PK Module 4: Practice Questions Part 2

PK Module 4: Practice Questions Part 2

University

10 Qs

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PK Module 4: Practice Questions Part 2

PK Module 4: Practice Questions Part 2

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Quiz

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University

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Marcy Hernick

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10 questions

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1.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 2.4 kg neonate presents with seizure activity, and the neurologist would like to start the baby on phenobarbital sodium (S = 0.9) IV. Calculate an IV bolus loading dose in mg to achieve a concentration of 25 mg/L. Round to the nearest mg.

V = 0.96 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

V = 0.96 L/kg x 2.4 kg = 2.304 L

LD = (C x V)/S

LD = (25 mg/L x 2.304 L)/0.9

LD = 64 mg

2.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 2.4 kg neonate presents with seizure activity, and the neurologist would like to start the baby on phenobarbital sodium (S = 0.9) . After receiving 64 mg loading dose to achieve a concentration of 25 mg/L, the patient's level comes back at 15 mg/L. Calculate an additional dose of phenobarbital to give to achieve a C = 25 mg/L Round to the nearest mg.

V = 0.96 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

V = 0.96 L/kg x 2.4 kg = 2.304 L

LD = (ΔC x V)/S

LD = ((25 mg/L-15 mg/L) x 2.304 L)/0.9

LD = 25.6 mg round to 26 mg

3.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 2.4 kg neonate presents with seizure activity, and the neurologist would like to start the baby on phenobarbital sodium (S = 0.9) IV. Calculate a MD in mg to be administered every 12 hours (given with an IV bolus loading dose) to achieve a Css of 25 mg/L. Round to the nearest tenth.

V = 0.96 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

Cl = 0.0047 L/hr/kg x 2.4 kg = 0.0113 L/hr

MD = (Cssavg x Cl x τ)/S

MD = (25 x 0.0113 x 12)/0.9

MD = 3.39/0.9 = 3.8 mg

4.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 4.5 kg infant presents with seizure activity, and needs to receive phenobarbital sodium (S = 0.9) IV every 12 hours with a desired Css of 20 mg/L. Calculate an IV bolus loading dose in mg to achieve a concentration of 20 mg/L. Round to the nearest mg.

V = 0.63 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

V = 0.63 L/kg x 4.5 kg = 2.835 L

LD = (C x V)/S

LD = (20 mg/L x 2.835 L)/0.9

LD = 63 mg

5.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 4.5 kg infant presents with seizure activity, and needs to receive IV phenobarbital sodium (S = 0.9) with a desired Css of 20 mg/L. After receiving a 63 mg loading dose, the patient's level comes back at 10 mg/L. Calculate an additional phenobarbital dose in mg to achieve a concentration of 20 mg/L. Round to the nearest mg.

V = 0.63 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

V = 0.63 L/kg x 4.5 kg = 2.835 L

LD = (ΔC x V)/S

LD = ((20 mg/L-10 mg/L) x 2.835 L)/0.9

LD = 31.5 mg round up to 32 mg

6.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 4.5 kg infant presents with seizure activity, and needs to receive phenobarbital sodium (S = 0.9) IV every 12 hours with a desired Css of 20 mg/L. Calculate the maintnance dose in mg to to be given every 12 hours to achieve a Css of 20 mg/L. Round to the nearest tenth.

V = 0.63 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

Cl = 0.0047 L/hr/kg x 4.5 kg = 0.02115 L/hr

MD = (Cssavg x Cl x τ)/S

MD = (20 x 0.02115 x 12)/0.9

MD = 5.076/0.9 = 5.6 mg

7.

MATH RESPONSE QUESTION

2 mins • 1 pt

A 4.5kg infant presents with seizure activity, and the neurologist would like to start the baby on phenobarbital sodium (S = 0.9) IV. What is the expected half-life of phenobarbital in this patient in hr? Round to the nearest tenth.

V = 0.63 L/kg, Cl = 0.0047 L/hr/kg

Mathematical Equivalence

ON

Answer explanation

V = 0.63 L/kg x 4.5 kg = 2.835 L

Cl = 0.0047 L/hr/kg x 4.5 kg = 0.02115 L/hr

K = Cl/Vd

K = 0.02115/2.835 = 0.00746 hr-1

t1/2 = 0.693/K

t1/2 = 0.693/0.00746 = 92.9 hr

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