Evolution Question: The following are all equivalent definitions of the coeffici

Question

Evolution Question:

The following are all equivalent definitions of the coefficient of linkage disequilibrium

D = gAB - ps

D = pt - gAb

D = qs - gaB

D = gab - qt

where p is the frequency of the A allele, q is the frequency of the a allele, s is the frequency of the B allele, and t is the frequency of the b allele. For a given set of allele frequencies (p, q, s, t), what is the maximum attainable value of D (hint: it is not 1/4 and can be expressed as the minimum of teo values). If p=0.7 and s=0.3, what are the eqpected chromosome frequencies under linkage equilibrium? What are the chromosome frequencies when D is at half of its maximum attainable value?

Explanation / Answer

The frequency of any particular allele at one locus does not vary with the alleles present at the other locus,

The frequency of any chromosome haplotype can be calculated by multiplying the frequencies of the separate alleles, and:

D, the coefficient of linkage disequilibrium, is zero. D is calculated as: gABgab = gAbgaB, where gAB is the frequency of chromosomes carrying alleles A and B, and so on.

The frequency of B on chromosomes carrying allele A is: (0.2304*2 + 0.0576 + 0.1536 + 0.0576 + 0.1536) / (0.2304*2 + 0.0576*2 + 0.1536 + 0.0576*2 + 0.1536) = (0.8832/0.9984) = 0.88. The frequency of B on chromosomes carrying allele a is: (0.1536 + 0.1536) / (0.1536 + 0.1536) = (0.3072/0.3072) = 1.00.

Since 0.88 is not equal to 1.00, the population fails criterion #2 for linkage equilibrium.

B) D = (0.2304*2 + .0576 + 0.1536 + 0.0576 + 0.1536)/1.3056 * (0)/1.3056 – (0.0576 + 0.0576)/1.3056 * (0.1536 + 0.1536)/1.3056 = (0.676)(0) - (0.088)(0.235) = -0.021

This is not zero, so the population fails criterion #3 for linkage equilibrium.

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