In a population at Hardy-Weinberg equilibrium the dominant allele has frequency $p = 0.7$. What percentage of the population are heterozygotes?

42 (answers within ±1 count)

$q = 0.3$, so $2pq = 2 \times 0.7 \times 0.3 = 0.42$, or 42 %. Heterozygotes peak at 50 % when both alleles are equally common, and fall away as either allele becomes rare.