Acute Kidney Injury and Kidney Replacement Therapy in the Critically Ill Patient

Answer: C

Both RIFLE class “F” and AKIN stage 3 are met

using similar SCr and UOP criteria. For this case, the

SCr increased by at least 3-fold above baseline, and

the patient’s UOP was less than 0.3 mL/kg/hour for 24

hours. Although A and B are true, the worse values in

the RIFLE class and AKIN staging systems should be

chosen when determining the class and stage of AKI.

Answer D is incorrect because his renal indices have not

been present for at least 3 months. Collectively, both the

RIFLE and the AKIN criteria have been included in the

KDIGO criteria, which is the current standard definition.

This patient has severe sepsis that led to decreased renal

perfusion, causing AKI. Enalapril likely contributed to

the injury by altering renal hemodynamics (Answer A is

correct). Neither glipizide nor acetaminophen is likely to

cause AKI (Answers B–D are incorrect).

Answer: A

Acute kidney injury cannot currently be diagnosed

with a specific blood test or imaging study (Answer C

is incorrect). The degree of kidney dysfunction and the

patient’s symptoms and coexisting diseases may provide

clues regarding the etiology of the patient’s kidney injury

(Answer A is correct). Drug-induced AKI is most com-

mon in the setting of additional insults and may occur

even after the drug is discontinued. Given a patient’s

history and presentation as well as a patient’s physical

examination, a clinician may pursue additional test-

ing to find the DIKD phenotype. Serum markers (e.g.,

SCr, UOP) are often nonspecific (Answer C is incor-

rect); kidney biopsy is only warranted for certain DIKD

phenotypes (e.g., AIN, glomerular diseases) and is often

not pursued in the acute setting (Answer D is incorrect).

Renal ultrasonography would be helpful for nephroli-

thiasis/crystalluria phenotypes or evidence of postrenal

AKI not alleviated with urinary catheter placement

(Answer B is incorrect).

Answer: C

Intermittent hemodialysis is often used in critically ill

patients because many physicians are familiar with this

therapy; however, about 20%–30% of patients receiving

IHD become hypotensive and require discontinuation or

a switch to an alternative therapy. Although prolonged

intermittent KRT methods like SLED or EDD are an

option, some form of continuous KRT would likely be

chosen because of this patient’s unfavorable hemody-

namics. Continuous renal replacement therapies such as

CVVH, CVVHD, and CVVHDF are often used because

they allow for slower flow rates and improved hemo-

dynamics. However, there is no clear benefit with one

therapy over another.

Answer: D

Protein binding primarily determines drug clearance

during KRT. Other factors that contribute to drug clear-

ance include low MW and low Vd. Drug charge leading

to dialyzer adhesion has only a limited effect on drug

clearance during continuous KRT and has been poorly

studied. When available, SC should be identified from

the package insert information or the primary literature

or calculated from the concentration of the solute in the

ultrafiltrate divided by the concentration of the solute in

the plasma. When unavailable, SC can be estimated as

SC = 1 − fb. In this case, the SC = 1 – 0.12 = ~0.88

(Answer D is correct; Answers A–C are incorrect).

Answer: B

The general approach to calculating clearance during

CVVH multiplies the SC (the degree to which a sub-

stance is able to pass through the membrane) by the

ultrafiltration rate. If predilution (i.e., before the fil-

ter) fluids are used, clearance across the membrane is

reduced, which must be accounted for in the formula.

Clearance in that case can be estimated using the follow-

ing equation: CVVHpre-dilution = QUF x SC x [Qb/(Qb

+ Qrf)] (where QUF is ultrafiltration flow rate (liters per

hour), Qb is blood flow rate [converted to liters per hour

for consistent units], and Qrf is predilution replacement

fluid flow rate, as stated earlier). In this case, clearance

= 2 L/hr x 0.88 x [12 L/hr/(12 L/hr + 2 L/hr)] = 1.5 L/hr

(Answer B is correct; Answers A, C, and D are incorrect).