Arosa Hya
01-14-2015, 04:07 AM
Determining Linkage/Independence
Suppose you want to map the distance between genes "N" and "P." Each has two alleles (N and n, P and p). Both pairs of alleles have a complete dominance relationship; we will identify the phenotypes simply by the letters of the alleles.Your mating will be: NnPp x nnppIf your genes are independent, then your first parent's four possible gametes (NP, Np, nP, and np) will be produced in equal numbers, and thus you will get four offspring classes, also with equal numbers, like so:
NP offspring: 100
Np offspring: 100
nP offspring: 100
np offspring: 100
These equal numbers are the evidence that these two genes are independent. However, if you get numbers like:
NP offspring: 150
Np offspring: 50
nP offspring: 50
np offspring: 150
then you have established that the genes are not independent. Moreover, you know that in your original heterozygous parent, one chromosome had N and P on it and the other had n and p on it (as opposed to N and p on one and n and P on the other). In other words, your parental linkage is N linked with P and n linked with p.
In this example, the parental offspring are the NP and np offspring; the recombinanant offspring are the Np and nP offspring.
http://www.cod.edu/people/faculty/fancher/genetics/GenSpSm.jpgTwo-point Cross
Using the example above, the linkage map distance between the N and P gene loci is calculated by calculating the percentage of recombination between the two genes. This is done by adding the number of recombinant offspring, dividing by the total number of offspring, and multiplying by 100 (to convert to a percentage.
In this example, we have a total of 100 recombinant offspring (the Np and nP offspring added together), and an overall total of 400 offspring. So...
(100/400) x 100 = 25%25% of our offspring are recombinants, thus the linkage map distance between these two genes is 25 LMU.
Suppose you want to map the distance between genes "N" and "P." Each has two alleles (N and n, P and p). Both pairs of alleles have a complete dominance relationship; we will identify the phenotypes simply by the letters of the alleles.Your mating will be: NnPp x nnppIf your genes are independent, then your first parent's four possible gametes (NP, Np, nP, and np) will be produced in equal numbers, and thus you will get four offspring classes, also with equal numbers, like so:
NP offspring: 100
Np offspring: 100
nP offspring: 100
np offspring: 100
These equal numbers are the evidence that these two genes are independent. However, if you get numbers like:
NP offspring: 150
Np offspring: 50
nP offspring: 50
np offspring: 150
then you have established that the genes are not independent. Moreover, you know that in your original heterozygous parent, one chromosome had N and P on it and the other had n and p on it (as opposed to N and p on one and n and P on the other). In other words, your parental linkage is N linked with P and n linked with p.
In this example, the parental offspring are the NP and np offspring; the recombinanant offspring are the Np and nP offspring.
http://www.cod.edu/people/faculty/fancher/genetics/GenSpSm.jpgTwo-point Cross
Using the example above, the linkage map distance between the N and P gene loci is calculated by calculating the percentage of recombination between the two genes. This is done by adding the number of recombinant offspring, dividing by the total number of offspring, and multiplying by 100 (to convert to a percentage.
In this example, we have a total of 100 recombinant offspring (the Np and nP offspring added together), and an overall total of 400 offspring. So...
(100/400) x 100 = 25%25% of our offspring are recombinants, thus the linkage map distance between these two genes is 25 LMU.