The ability to taste PTC is due to a single dominant allele "T". You sampled 215 evolution students, and determined that 150 could detect the bitter taste of PTC and 65 could not. Assuming this trait is in Hardy-Weinberg equilibrium in this population and that there are only two alleles, calculate the number of heterozygotes for this trait there are in the population.

  • $\begingroup$ How do I formulate this into hardy-weinberg eqbm form need help. $\endgroup$ – krushna Feb 23 '14 at 19:44
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    $\begingroup$ This looks like homework. What have you tried so far? $\endgroup$ – kmm Feb 23 '14 at 20:39
  • $\begingroup$ Total 215 Calculations Can taste PTC 150 Cam't taste PTC 65 For recessive 0.3023255814 Sqrt of recessive q 0.5498414148 p=1-q 0.4501585852 2pq 0.4950316667 $\endgroup$ – krushna Feb 23 '14 at 21:21
  • $\begingroup$ Maybe just edit your question to add that information. $\endgroup$ – kmm Feb 23 '14 at 21:28
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    $\begingroup$ @krushna this site is not a homework answering service. I suggest you read this post about homework questions. meta.biology.stackexchange.com/questions/266/… $\endgroup$ – rg255 Feb 26 '14 at 10:23

Hardy-Weinberg law gives the frequencies of genotypes from allele frequencies: $p^2 + 2pq + q^2$

Because the allele "T" is dominant, the frequency you're directly measuring through the experiment is the addition of $p^2$ and $2pq$.

$$p^2 + 2pq = 150/215 ≈ 0.6976$$

solving for $p$ (and replacing $q$ by $1-p$) gives:

$$p = 1 - \sqrt{\frac{13}{42}} ≈ 0.45 $$

The frequency of heterozygotes is given by:

$$2p(1-p) ≈ 0.4950 $$

And the number of heterozygotes in a population of 215 individuals is given by:

$$215 \cdot 0.4950 = 106.43$$

That sounds very alike what you wrote in the comments!

  • $\begingroup$ yeah I calculated the heterozygotes frequency but I was clueless how to find the number...of heterozygotes...but thanks for your help.. $\endgroup$ – krushna Feb 24 '14 at 3:12

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