Title: I need help with the Hardy-Weinberg principle please.? Post by: smile on Jan 31, 2012 4) In a population of 100 rock pink plants, 84 individuals have red flowers while 16 have white flowers. Assume that white petals are inherited as a recessive trait(a) and red petals as a dominant trail (A). What are the frequencies p and q? In the next generation, what will be the equilibrium genotypic frequencies?
I get that p^2 = 84/100 = .84 p = .9165 q^2 = 16/100 = .16 and q = .4 but that doesn't make sense. using P + Q = 1, that doesn't add up yet the example I am following does work out. What am I doing improperly? Title: I need help with the Hardy-Weinberg principle please.? Post by: firemonkey on Jan 31, 2012 You're forgetting the Heterozygotes 2pq
The full formula is p^2+2pq+q^2 = 1 We know that 2/3 of the Red flowers are heterozygotes p^2 = 28 AA Red 2pq = 56 Aa Red q^2 = 16 aa White So the allele frequencies p are: 2 A's from the homozygous = 28 + 1 A from the heterozygous = 56/2 = 28 = 56/100 = .56 So the allele frequencies q are: 1-.56 = .44 Title: I need help with the Hardy-Weinberg principle please.? Post by: rivertube on Feb 1, 2012 Always work with the recessive phenotype first. It seems you have forgotten to take the sqrt of the known genotype, which is 16/100. If 16/100 have white flowers, that's a frequency of 0.16. Then, the frequency of the recessive homozygote is 0.16 (using Hardy-Weinberg notation, q?2 = 0.16). The frequency of the recessive allele, q, is sqrt(0.16) = 0.4. According to H-W equation #1, p + q = 1, so the frequency of the dominant allele, p, is 1 - 0.4 = 0.6.
If equilibrium is maintained, p will still be 0.6 and q will still be 0.4. Just to finish it off: The frequency of the dominant homozygote, AA = p?2 = 0.36; the frequency of the heterozygotes, Aa = 2pq = 0.48, and the frequency of the recessive homozygotes, aa = q?2 = 0.16. Add up the frequencies and you get 1. Title: I need help with the Hardy-Weinberg principle please.? Post by: firemonkey on Feb 2, 2012 Asst Prof is correct. In a population, you MUST start with the recessive phenotype; it's the only genotype you can be sure of. In this case, the frequency of the recessive genotype, aa, is 0.16, so the frequency of a is the squareroot of 0.16, which is 0.4. that's called q. Since p + q =1, p must be 0.6.
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