Medchem Exams > Exam_2_key

Basic Principles of Medicinal Chemistry Lab

PHR 143P

SUMMARY EXAM #2 

December 3, 2003

150 Total Points NAME__________KEY_________________  UT­EID________________ Fill in your answers to the questions below ON THIS SHEET.  Keep your  answers brief.  ONLY ANSWERS WITHIN THE BOXED AREAS WILL BE  GRADED.

HPLC Lab Questions (Dr. Kerwin) 1. (7 points) A certain barbiturate is 10% ionized in a 51:49 mixture of methanol/phosphate buffer (pH 7.2). What is the pKa of this barbiturate?

(2 pts)

pH = pKa + log {[A-]/[HA]} or fu = 1/{1-10^(pH – pKa)}

(3 pts) [A-]/[HA] = 0.1/0.9

or

fu = 0.9

(2 pts) pKa = 7.2 – (-0.954); pKa = 8.15

2. (8 points) A. Label the TLC plate shown below to indicate: • The origin • The solvent Front B. Draw on the TLC plate three spots of the following approximate Rf values (you can estimate the position on the plate – you do not need a ruler): • Rf = 0.9 • Rf = 0.5 • Rf = 0.1 C. Assume that a TLC experiment was performed to separate a mixture of the three compounds shown below using silica gel as the stationary phase and 1% methanol (CH3OH) in chloroform (CHCl3) as the mobile phase. These three compounds give rise to three separate spots on the TLC plate, one for each compound, corresponding to the three spots that you drew in part B above. Label each of the three spots that you drew with its Rf value and the compound from the list below that would most likely give rise to that spot. CO2H O CO2H Compound X

O Compound Y

Compound Z

solvent front Rf = 0.9, Z

Rf = 0.5, Y

Rf = 0.1, X origin

3. (7 points) You have completed the HPLC lab by determining the reverse-phase HPLC retention time for a series of barbiturates using an aqueous mobile phase, pH 7.4. You have plotted these retention times versus the apparent partition coefficients (logPapp) for these drugs (see the plot below). Use the graph below to answer the following question: A barbiturate with a pKa of 7.4 gives a retention time of 7.0 minutes. What is the intrinsic partition coefficient (Log Po) for this barbiturate?

2.8 2.6 2.4

y = 0.1645x + 0.2483 R2 = 0.9964

2.2

LogPapp

2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

LogPapp = 1.4

LogPapp = LogPo + log(fraction unionized) 1.4 = LogPo + (log(0.5)) LogPo = 1.7

6

7

8

9

10

11

Retention Time (min )

12

13

14

15

16

Visualization Lab Questions (Dr. Dalby) 4. (8 points) Starting at the arrow name this cyclic peptide? H2N CH2 CH2 CH2

H N

O

CH

C

HN H2C

CH2 NH O

C

CH

CH O

OH CH

C

CH3 NH

HN CH CH2

H N

C O

C

CH CH3

O

H2C C HO

O

Trp-Glu-Ala-Thr-Lys (1 pt each, in any order) (3 pts for correct order)

5. (8 points) What amino acids is this? = Carbon = Nitrogen = Oxygen

Asn

6. (8 points) What amino acid is this?

Pro

7. (16 points) What amino acids are present in the structure below? Is it a typical peptide? Explain your answer.

Lys and Asn Not a typical peptide because peptide (amide) bond is formed by the side chain amino group of Lys, not the alpha-amino group.

Liver Game and Metabolism Labs (Dr. Davis) Attached you will find the composite metabolic tree that was developed in the liver game. The following three questions relate to that scheme. Circle the letter corresponding to the single best answer on this exam paper – do not  use the Scantron. 23. (7 points) On the upper right part of the diagram, we see a product resulting from omega-hydroxylation and then an acid product generated from that alcohol. Why is there no aldehyde ‘intermediate’ structure shown where the question mark (“?”) is? A. You cannot form an aldehyde from a primary alcohol. B. The production of an acid from a primary alcohol does not go through an aldehyde intermediate. C. While the production of an acid from a primary alcohol usually goes through an aldehyde intermediate, there are other metabolic pathways that do not require this (i.e., oxidation of the alcohol in one step to the acid). That’s what we’re seeing here. D. There was probably aldehyde formed, but apparently it went to the sulfate conjugate (in levels of <1% so it’s not shown). E. None of the above. 24. (7 points) Why is benzylic hydroxylation not depicted in this scheme? A. What are you talking about? Benzylic hydroxylation is clearly shown in this scheme? B. There are no positions on this molecule that would be considered ‘benzylic’. C. While there are benzylic positions on this molecule, hydroxylation cannot occur at these centers because the benzylic carbon atoms have four substituents. D. While benzylic hydroxylation could occur on this molecule, it must be that none of the students were assigned this enzyme activity. E. None of the above. 25. (7 points) At the bottom of the page we see O-dealkylation to form a phenol, and from that phenol, both sulfate and glucuronide conjugates are generated. Why are both generated? A. The students who were passed the ‘phenol’ must have been assigned phase-2 reactions. For any one of those students, that student would have had the option of carrying out sulfate conjugation or glucuronide formation (their choice), and so a mixture resulted. B. Some students who were passed the ‘phenol’ must have been assigned glutathione transferase, and some students were assigned UDP-glucuronyl transferase. C. Some students who were passed the ‘phenol’ must have been assigned UDPglucuronyl transferase, and some students were assigned aryl sulfotransferase. D. Some students who were passed the ‘phenol’ must have been assigned aryl sulfotransferase, and some students were assigned glutathione transferase. E. None of the above.

26. (7 points) At the beginning of the lab session, you and your lab partner were given the molecule shown below and asked to complete as much of a metabolic tree as you could. After the Flash® tutorials, you were asked to return to the scheme you started and add any additional routes of metabolism that you could. In this second step, how many additional metabolites did you and your partner add to this scheme (Note: yes, this is really a test question. Be honest; do not inflate this number).

A. B. C. D. E.

Zero (no additional metabolites) 1-2 additional metabolites 3-5 additional metabolites 6-10 additional metabolites >10 additional metabolites

27. (7 points) On the tutorial for acetaminophen (structure shown below) a number of phase-2 metabolites were generated in the metabolic scheme. For the statements that follow, when is the first time an incorrect statement is made?

A. The parent drug could undergo phase-2 to form a glucuronide conjugate. B. The parent drug could undergo phase-2 to form glucuronide and sulfate conjugates. C. The parent drug could undergo phase-2 to form glucuronide and sulfate conjugates; in addition, the parent drug could undergo amide cleavage, and the resultant phase-1 metabolite could then be conjugated to form a glucuronide on the amine. D. The parent drug could undergo phase-2 to form glucuronide and sulfate conjugates; in addition, the parent drug could undergo amide cleavage, and the resultant phase-1 metabolite could then be conjugated to form glucuronide and sulfate conjugates on the amine. E. None of the above. Statements A-D are all correct.

28. (7 points) What did we learn from the cocaine tutorial?

A. Phase-2 metabolism (glucuronidation ONLY) can occur on cocaine. B. Phase-2 metabolism (sulfate formation ONLY) can occur on cocaine. C. Both Phase-2 reactions (glucuronidation AND sulfate formation) can occur on cocaine. D. We must first conduct a phase-1 reaction on cocaine before a phase-2 reaction will be possible. E. None of the above.