Allele Frequency/Hardy Weinberg help please?

If fr(DD) = 0.49, then fr(D) = 0.7. Therefore, fr(d) = 0.3. Fr(Dd) = 2(0.7)(0.3) = 0.42.

Yes. In this case, fr(DD) = 0.49, fr(Dd) = 0.42, and fr(dd) = 0.09.

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