If the albino individuals are present at a frequency of about 1 in 10,000 (or 0.0001); then;
aa = 0.0001
a = 0.01
Using Hardy-Weinberg principle of p + q = 1
p + 0.01 = 1
p = 1 – 0.01 = 0.99
Using Hardy-Weinberg equation; p2 + 2pq + q2 = 1
Carriers of albino allele are the heterozygous population (2pq);
2*.099*0.01 = 0.0198
The frequency is 0.0198