<span><u>The answer is A. 72.25 percent.</u>
The Hardy-Weinberg principle can be used:</span>
<em>p² + 2pq + q² = 1</em> and <em>p + q = 1</em>
where <em>p</em> and <em>q</em> are the frequencies of
the alleles, and <em>p²</em>, <em>q²</em> and <em>2pq </em>are the
frequencies of the genotypes.
<span>The <em>r</em> allele (<em>q</em>) is found in 15% of the population:
q = 15% = 15/100
Thus, q = </span><span>0.15
To calculate the <em>R</em> allele frequency (<em>p</em>), the formula p + q = 1 is
used:
If p + q = 1, then p = 1 - q
p = 1 - 0.15
Thus, </span><span>p = 0.85
Knowing the frequency of the <em>R</em> allele (<em>p</em>), it is easy to determine the
frequency of the RR genotype (p²):
p² = 0.85² = 0.7225
Expressed in percentage, p² = 72.25%.</span>
Answer:
yes that is the male part of a plant.
Explanation: You are correct
Hope this helps
<span>Hepatic encephalopathy is a condition seen patients
with liver dysfunction and this condition describe a spectrum of reversible
neuropsychiatric abnormalities which implies that altered brain function which
is due to metabolic abnormalities. However, the administration of lactulose
causes reduction in the urea production rate which is consistent with the
reduced entry of ammonia into portal blood.</span>