Three genes A B C are located in sequence along a chromosome

Three genes A, B, C are located in sequence along a chromosome and have two allelic forms each (A,a), (B,b) and (C,c). The following frequences are observed for these genes in a large population:

ABC=1/4, AbC = 1/4, aBC = 0, abC = 0, ABc = 0, Abc = 0, aBc = 1/4, and abc = 1/4

a. Compute the pairwise gene frequences P_AB, P_Ab, P_aB, P_ab etc for the three types of two gene combinations.

b. Compute D\'_AB, D\'_AC, and D\'_BC. Which pair of genes are in high linkage disequilibrium, and which pairs are not linked?

Solution

a) Expression of most character depend on more than 1 locus.

Pair wise gene frequency can be given as-:

First we take combination of two alleles in form of Aa and Bb

So pair wise frequency for two pair Aa Bb can be express as

PAB= P_A x P_B

PAb= P_A x P_b

PaB=P_a x P_B

Pab= P_a x P_b

Gene frequencies =product of frequency of constituent alleles.

Also if,

P_AB =1/2

P_Ab = 0

P_aB = 0

P_ab = 1/2

Then P_A = P_a = P_B =P_b = 1/2

P_A.B =1/2 is not equal to P_A x P_B = 1/4

P_Ab = 0 not equal to P_A x P_b

From the above results we can say that population will approach linkage equilibrium.

We can calculate for other combination Aa Cc and Bb Cc by following above method.

b) If two loci are in linkage equilibrium then D=0

DAB = P_AB P_ab - P_Ab x P_aB

D\'AB =P\'_AB P\'_ab - P\'_Ab x P\'aB

So if D=0

P_AB = P_A x P_{B ,} P_Ab = P_A x P_B

D= 1/2 x1/2 - 0 = 1/4

D is not equal to zero so these pair are in linkage disequlibrium.

D\'= (1- r) D

Where r is the rate of combination between locus A and B

If loci are very close on a chromosome r= 0 ( tightly linked)

If loci are far apart on different chromosome r=1/2 ( loosely linked).

The pairs are tightly linked as D is not equal to zero.

As if r is greater than 0 D tends to zero.

Similarly we can calculate for D\'AC and D\'BC.