Answer:
49.14%
Explanation:
The frequency of the heterozygous individuals in the population is therefore: 64/400 = 0.32
Using the Hardy Weingburg equation of a population is equilibrium = p2 + 2pq + q2 = 1
q2 = 0.32; Therefore q = 0.5657 –> alellic frequencies
If p + q = 1; Then p 1 – q = 1 – 0.5657 = 0.4343
Heterozygous individuals are represented by 2pq = 2 * 0.4343 * 0.5657 = 0.4914
0.4914 * 100 = 49.14% of the population which translates to 0.4914 * 400 = 196.56 (197) individuals
Absolutely true :)
this is to ensure that the cell performs the task effectively
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