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Lera25 [3.4K]
3 years ago
13

A nurse is transfusing multiple units of packed red blood cells. After the second unit is transfused, the nurse auscultates bila

teral crackles at the bases of the client's lungs and the client reports dyspnea. The nurse telephones the health care provider and provides an SBAR report. Which statement represents the final step in this type of communication?
Biology
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

a. "I think the client would benefit from intravenous furosemide (Lasix)."

Explanation:

Communication between doctor and nurses regarding patients should follow the order: situation, background, evaluation and recommendations. Based on this, since the nurse has already reported the first three topics, the final communication step should be a recommendation on how to solve the patient's problem.

In the above case, the patient was diagnosed with dyspnea which is difficulty breathing, a condition usually related to heart and respiratory disease. A recommendation for this case is the use of intravenous furosemide to solve the patient's problem. In this case, the nurse should close the communication with the doctor with the phrase: "I think the client would benefit from intravenous furosemide (Lasix)."

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If the frequency of the recessive allele for a gene is 0.3, calculate the expected frequency of heterozygotes in the next genera
Kisachek [45]
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!
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brilliants [131]

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4 0
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Area B of the graph shows the activation energy required if an enzyme was not present

Explanation:

Reactions with high activation energy cannot occur spontaneously. Enzymes are responsible for lowering this activation energy and enabling reactions to occur at a faster pace than natural. An example is carbonic anhydrase enzyme that enables increased rates of carbon dioxide dissolving in and out of blood plasma.

Enzymes distort the bond of reactants such that they become unstable ( this raises the reactants Gibbs free energy). The bonds therefore break and rearrange to form the products of lower and stable energy states.

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