Investigating the linked segregation of genes situated at different loci is a way of testing the independence of their transmission. This concept of independence is also reflected in the recombination fraction, q, which is the percentage of the gametes transmitted by the parents to be recombined. If they are transmitted independently, there will be the same number of recombined gametes as there are parental gametes, and so q= 1/2. If they are not transmitted independently, then the parenteral gametes are transmitted preferentially to the recombined gametes, and 0 q< 1/2. In this case, there is said to be "linkage" between the two loci.
Let us consider the caseof two loci, A and B, with two codominant alleles at each of these loci, A1, A2 and B1, B2 respectively. Such an individual can produce four types of gamete:
A1B 1A2B 1A1B2 A2B 2
Two situations are possible:
Assuming that the crossing-over event for a pair of chromosomes follows Poisson’s law, and knowing that a parental gamete has zero or an even number of crossings-over, whereas a recombined gamete has an odd number, we can show that the frequency of recombined gametes is always equal to or lower than that of the parenteral gametes and so
0 q < 1/2
If q = 1/2, then all the gamete types have the same probability and the alleles at the loci A and B loci are transmitted independently. Loci A and B are therefore said not to enhibit genetic linkage. This is the situation if A and B are on different pairs of chromosomes, and also if A and B are one the same pair, but at some distance from each other. However, if q < 1/2, then the two loci are genetically linked.
For a couple of which the genotypes at the A and B are known, the probability of observing the genotypes of the offspring depends on the value of q. Let us assume the following crossing:
Therefore, such a couple can have 4 types of offspring
Assuming that there is gamete equilibrium at the A and B loci, in parent 1 there is a probability of 1/2 that alleles A1 and B1 will be coupled, and a probability of 1/2 that they will be in repulsion.
So in the end, with no additional information about the A1 and B1 phase , and assuming that the alleles at the A and B loci are in a state of coupling equilibrium, the probability of finding n1, n2, n3 and n4 children in categories (1), (2), (3), (4) is: p(n1,n2,n3,n4/q)=1/2{[(1 -q/2]n1+n2 x (q/2)n3+n4 + (q/2) n1+n2 x [(1-q/2] n3+n4}
So the liklihood of q for an observation n1, n2, n3, n4 can be written : L(q/n1,n2,n3,n4)=1/2 {[(1-q)/2]n1+n2 (q/2)n3+n4 + (q/2) n1+n2 [(1-q)/2] n3+n4}
Special case: number of children n= 1 Regardless of the category to which this child belongs L(q) = 1/2 [(1-q)/2] + 1/2 [q/2] = 1/4 The likelihood of this observation for the family does not depend on q. We can say that such a family is not informative for q.
Informative families An "informative family" is a family for which the liklihood is a variable function of q. One essential condition for a family to be informative is, therefore, that it has more than one child. Furthermore, at least one of the parents must be heterozygotic. Definition: if one of the parents is doubly heterozygotic and the other is
Take a family of which we know the genotypes at the A and B loci of each of the members. Let L(q) be the liklihood of a recombination fraction 0 q < 1/2 L(1/2) be the liklihood of q= 1/2, that is of independent segregation into A and B. The lod score of the family in q is: Zq = log10 [L(q)/L(1/2)]
Z can be taken to be a function of q defined over the range [0,1/2]. Lod score of a sample of families The liklihood of a value of q for a sample of independent families is the product of the liklihoods of each family, and so the lod score of the whole sample will be the sum of the lod scores of each family.
Several methods have been proposed to detect linkage: the U scores, the sib pair test, the likelihood ratios, the lod score method. The lod score method is the one most commonly used at present. The test procedure in the lod score method is sequential. Information, i.e. the number of families in the sample, is accumulated until it is possible to decide between the hypotheses H0 and H1 :
The lod score of the q1 sample
The decision thresholds of the test are usually set at -2 and +3, so that if:
In fact, what is being tested is not a single value of q1 relative to q = 1/2, but a whole set of values between 0 and 1/2, with a step of various size (0.01 or 0.05). If there is a value of q1 such that Z(q1) 3: linkage is concluded to exist.
If there is a value of q1 such that Z(q1) = -2 The linkage is excluded for any q q1
If q-2 < Z(q) < 3, no conclusion can be drawn, the sample is not sufficiently informative.
The proposed test has the advantage of being very simple, and of providing protection against falsely concluding linkage. However, some criticisms can be levelled, not only against the criteria chosen, but also against the entire principle of using a sequential procedure. The number of families typed is, indeed, rarely chosen in the light of the test results.
If the test, on a sample of the family, has demonstrated linkage between the A and B loci, then one may want to estimate the recombination fraction for these loci. The estimated value of q is the value which maximizes the function of the lod score Z, and this is equivalent to taking the value of q for which the probability of observing linkage in the sample is greatest.
Let us assume we are dealing with a disease carried by a single gene, determined by an allele, g0, located at a locus G (g0 : harmful allele, G0 : normal allele). We would like to be able to situate locus G relative to a marker locus T, which is known to occupy a given locus on the genome. To do this, we can use families with one or several individuals affected and in which the genotype of each member of the family is known with regard to the marker T. In order to be able to use the lod scores method described above, what is needed
It will often happen that the information available for the marker is not also genotypic, but phenotypic in nature. Once again, all possible genotypes must be envisaged. As a general rule, the information available about a family concerns the phenotype. To calculate the likelihood of q, we must envisage all the possible genotype configurations at each of the loci, for this family, writing the likelihood of q for each configuration, weighting it by the probability of this configuration, and knowing the phenotypes of individuals in A and B. Knowledge of the genetic parameters at each of the loci (gene frequency, penetration values) is therefore necessary before we can estimate q. It is obvious that calculating the lod scores, despite being simple in theory, is in fact a lengthy and tedious business; specific software have been designed for linkage analysis. Analysis of gene linkage has made it possible to construct a gene map by locating the new polymorphisms relative to one other on the genome. The measurement used on the gene map is not the recombination fraction, which is not an additive datum, but the gene distance, which we will define below.
Clerget-Darpoux F
Atlas of Genetics and Cytogenetics in Oncology and Haematology 2002-05-01
Genetic Linkage Analysis
Online version: http://atlasgeneticsoncology.org/teaching/30031/genetic-linkage-analysis