Multiple allele problem:

3977 surveyed Swiss people were found with the 
following allele frequencies at the ABO blood groups:

IA = 0.27
IB = 0.06
i = 0.67

calculate the genotype frequencies for the four 
blood types

freq(IA IA): (0.27)(0.27) = 0.0729
freq(IA i): 2(0.27)(0.67) = 0.3618

freq(IB IB): (0.06)(0.06) = 0.0036
freq(IB i): 2(0.06)(0.67) = 0.0804

freq(IA IB): 2(0.27)(0.06) = 0.0324

freq(ii): (0.67)(0.67) = 0.4489


calculate the phenotype frequencies for the four 
blood types

phenotype frequency of type A = 0.0729 + 0.3618 = 0.4347

phenotype frequency of type B = 0.0036 + 0.0804 = 0.0840

phenotype frequency of type AB = 0.0324

phenotype frequency of type O =  0.4489

calculate the numbers of individuals expected to have 
each genotype and phenotype if this population were in 
Hardy-Weinberg equilibrium.

multiply the frequency by the number of people in the population:
e.g., to calculate the numbers of individuals expected to have
the genotype IA IA, multiply 0.0729 x 3977 = 290

to calculate the numbers of individuals expected to have
the phenotype of type A blood, multiply 0.4347 x 3977 = 1729.