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49 Cards in this Set
- Front
- Back
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Which of the following is not a side effect of the Ace Inhibitor (Captopril)?
A. Rash B. Angioedema C. Cough D. Congestion |
D. Congestion
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Which of the following is not a side effect of the Vasodilator (Nifedipine)?
A. Nausea B. Flush appearance C. Vertigo D. Sexual dysfunction |
D. Sexual dysfunction
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Which of the following is not a side effect of the Sympathoplegics (Clonidine)?
A. Hypertension B. Asthma C. Dry oral cavity D. Lethargic behavior |
B. Asthma
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Which of the following is not a side effect of the Dieuretics (Loop dieuretics)?
A. Alkalosis B. Nausea C. Hypotension D. Potassium deficits |
B. Nausea
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Which of the following is not an effect of the drug (Clozapine)?
A. Agranulocytosis B. Antipsychotic C. Used for Schizophrenia D. Increased appetite |
D. Increased appetite
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Which of the following is not treated with (Epinephrine)?
A. Renal disease B. Asthma C. Hypotension D. Glaucoma |
A. Renal disease
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Which of the following is not treated with (Ephedrine)?
A. COPD B. Hypotension C. Congestion D. Incontinence |
A. COPD
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Which of the following are not treated with Barbiturates?
A. Seizures B. Hypotension C. Insomnia D. Anxiety |
B. Hypotension
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Which of the following are not treated with Nifedipine?
A. Angina B. Arrhythmias C. Htn D. Fluid retention |
D. Fluid retention
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Which of the following are not treated with Methotrexate?
A. Sarcomas B. Leukemias C. Ectopic pregnancy D. Rheumatic fever |
D. Rheumatic fever
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Which of the following are not treated with Prednisone?
A. Cushing's disease B. Testicular cancer C. Lympthomas D. Chronic leukemias |
B. Testicular cancer
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Which of the following are not treated with Dexamethasone?
A. Inflammation B. Asthma C. Addison's disease D. Wilson's disease |
D. Wilson's disease
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Which of the following is the primary site of activity for the drug Warfarin?
A. Kidney B. Liver C. Blood D. Heart |
B. Liver
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Which of the following drugs is associated with the reaction of hepatitis?
A. Valproic acid B. Quinidine C. Isoniazid D. Ethosuximide |
C. Isoniazid
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A drug ending in the suffix (pril) is considered a ______.
A. H B. ACE inhibitor C. Antifungal D. Beta agonist |
B. ACE inhibitor
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A drug ending in the suffix (azole) is considered a ______.
A. H B. ACE inhibitor C. Antifungal D. Beta agonist |
C. Antifungal
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A drug ending in the suffix (tidine) is considered a ______.
A. Antidepressant B. Protease inhibitor C. Beta antagonist D. H2 antagonist |
D. H2 antagonist
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A drug ending in the suffix (navir) is considered a ______.
A. Antidepressant B. Protease inhibitor C. Beta antagonist D. H2 antagonist |
B. Protease inhibitor
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Which of the following drugs is associated with the reaction of extreme photosensitivity?
A. Digitalis B. Niacin C. Tetracycline D. Fluoroquinolones |
C. Tetracycline
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Which of the following is not related to a drug toxicity of Prednisone?
A. Cataracts B. Hypotension C. Psychosis D. Acne |
B. Hypotension
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Which of the following is considered a class IA Sodium Channel blocker?
A. Mexiletine B. Aminodarone C. Quinidine D. Procainamide |
B. Aminodarone
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Which of the following is considered a class IA Sodium Channel blocker?
A. Propafenone B. Disopyramide C. Aminodarone D. Quinidine |
A. Propafenone
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Potassium sparing diuretics have the primary effect upon the _____ found in the kidney.
A. Proximal convoluted tubule B. Loop of Henle C. Collecting duct D. Distal convoluted tubule |
D. Distal convoluted tubule
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Which of the following is not directly related to a drug toxicity of Nitroglycerin?
A. Headaches B. Tachycardia C. Dizziness D. Projectile vomiting |
D. Projectile vomiting
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Which of the following cannot be refilled under any circumstances?
a. Metoprolol b. Methylphenidate c. Mitomycin d. Hydroxyurea |
(b) Methylphenidate is a controlled class II drug. It is indicated for treatment of attention deficit hyperactivity disorder (ADHD). It cannot be refilled under any circumstances.
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Valproic acid syrup is available as 250 mg/5 cc. If a patient is taking 1000 mg in the morning and 750 mg in the evening, how many cc of syrup will you dispense for a 30 day supply ?
a. 1500 cc b. 1050 cc c. 500 cc d. 480 cc |
(b) 1050cc. In this type of calculation first find out number of cc required for 1 day.
Morning dose Drug ml of solution 250 mg present 5 cc of solution 1000 mg ? 1000 x 5cc / 250 = 20 cc of solution. Evening dose Drug ml of solution 250 mg present in 5 cc of solution 750 mg ? 750 x 5cc / 250 = 15 cc of solution. For a 30 day supply of Valproic acid (20 cc (am) + 15 cc (pm) ) x 30 = 1050cc. |
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If a patient was taking Glyburide for type II diabetes and a physician prescribed Chlorpropamide, the pharmacy technician will :
a. Fill the prescription. b. Notify the pharmacist about duplication of therapy. c. Notify the pharmacist about any interaction. d. All of the above. |
(b) Glyburide and Chlorpropamide are oral sulfonylurea agents indicated for treatment of diabetes. Pharmacy technician will notify the pharmacist about duplication of therapy.
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All of the following drugs should be carefully prescribed with aspirin EXCEPT
a. Enoxaprine b. Coumadin c. Heparin d. Metoclopropamide |
(d) Aspirin is a blood thinning agent indicated for prevention of heart stroke. It should be carefully prescribed with other blood thinning agents because of risks of bleeding. Lovenox (Enoxaprine), Coumadin (Warfarin), Heparin, Plavix (Clopidrogel), Ticlid (Ticlopidine), Depakene (Valproic acid), Persantine (Dipyridamole), Mandol (Cefamandole), Cefotan (Cefotetan), Cefobid (Cefoperazone) and Moxam (Moxalactam) are agents that increase bleeding tendency in patients and should be carefully prescribed with other blood thinning agents.
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Overdose of Coumadin can be treated by
a. Vitamin K b. Acetylcysteine c. Mesna d. Protamine sulfate |
a) The overdose of Coumadin can be treated by administration of Vitamin K1 (Mephyton).
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Prochlorperazine can be classified as a
a. Antiemetic b. Antidepressant c. Anti-anxiety d. Antihypertensive |
(a) Compazine (Prochlorperazine) is indicated for treatment of nausea and vomiting and therefore is classified as an anti-emetic (agent that prevent nausea and vomiting).
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A patient is taking Amoxicillin 500 mg by mouth three times a day for 7 days. How many capsules of 500 mg of Amoxicillin will you dispense?
a. 42 b. 21 c. 7 d. 30 |
b. 21
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The cost of 100 tablets of Olanzapine is $420.00. The percentage mark up on the prescription is 15%. What would be the retail cost of 30 tablets of the above drug ?
a. $200 b. $145 c. $300 d. $450 |
(b) The cost of 100 tablets of Olanzapine is $420 and therefore the cost of one tablet of Olanzapine would be $420/100 = $ 4.20.
The % mark-up on the prescription is 15% Cost of prescription mark up on rx $ 100 $ 15 $ 4.20 ? 15 x 4.20 / 100 = $ 0.63, therefore dispensing cost of each prescription would be $ 4.20 + $ 0.63 = $ 4.83 The cost of 30 tablets would be $ 4.83 x 30 = $145 |
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How much Lidocaine is required to prepare 1 : 1000, 30 cc of solution of Lidocaine?
a. 10 mg b. 0.03 mg c. 30 mg d. 300 mg |
(c) 1: 1000 generally interprets as 1 gm in 1000cc solution. The amount of lidocaine in 30cc of 1:1000 solution can be calculated as follows:
= 30 x 1/1000 = 0.03 gm = 30 milligrams. |
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How many cc of 75 % alcohol should mix with 10 % of 1000cc alcohol to prepare 30% of 500cc alcohol solution?
a. 346.16 cc b. 234.43 cc c. 153.84 cc d. 121.12 cc |
(c) To solve this type of problem, we need to use alligation method.
75 20 (75%) 30 10 45 (10%) Total parts 65 (30%) To prepare 65 (30%) 20 parts (75%) need To prepare 500 (30%) ? = 500 x 20/65 = 153.84cc (75%) alcohol If we mixed 153.84 cc of 75% alcohol with 346.16cc [500cc - 153.84] of 10% alcohol, then we can get 500cc of 30% alcohol solution. |
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If 60 gm of 1% hydrocortisone is mixed with 80 gm of 2.5% of hydrocortisone, what is the % of hydrocortisone in final mixture?
a. 2.2 % w/w b. 1.85 % w/w c. 0.25 % w/w d. 1.75 % w/w |
(b) Amount of Hydrocortisone in 60 gm, 1%
= 60/100 = 0.6 gm of hydrocortisone. Amount of hydrocortisone in 80 gm, 2.5% = 80 x 2.5/100 = 2 gm hydrocortisone % amount of hydrocortisone in final mixture = 100 x 2.6 (2gm + 0.6gm)/140 (80gm + 60gm) = 1.85% w/w. |
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If 1000 tablets of Risperdal 1 mg cost $ 2250 and % mark up on prescription is 20, what would be the retail price of 30 tablets?
a. $ 150 b. $ 17 c. $ 500 d. $ 81 |
(d) 1000 tablets of Risperdal 1mg cost $ 2250. The % mark up on prescription is 20%.
Therefore retail price of 1000 tablets would be = 120 x 2250/100 = $ 2700 ** For each $ 100 cost = $120 retail cost** Price for 30 tablet would be = 30 x 2700/1000 = $ 81 |
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If the ratio of ionized to unionized species of drug is 103 and PKa = 2.2, what is the PH of the solution?
a. 2.2 b. 0.8 c. 5.2 d. 3.0 |
(c) A pH of the solution can be found by the following formula:
pH = pKa + log ionize/unionize = 2.2 + log 103 = 2.2 + 3 = 5.2 |
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If dropper is calibrated to deliver 325 mg of iron sulfate in 0.6 cc and adult dose of drug is 325 mg , what is the dose of a drug in cc for a 15 months-old infant?
a. 1.2 cc b. 0.3 cc c. 0.06 cc d. 0.01 cc |
(c) 0.06cc. According to Fried’s rule
= age in months/150 x adult dose = 15 x 325/150 = 32.5 mg The dropper is calibrated to deliver 325 mg of Iron sulfate in 0.6 cc, therefore = 0.6 x 32.5/325 = 0.06cc |
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If the dose of a drug is 10 mg/kg/day, how many 250 mg/100cc ready infusion-bags require to fill above order? Patient's weight is 156 lbs.
a. 1 bag b. 2 bags c. 3 bags d. 5 bags |
(c) Patient weight is 156 lbs, therefore weight in Kg would be 156/2.2 = 70.9 kg
A normal therapeutically recommended dose of drug is 10mg/kg/day, therefore dose in above patient = 10 x 70.9 = 709 mg Each ready-infusion-bag contains 250 mg of drug, so number of bags require to fill order would be = 709/250 = 2.83 @ 3 bags. |
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How much of atropine is required to prepare 240cc in such way that when 1 teaspoonful of the solution is diluted to 1 pint gives 1 in 500 solution?
a. 2.25 gm b. 46.08 gm c. 35.15 gm d. 25.35 gm |
(b) To solve this kind of problem, we must first find out the amount of drug present in final solution.
Amount of atropine in 1 pint, 1 in 500 soln . = 480 x 1 /500= 0.96 gm of atropine. Now, 0.96 gm of drug must be present in 1 teaspoonful of drug solution, therefore we can say 5cc (1 teaspoonful) contains 0.96 gm 240cc solution requires? |
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How much sodium bicarbonate powder is required to prepare 240cc of 0.10 N solution of sodium bicarbonate?
a. 1.35 gm b. 3.25 gm c. 4.81 gm d. 2.016 gm |
(d) Gram equivalent weights of solute in 1 liter of solution is defined as normality, therefore 1N solution of sodium bicarbonate will contain 84 gms in 1000 cc. We want to find quantity of sodium bicarbonate in 240cc, 0.1 N solution,
1 N solution contains 84 gm 0.1 N solution contains ? = 0.1 x 84 = 8.4 gms/1000cc. 240cc solution will contain: = 240 x 8.4/1000 = 2.016 gm of NaHCO3 |
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How many meq of Na+ are present in 0.9% 250cc normal saline solution? [Na+ = 23, Cl- = 35.5]
a. 23.12 meq b. 15.17 meq c. 53.15 meq d. 38.46 meq |
(d) An amount of sodium chloride presents in 250 cc of 0.9% NaCl,
= 250 x 0.9/100 = 2.25 gm NaCl Total equivalents Na+ = weight in gm equivalent wt = 2.25/58.5 = 0.03846 equivalents |
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If the probability of success in Null hypothesis is 0.6, what is the probability of failure?
a. 0.3 b. 0.9 c. 0.6 d. 0.4 |
(d) The sum of probability of success and failure would be equal to 1 in Null hypothesis and can be expressed by following formula:
p + q = 1, where p = probability of success q = probability of failure q = 1 - p = 1- 0.6 = 0.4 |
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How many grams of cocabutter are required to dispense 12 suppositories of tannic acid each weighing 2 gm and contain 400 mg of tannic acid?
a. 21.23 gm b. 18.67 gm c. 14.12 gm d. 13.25 gm |
(b) We want to dispense 12 suppositories each weighing 2gm and containing 400mg of tannic acid,
Amount of coca butter = 2 gm x 12 = 24 gm Amount of tannic acid = 0.4 gm x 12 = 4.8 gm. Displacement value of tannic acid is 0.9, therefore = 4.8/0.9 = 5.33 gm of base will displace 4.8 gm tannic acid = 5.33 gm cocabutter Amount of coca butter = 24 gm - 5.33 gm = 18.67 gm |
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What is the "mode" or "median" of the following values?
120, 135, 140, 118, 175, 105, 115, 190 a. 135 b. 118 c. 127.5 d. 175 |
(c) Median or Mode is generally expressed as a middle value of experiment, if number of values are even, then average of middle values should be considered. To find median or mode of experiment data, one should first arrange the data in ascending or descending order.
In our example, 105, 115, 118, 120, 135, 140, 175, 190 = (120 + 135)/2 = 127.50 |
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If the concentration of reactant M is half in a reaction that is third order in M, by what factor will rate of reaction change?
a. 1/8 times b. 1/4 times c. 8 times d. 4 times |
(a) The concentration of reactant M is half in a reaction that is third order in kinetic,
dx/dt = k (a-x) (b-x) (c-x) = k (a-x)3 where a=b=c = k (M)3 now M = M/2 = k (M/2)3 = 1 /8 k (M)3 |
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What is the rate of constant after 90 minutes if the initial concentration of drug is 500mg/cc and 50mg/cc after 90 minutes? (First Order Kinetic)
a. 0.051 min-1 b. 0.025 min-1 c. 0.35 min-1 d. 0.86 min-1 |
(b) 0.025 min-1
For the first order kinetic, K = 2.303/t x log Co/C = 2.303/90 x log 500/50 = 2.303/90 x log 10 = 0.025 min-1 |
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If the total body clearance of the patient is 2100cc/hr and hepatic clearance is 300cc /hr, what is the status of renal function in a patient?
a. Excellent b. Normal c. Moderately impaired d. Severely impaired |
(d) The status of renal function impairment can be expressed by creatinine clearance.
ClT = ClH + ClR where, ClT = Total body clearance ClH = Hepatic clearance ClR = Renal clearance 2100 = 300 + ClR ClR = 1800 ml/hr = 30 ml/min The normal creatinine clearance generally lies between 80 to 120 ml/min. A creatinine clearance in patient is 30 ml/min which will be considered severely impaired. |
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How many grams of sodium chloride are required to prepare 250cc of 1% boric acid solution to isotonic with eye tears?
FP of 1% boric acid = -0.29o C FP of 1% sodium chloride = - 0.58oC FP of blood = -0.52o C a. 250 mg b. 325 mg c. 991 mg d. 1221 mg |
(c) Blood serum freezes at 0.520 C, and all solutions having this freezing point are isotonic with blood serum. 0.9% sodium chloride have the same freezing point that of blood serum.
FP provides by 1% Boric acid = -0.29o C FP of blood = - 0.52o C FP (needed) by NaCl = (0.52-0.29) = 0.23o C Now as we know that FP provides by 1% NaCl would be -0.58o C therefore one can say, For FP 0.58o needs 1 % NaCl For FP 0.23o needs ? = (0.23 x 1)/0.58 = 0.396 % NaCl 0.396 gm of NaCl/ 100cc. The amount of NaCl needed for 250cc, = (250 x 0.396)/100 = 0.991 gm NaCl = 991 mg |
