Test 4: 46 principles part 2

Question 1

Solubility in the blood depends large on the — partition coefficient

Answer 1

blood/gas

Question 2

what routes of exposure is the fastest/most effective

Answer 2

IV > inhalation > IP > SQ > IM > ID > oral > dermal

Question 3

— the ratio of un-ionized:ionized, the greater the potential for absorption across a —

Answer 3

Higher

lipid membrane

simple diffusion/active transport: Un-ionized (uncharged) molecules have greater lipid solubility and thus can traverse phospholipids bilayers with greater ease than charged (ionized) molecules

Question 4

when water flows in bulk across a porous membrane, any solute small enough to pass through the pores flows with it

Answer 4

filtration

Question 5

saturable process not driven by concentration gradients, instead moving xenobiotics against concentration gradients

Answer 5

active transport

Question 6

Quantity or percentage portion of total chemical that was absorbed and available to be processed

Answer 6

Bioavailability (F) = Systemic Availability

F=100 means 100% of drug made it into the blood

Question 7

While —, toxicants cannot cross capillary walls to be distributed into the extravascular space or filtered by the kidneys

Answer 7

bound (reversibly) to plasma proteins

Question 8

4 reasons toxin do not enter CNS

Answer 8

• endothelial cells are tightly joined- do not let many things through
• ATP dependent multidrug resistant proteins move chemicals out of brain
• astrocytes block things through capillaries
• lower protein in CNS- prevent protein bound water insoluble compounds

Question 9

— lipid solubility enhances penetration into the CNS, while — diminishes it

Answer 9

Increased

ionization

Question 10

ideal metabolic system should produce metabolities that are — soluble and —. Enzymes used should be —

Answer 10

water

nontoxic

broad enough substrate specificity to properly handle any newly encountered substances

Question 11

types of phase 1 metabolism

Answer 11

Hydrolysis
Oxidation – alcohol dehydrogenase, aldehyde dehydrogenase, monoamine oxidase, cytochrome P-450
Reduction

Phase I metabolism converts apolar, lipophilic chemicals into more polar & more hydrophilic metabolites via introduction or liberation of functional groups

Question 12

— metabolism converts apolar, lipophilic chemicals into more polar & more hydrophilic metabolites via introduction or liberation of functional groups

Answer 12

Phase I

Hydrolysis
Oxidation- cytochrome P450
reduction

Question 13

— conjugates either the chemical itself or its metabolite formed during phase I with a functional group that results in a multifold increase in water solubility

Answer 13

Phase II

Glucuronidation
Sulfation
Amino acid conjugation
Acylation
Methylation

Question 14

phase II metabolism works by

Answer 14

conjugates either the chemical
itself or its metabolite formed during phase I with a functional group that results in a multifold increase in water solubility

Glucuronidation
Sulfation
Amino acid conjugation
Acylation
Methylation

Question 15

what happens with cats and acetaminophen

Answer 15

cats lack enzyme to metabolize by glucoronidation

cytochrome P-450 metabolizes it into N- acetybenzoquinoneimine, which binds to hepatic proteins & leads to centrilobular necrosis (liver necrosis)

Question 16

zero order kinetics

Answer 16

Only a finite amount of chemical can be
eliminated per unit of time

pathway becomes saturated and drug starts to accumulate (alcohol)

Question 17

1st order kinetics

Answer 17

Rate of elimination is proportional to the amount of chemical in the body at that time

Not saturable pathway

Question 18

1 ppm= — mg/kg

Answer 18

1 part per million (ppm)
= 1 mg/kg
= 0.91 grams/ton
= 1 microgram (µg)/mL

Question 19

1 ppm= — grams/ton

Answer 19

1 part per million (ppm)
= 1 mg/kg
= 0.91 grams/ton
= 1 microgram (µg)/mL

Question 20

1 ppm= — (μg)/mL

Answer 20

1 part per million (ppm)
= 1 mg/kg
= 0.91 grams/ton
= 1 microgram (µg)/mL

Question 21

1 ppm to %

Answer 21

1 ppm = 1 mg/kg = 0.0001%

So moving the decimal point 4 places to the right converts % to ppm

Question 22

10% = — mg/mL

Answer 22

10% = 100 mg/mL

moving the decimal point 1 place to the right converts % to mg/mL

Question 23

So 1 ounce, fluid or dry, is
approximately equal to — g or — mL

Answer 23

30
30

Question 24

1 teaspoon = — mL

Answer 24

5 mL

Question 25

1 tablespoon = — mL

Answer 25

15 mL

Question 26

2 tablespoons= — mL

Answer 26

30= 1 fluid oz

Question 27

The label for a medicated poultry feed states that Drug A is present at 50 parts per million (ppm)

What is the concentration of Drug A in mg/kg and g/ton?

Answer 27

1 ppm = 1 mg/kg = 0.91 g/ton = 0.0001% = 1 μg/mL

So, 50 ppm = 50 mg/kg = 45.5 g/ton

Question 28

100 mg of Drug B causes hepatic necrosis in dogs and 700 mg causes CNS depression

If a dog eats 12 pounds of food containing 25 mg/kg of Drug B, what effects can be expected?

Answer 28

1 lbs = 0.454 kg
1 kilogram = 2.2 pounds

12 lbs ➗ 2.2 = 5.4 kg
5.45 kg x 25 mg/kg Drug B = 136 mg Drug B

Hepatic necrosis would be expected, but not CNS depression

Question 29

A full 8 ounce bottle of weed killer containing 15% Chemical C was chewed open by a goat

Now the bottle is completely empty

What is the maximum dose of Chemical C that the goat may have ingested?

Answer 29

1% solution = 10 mg/mL
1 fluid oz = 29.6 mL or ≈ 30 mL

15% = 150 mg/ml
8 ounces × ~30 ml/fluid oz ≈240 ml
240 ml × 150 mg/ml = 36,000 mg

Question 30

65-pound Labrador retriever has eaten a homemade chocolate cake that contained among other ingredients, 4 oz of unsweetened chocolate (393 mg caffeine per ounce) and 1.5 cups of strong brewed coffee (30 mg caffeine per ounce)

What total dosage of caffeine has this dog ingested?

Answer 30

4 oz choc x (393 mg caff/oz)= 1572 mg

65lbs dog ➗ 2.2= 29.5 kg dog

1.5 cups x 8 oz/cup × 30 mg/oz = 360 mg

Total amount caffeine ingested = 1572 mg + 360 mg = 1932 mg

Total dosage of caffeine = 1932 mg ÷ 29.5 kg = 65.5 mg/kg

Question 31

A 350-gram female rat needs to receive 50 mg/kg of Drug D twice a day

The drug solution contains 45 mg/mL of Drug D

How much of the drug solution
should the rat receive each day?

How much drug solution should be
dispensed for a 21- day course of
treatment?

Answer 31

350 grams × 1 kg/1000 grams = 0.35 kg rat

0.35 kg × 50 mg/kg = 17.5 mg of Drug D per dose

45 mg/1 mL = 17.5 mg/x
solve for x
x = 0.39 mL

Because there are 2 dose periods, 0.39 × 2 = 0.78 mL/day

So 0.78 mL/day × 21 days = 16.33 mL drug solution needed

Question 32

Your dog has recently been placed on a sodium- restricted diet but hates her new sodium-free
dog food

You want to try mixing Brand E chicken soup containing 950 mg of sodium per cup into her food to make it more palatable

If she can have no more than 1200 mg of sodium per day and is
fed twice a day, how much broth can you add to each meal in cups?

Answer 32

1200 mg of sodium per day ÷ 2 meals/day = 600 mg sodium per meal

600 mg ÷ 950 mg/cup ≈ 0.63 cups per meal

0.63 cups × 8 oz/cup = 5.04 oz per meal

5 oz ÷ 8 oz/cup = 5/8 cup

Question 33

Disinfectant F is sold as a 45% solution and has to be diluted to a 5% solution in water before use. To prepare 1 gallon of 5% solution, how much concentrated solution and water should be mixed together?

Answer 33

5% = 50 mg/mL
1 gallon = 3.785 L = 3785 mL

50 mg/mL = x/3785 mL
x= (50 x 3785)

x = 189,250 mg of Disinfectant F needed to make 1 gallon of 5% solution

189,250 mg ÷ 450 mg/mL = 421 mL of concentrated 45% solution

421 mL + x mL water = 3785 mL total

x = 3364 mL water = 3.364 L