<span>The answer is B. 72.25 percent.
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>p</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>P</em> allele frequency (<em>p</em>), the formula <em>p + q = 1</em> can be 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>P</em> allele (<em>p</em>), it is easy to determine the frequency of the <em>PP </em>genotype (<em>p²</em>):
p² = 0.85² = 0.7225
Expressed in percentage, p² = 72.25%.</span>
The answer to your question is homeostasis
Explanation:
Noble gases have a full outer energy level and are inactive. These atoms are considered inert.
Hope this helps
Good Luck
Typhitis, also called neutropenic enterocolitis, is an infection that often develops in cancer patients who undergo chemotherapy. In many cases, surgical intervention is required.
Without surgical intervention, the patient would be transferred to an ICU (Intensive Care Unit) for monitoring, and the nurse would perform some or all of these emergency actions:
1. Bowel rest and nasogastric suction,
2. Serial abdominal examinations,
3. Providing intravenous fluids, blood, and platelet transfusions when needed,
4. Using antibiotics to fight the infection, and obtaining cultures to determine if the antibiotic is working,
5. Not administering medication that could worsen the situation.
Answer:
The Answer would be AATGCG.
Explanation: The reason is because A pairs with T and C pairs with G.
