Why does the Hardy-Weinberg principle not work in some cases?

For example, if you are given that there are 142 ';AA'; individuals, 303 ';Aa'; individuals, and 127 ';aa'; individuals, you can count the alleles to eventually calculate what q^2 equals. Why then, does q^2 not equal 127/(142+303+127)?





Note: If you use very simplistic numbers for the number of individuals, such as 1:2:1, then q^2 does equal 1/(1+2+1).Why does the Hardy-Weinberg principle not work in some cases?
Two reasons: experimental error/sampling error due to the fact that your population is very small and/or the fact that q^2 predicts the frequency of recessive homozygotes in the *next* generation, not the current generation (unless the population is already in Hardy-Weinberg equilibrium).





You noted that you can get q^2 to equal what you expect if you choose numbers carefully. In your simple example, if the number of individuals in the population is 4 and there is 1 dominant homozygote, 2 heterozygotes, and 1 recessive homozygote. Therefore, for the allele frequencies, the recessive allele has a frequency of 4/8 = 1/2, thus the dominant allele also has a frequency of 1/2. Mathematically, this will work out easily to give a value of q^2 = 1/4, 2pq = 1/2 and p^2 = 1/4, since the values were carefully chosen.





However, in a real-world sample, the numbers might not be so easily chosen and so you have to go with the numbers and see that they are fairly close and be okay with that and just assume sampling error, OR, the numbers may be exact for the current generation and q^2 may give you a different number because it is based on the probability that two recessive alleles come together in an individual and thus, the frequency of recessive homozygous individuals may be different in the next generation after a round of random mating. The situation depends on whether or not you believe it was the small sample size that gave you the numbers or whether it was non-random mating that gave you numbers that don't fit the Hardy-Weinberg equation, and that the Hardy-Weinberg equation numbers will come out as expected in the next generation after a round of random mating.





So, to take a look at the numbers:





142 AA


303 Aa


127 aa





So, the frequency of the a allele, q, is:





q = ((127 x 2) + 303)/((127 x 2) + (303 x 2) + (142 x 2)) =


557/(254 + 606 + 284) =


557/1144 = 0.4869





The frequency of the A allele, p, is:





p = ((142 x 2) + 303)/((127 x 2) + (303 x 2) + (142 x 2)) =


587/(254 + 606 + 284) =


587/1144 = 0.5131





Note that p + q = 1





So, the question is, if


q^2 = 0.4869 x 0.4869 = 0.2371


and the frequency of the recessive homozygotes in the population is:


127/(142 + 303 + 127) = 0.222, then why aren't these numbers equal?





The reason in this case could easily be sampling error due to small sample size, but it is just as likely that the calculation of q^2 is giving you the probability of recessive homozygotes in the next generation rather than the current generation. You might also conclude that a slight amount of non-random mating or genetic drift, etc. is giving you slightly skewed numbers in the current population.