Five thousand Thai individuals were tested for their MN blood group, which exhibits incomplete dominance. The phenotype distribution revealed 600 with type M, 4000 with type MN, and 400 with type N. What is the frequency of the N allele?

To determine the frequency of the N allele, let's use the Hardy-Weinberg equation for allele frequencies in a population exhibiting incomplete dominance.

Given:

• 600 individuals with type M
• 4000 individuals with type MN
• 400 individuals with type N

The MN blood group exhibits incomplete dominance, where individuals with the MN phenotype have an intermediate phenotype between types M and N.

Let's denote:

• M as the dominant allele (M)
• N as the recessive allele (N)

We know that individuals with the MN phenotype have both the M and N alleles.

Given:

• 600 individuals with type M (genotype: MM)
• 4000 individuals with type MN (genotype: MN)
• 400 individuals with type N (genotype: NN)

The individuals with the MN phenotype (4000) have both the M and N alleles. So, in terms of allele count:

• The number of M alleles contributed by MN individuals is 4000 (since they have both M and N alleles).
• The number of N alleles contributed by MN individuals is also 4000 (again, they have both M and N alleles).

The total count of N alleles in the population:

• From individuals with type N: 400 (each having two N alleles, as they have the NN genotype)
• From individuals with type MN: 4000 (each carrying one N allele)

Total count of N alleles = (400 * 2) + 4000 = 800 + 4000 = 4800 N alleles

Total count of alleles in the population = (number of individuals * 2 alleles per individual)

Total count of alleles = (600 * 2) + (4000 * 2) + (400 * 2) = 1200 + 8000 + 800 = 10000 alleles

Now, to find the frequency of the N allele:

Frequency of N allele = (Total count of N alleles) / (Total count of alleles)

Frequency of N allele = 4800 / 10000 = 0.48

Therefore, the frequency of the N allele in the population is 0.48 or 48%.