Given the haplotype frequencies AB: 0.15, Ab: 0.35, aB: 0.25, ab: 0.25, calculate the coefficient of linkage disequilibrium (r2).

Let the two loci have alleles A/a and B/b.
From the table:

• Haplotype AB frequency ($P_{AB}$) = 0.15
• Haplotype Ab frequency ($P_{Ab}$) = 0.35
• Haplotype aB frequency ($P_{aB}$) = 0.25
• Haplotype ab frequency ($P_{ab}$) = 0.25

Calculate the allele frequencies:

• Frequency of allele A ($P_A$) = $P_{AB} + P_{Ab} = 0.15 + 0.35 = 0.50$
• Frequency of allele a ($P_a$) = $P_{aB} + P_{ab} = 0.25 + 0.25 = 0.50$
• Frequency of allele B ($P_B$) = $P_{AB} + P_{aB} = 0.15 + 0.25 = 0.40$
• Frequency of allele b ($P_b$) = $P_{Ab} + P_{ab} = 0.35 + 0.25 = 0.60$

The formula for D is $D = P_{AB} - P_A P_B$.

Substitute the values:

$$D = 0.15 - (0.50 \times 0.40)$$ $$D = 0.15 - 0.20$$ $$D = -0.05$$

The formula for $r^2$ is $r^2 = \frac{D^2}{P_A P_a P_B P_b}$.

Substitute the calculated values:

$$r^2 = \frac{(-0.05)^2}{(0.50)(0.50)(0.40)(0.60)}$$ $$r^2 = \frac{0.0025}{(0.25)(0.24)}$$ $$r^2 = \frac{0.0025}{0.06}$$ $$r^2 \approx 0.041666...$$

Rounding to four decimal places:

$$r^2 \approx 0.0417$$