DIT review - Pharm basics Flashcards
Describe the difference between zero-order and first-order velocity in enzyme kinetics (comparing velocity of reaction to concentration of substrate)
- At first, as you increase substrate concentration the reaction velocity increases
- First order velocity – reaction velocity is directly proportional to substrate concentration
- As you add more substrate the enzyme becomes saturated and you cannot increase velocity of reaction
- Zero order velocity – reaction velocity is independent of substrate concentration
Vmax is proprtional to the concentration of what?
- Proportional to the enzyme concentration
- After enzyme is saturated, the only way to increase velocity (Vmax) of the reaction is to increase enzyme
What is Km and what is it affected by
- Is the concentration of substrate that brings reaction velocity to half of Vmax
- Is inversely related to the affinity of the enzyme for its substrate
- If enzyme has high affinity for substrate, the slope of the graph will be steeper and Km will be a smaller number
- If enzyme has low affinity for substrate, the slope will be less steep and Km will be a larger number
Label the points on a Lineweaver-Burk plot (Y-intercept, X-intercept, and Slope)
- Y-intercept = 1/Vmax
- X-intercept = 1/-Km
- Slope = Km/Vmax
Using a Lineweaver-Burk plot, interpret what it means when the y-intercept decreases
- Decreased y-intercept = decreased 1/Vmax = increased Vmax, which means that more enzyme was added
Using a Lineweaver-Burk plot, interpret what it means when the x-intercept is moved to the right
X-intercept moved to the right = 1/-Km moving closer to 0 = Km increasing = decreased enzyme affinity for substrate
Describe the changes of a Michaelis-Mentin graph in the case of a competitive and noncompetitive inhibitor
(Changes in Km and Vmax)
- Competitive inhibitor (reversible)
- Resembles substrate and competes for active site on enzyme
- Can be overcome by increasing amount of substrate
- Effects:
- Increases Km – more substrate needed in order to achieve 1/2Vmax
- Does not affect Vmax – can still reach Vmax with increased [S]
- Resembles substrate and competes for active site on enzyme
- Noncompetitive inhibitor (irreversible)
- Binds to a separate site on the enzyme, causing a conformational change so the substrate can no longer bind
- Cannot be overcome with increased substrate
- Effects:
- Decreases Vmax
- Does not affect Km – changing substrate concentration will not help
- Km increases by default because 1/2Vmax is a new number
- Binds to a separate site on the enzyme, causing a conformational change so the substrate can no longer bind
Describe the changes of a Lineweaver-Burk plot in the case of a competitive and noncompetitive inhibitor
(Changes in Km and Vmax)
- Competitive inhibitor (reversible)
- Increases Km – more substrate needed in order to achieve 1/2Vmax
- X-intercept moves to the right
- Does not affect Vmax – can still reach Vmax with increased [S]
- Y-intercept remains the same
- Increases Km – more substrate needed in order to achieve 1/2Vmax
- Noncompetitive inhibitor (irreversible)
- Decreases Vmax
- Y-intercept increases
- Does not affect Km – changing substrate concentration will not help
- X-intercept remains the same
- Decreases Vmax
What is the equation for volume of distribution
- Vd = (amount of drug in body) / (plasma drug concentration)
What is the equation for clearance
- CL = (rate of elimination of drug) / (plasma drug concentration)
- CL = Vd x Ke (elimination constant)
- Ke = 0.7 / (half-life of drug)
What is the equation for loading dose
- LD = (concentration at steady state) x (volume of distribution)
- Concentration at steady state (Css) = desired concentration
What is the equation for maintenance dose
- MD = Steady state concentration x Clearance
When determining how many half-lives it takes to reach steady state, what is an important number to remembr
In first-order kinetics, a drug infused at a constant rate takes 4 half-lives to reach 94% of steady state
What is the equation for half life in first order kinetics
- Half life = (0.7 x Vd) / CL
Describe the difference between efficancy and potency of a drug
- Efficacy = maximum effect that a drug can produce
- Increased Vmax = increased efficacy
- Partial agonists have less efficacy than full agonists
- Potency = amount of drug needed for a given effect
- Decreased EC50 = increased potency = less drug needed
Will efficacy or potency be affected by:
- Competitive antagonists
- Non-competitive antagonists
- Efficacy will be decreased by non-competitive antagonists (decrease Vmax)
- Competitive antagonists will have no effect on efficacy (do not change Vmax)
- Potency will be decreased (curve shift to R) by competitive antagonists (increase Km)
- Non-competitive antagonists have no effect on potency (do not change Km)
What is the equation for therapeutic index
- Therapeutic index:
- TD50 / ED5
- TD50 = drug dose that toxic to 50% of population
- ED50 = drug dose that is effective in 50% of population
- THINK: TITE à _T_herapeutic _I_ndex = _T_D50 / _E_D50
- Safer drugs have higher therapeutic index
- TD50 / ED5
Which type of drug metabolism (phase 1 vs phase 2) occurs via CYP450 enzymes
Phase I
What reactions fall under the category of Phase I reaction
- Hydrolysis, Oxidation, Reduction
What are the characteristics of metabolites resulting from phase I reactions
- Result in metabolites that are:
- Slightly polar
- Water-soluble
- Still active