Ask your own question, for FREE!
Biology
OpenStudy (anonymous):

Explain why p^2-2pq+q^2=1

7 years ago
OpenStudy (anonymous):

This is irrelevant.

7 years ago
OpenStudy (anonymous):

Because this P and Q are unknowns, and therefore must be simplified / evaluated through simple Algebra laws. Also , this is meant to be in the Maths section . not the Biology section ==

7 years ago
OpenStudy (anonymous):

\[P ^{2}-2PQ+Q ^{2}\] is an emphasis on (\[α-β)^{2}\]

7 years ago
OpenStudy (luke):

well i can see nobody studied their hardy-weinberg equilibrium problems. first off cyter, P and Q are percentages of two alleles in a populatioin in hardy-weinberg equilibrium. in such a populations, the gene ratios are remaining the same, and these two alleles are the only two for that gene. the other two hardy-weinberg equations are P=1-Q and Q=1-P. they are used when the information about the frequency of a genotype or genotypes in a hardy-weinberg population is given and other genotypes are to be solved for. for instance population A has organisms with a gene x, gene X has two alleles in hardy-weinberg equilibrium in a population of 100. allele P is dominant to allele Q(dominant always given as P). sixty four organisms have the trait given by P. what are the percentages of the P and Q alleles in the population? someone who doesn't know Hardy weinberg says 64 percent have gene P and 36% have gene Q. however that is not the case. 64 present with P, so 36 present with Q, which means q^2=36% or .36, thus q=.6 using P=1-Q we find that P is in fact .4, not .64 this accounts for the heterozygous organisms for gene x while the common sense answer does not.

7 years ago
Can't find your answer? Make a FREE account and ask your own question, OR you can help others and earn volunteer hours!