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Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
p (dominant). Therefore all of the p and q present for this gene in a population must account for 100% of this gene's alleles. And 100% = 1.00.
So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
So multiplying out (p+q)(p+q) = 1, we get: p^2+2pq+q^2 = 1 (all genotypes), where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
2pq = 2(0.21) = 0.42, or 42% of the population.
Hope that helps you to understand how to solve population genetics problems!
The correct option is A.
When an end of a tuning fork that is vibrating is placed against a table, it produces a sounder that is dramatically louder than the original sound produced by the vibrating tuning fork. At this stage, the tuning fork is forcing the table to vibrate at its own frequency. Thus, the statement in option A is not true.
The best and most correct answer among the choices provided by your question is The best and most correct answer among the choices provided by your question is the last choice.
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Kingdom Protista is not valid under evolutionary classification because they evolved before any other kingdom, and as such share a most recent common ancestor with each other, as opposed to any other group.</span>
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