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Use the following Intravenous Heparin protocol to answer the following questions:

\begin{tabular}{|c|c|c|c|}
\hline
aPTT (sec) & \begin{tabular}{c}
Bolus \\
Dose \\
(units/kg)
\end{tabular} & \begin{tabular}{c}
Stop \\
Infusion \\
(min)
\end{tabular} & Rate Change (units/kg/h) \\
\hline
Initial dose & 80 & & 18 (initial rate) \\
\hline
[tex]$\ \textless \ $[/tex]34 & 80 & & Increase by 4 \\
\hline
[tex]$35-59$[/tex] & 40 & & Increase by 2 \\
\hline
[tex]$60-90$[/tex] & No & & No change \\
\hline
[tex]$91-110$[/tex] & No & & Decrease by 1 \\
\hline
[tex]$111-212$[/tex] & No & & Decrease by 2 \\
\hline
[tex]$\ \textgreater \ $[/tex]212 & No & 60 & Decrease by 3 \\
\hline
\end{tabular}

The IV is infusing at [tex]$28 \text{ mL/h}$[/tex]. Calculate the rate change if the patient's aPTT was 110. What is the new rate?

A. [tex]$25 \text{ mL/h}$[/tex]

B. [tex]$29 \text{ mL/h}$[/tex]

C. [tex]$27 \text{ mL/h}$[/tex]

D. [tex]$30 \text{ mL/h}$[/tex]


Sagot :

To determine the rate change and calculate the new rate based on the given protocol and the patient's aPTT value, follow these steps:

1. Identify the current infusion rate:
- The initial infusion rate is given as [tex]\(28 \, \text{mL/h}\)[/tex].

2. Determine the patient’s aPTT value:
- The patient's aPTT value given in the problem is [tex]\(110\)[/tex].

3. Refer to the aPTT protocol table:
- According to the protocol:
- If the aPTT is [tex]\(91-110\)[/tex] seconds, the rate should decrease by [tex]\(1 \, \text{unit/kgh}\)[/tex].

4. Calculate the rate change:
- The rate change corresponding to an aPTT of [tex]\(110\)[/tex] is a decrease by [tex]\(1 \, \text{unit/kgh}\)[/tex].

5. Apply the rate change to the initial infusion rate:
- Initial rate: [tex]\(28 \, \text{mL/h}\)[/tex]
- Decrease by: [tex]\(1 \, \text{mL/h}\)[/tex]

6. Calculate the new rate:
- New rate = Initial rate - Rate change
- New rate = [tex]\(28 \, \text{mL/h} - 1 \, \text{mL/h} = 27 \, \text{mL/h}\)[/tex]

Thus, the new infusion rate will be [tex]\(27 \, \text{mL/h}\)[/tex].

Answer: [tex]\(\mathbf{27 \, \text{mL/h}}\)[/tex]
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