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

Question

The following are all equivalent definitions of the coefficient of linkage disequilibrium given in class D = gAB ps = pt gAb = qs gaB = 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 two values). If p = 0.7 and s = 0.3, what are the expected chromosome frequencies under linkage equilibrium? What are the chromosome frequencies when D is at half of its maximum attainable.

The last part is what i need help with

Explanation / Answer

Here,
Alleles -> A a B b
Allele frequencies -> p q s t

Q1)To find the theoretical maximum of D, which we will call Dmax, we do the following thing:

Multiply the frequencies of heterozygous alleles (i.e. p*t or q*s). The product which is less than 0.25 is the Dmax.
Therefore, if p*t < 0.25 then the value (p*t) is Dmax
if q*s < 0.25 then the value (q*t) is Dmax
(Note: if both pt and qs are < 0.25 then take the smaller number from the two as Dmax)

Q2)
p = 0.7 -> q = 1 - p = 0.3
s = 0.3 -> t = 1 - s = 0.7

in linkage equilibrium D = 0
Therefore, (Notation: f(AB) = frequency of AB)
f(AB) = f(A)*f(B) = ps = 0.7*0.3 = 0.21
f(Ab) = f(A)*f(b) = pt = 0.7*0.7 = 0.49
f(aB) = f(a)*f(B) = qs = 0.3*0.3 = 0.09
f(ab) = f(a)*f(b) = qt = 0.3*0.7 = 0.21
These are the chromosomal frequencies at linkage equilibrium.

Q3)
We have calculated the values of p, q, s and t in the previous question
p = 0.7, q = 0.3, s = 0.3, t = 0.7
Now to calculate Dmax, we do the steps that I have mentioned earlier,
p*t = 0.7*0.7 = 0.49 > 0.25, Hence 0.49 cannot be Dmax
q*s = 0.3*0.3 = 0.09 <0.25, hence our Dmax = 0.09

According to the formula mentioned in the question
D= f(AB) - ps = pt - f(Ab) = qs - f(aB) = f(ab) - qt
Substitute the values of p,q,s and t. Put D = Dmax/2 = 0.09/2 = 0.045
Therefore, (rearranging the above formula we get)
f(AB) = 0.045 + ps = 0.045 + 0.21 = 0.255
f(Ab) = pt - 0.045 = 0.49 - 0.045 = 0.445
f(aB) = qs - 0.045 = 0.09 - 0.045 = 0.045
f(ab) = 0.045 + qt = 0.045 + 0.21 = 0.255
These are the required frequencies at D = Dmax/2 = 0.045

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