrium for every AiAi homozygote and every AiAj: heterozygote will be:
In a population of a diploid species with separate sexes and separate generations, we are dealing with a triallele autosomal locus (three possible allele states: A1, A2 and A3).
A sample of 400 individuals is examined. The numbers of the various genotypes are as follows:
Kalmes R, Huret JL
Atlas of Genetics and Cytogenetics in Oncology and Haematology 2002-06-01
Online version: http://atlasgeneticsoncology.org/teaching/30039/consanguinity