LS1a Midterm Exam 1 Review Session Problems
1. An aqueous mixture of a weak acid and its conjugate base is often used in the laboratory to prepare solutions referred to as buffers. One commonly used acid is called TRIS, which has the chemical formula shown below.
+ (HOCH2)3CNH3 (acid)
(HOCH2)3CNH2 (conjugate base)
a. Draw the chemical structures for both the acid and conjugate base forms of TRIS. You may use any drawing convention you wish, but you MUST include all lone pairs of electrons and any net formal charges.
Acid Conjugate Base
b. Write the chemical reaction for the deprotonation of TRIS acid to its conjugate base using the chemical formulae provided above.
+ + (HOCH2)3CNH3 (HOCH2)3CNH2 + H
c. Write an expression for the equilibrium constant for the deprotonation reaction of the TRIS conjugate base to TRIS acid.
[(HOCH ) CNH ] [H+] K = 2 3 2 eq [(HOCH ) CNH +] 2 3 3
d. Is the equilibrium constant you have just defined the same as the acid dissociation constant Ka?
Yes
e. The pKa of TRIS is 8.3. If the concentration of protonated TRIS is 10 mM in a solution of pH 7.3, what is the concentration (in mM) of the conjugate base at the same pH?
The concentration of Tris base is 1 mM. 2. Pictured below is the small molecule thromboxane. This molecule plays a very important role in the process of blood clotting.
O OH pKa = 5
cis Thromboxane
O CH O 3 OH trans pKa = 15
a. Circle the chiral centers in the molecule.
b. Identify and label any cis- or trans- double bonds in the molecule.
c. Ignoring geometric isomers, how many different stereoisomers are possible for this molecule? 25= 32
d. How many geometric isomers does this molecule have? 22= 4
e. Using the pKa values provided above, draw the dominant form of thromboxane at the indicated pH.
i) pH = 3 O OH
O CH O 3 OH ii) pH = 9 O O-
O CH O 3 OH 3. The molecule shown below is R-Carvone
O O O
(R)-carvone (L)-carvone or
a. Draw the enantiomer of R–Carvone
O O O
(R)-carvone (L)-carvone or
b. Briefly describe one way you could distinguish these enantiomers from one another
By different interactions with another chiral molecule. [Or by rotation of plane-polarized light]
c. Write the correct molecular formula for Carvone
C10H14O
4. Shown below are two arrow-pushing mechanisms. Please provide the products.
H +H NH2 H3C I N + I- R R a. CH3
b. O O- S R R S S R' R' S
5. Triple-stranded DNA was first observed in 1957. Scientists later discovered that the formation of triple-stranded DNA involves a type of base pairing called Hoogsteen pairing, which differs from the Watson-Crick base pairing observed in normal double-stranded DNA. Below is an example of a Hoogsteen base pair.
NH2 O N 1 N HN 2 N N N O DNA DNA
a. Identify each of these bases by name.
1 = __ Adenine ______
2 = __Thymine______
b. Redraw these two base pairs showing all lone pair electrons.
H N H O N 1 N H N 2 N N N O DNA DNA
c. Draw the hydrogen bond interactions expected for this Hoogsteen base pair. d. Formation of a triple helix requires three bases to interact with each other to form a “base trio” rather than a “base pair”. One way in which this three-way interaction can take place is for the three bases to engage in both Watson-Crick base pairing and Hoogsteen base pairing. Show how two thymines and an adenine could interact with each other to form a “base trio” by using both Watson-Crick and Hoogsteen base pairing.
DNA N O N O H H N H O N N H N N N N O DNA DNA
6. Consider a DNA polymerization step during PCR:
3' 5'
A - T T - A DNA polymerase T - A dATP, dTTP, - G C dGTP,dCTP C - G A A 3' 72o G T 5'
a. Would the following reaction proceed? Why or why not?
3' 5' 3' 5'
A - T A - T T - A - DNA polymerase T A T - A T - A G - C dATP, dTTP, G - C C - G dGTP, dCTP C - G T A - T 5' T - 72o A T C G - C A T - A T A - T 5' 3' 3'
No, because DNA synthesis occurs in a 5’ to 3’ direction and requires a free 3'-OH group as well as a template following the 3'-OH group. At the 3’ ends of both red and blue strands, no remaining template exists for the polymerase to continue synthesizing DNA. b. Consider the reaction below; what would the result be if you forgot to add dCTP to the reaction mixture? Draw the resulting product (be sure to label the 3’ and 5’ end of your products).
3' 5'
A - T T - A DNA polymerase T - A dATP, dTTP, - G C dGTP C - G A ? A 3' 72o G T 5'
5 ’- TAACGTT - 3’
7. Shown below are three amino acids with the pKa values of their respective groups.
~9.8 O O O ~2.1 ~9.2 ~1.8 ~9.0 ~2.2 +H3N CH C OH +H3N CH C OH +H3N CH C OH CH2 CH2 CH2 C O CH2 HN OH CH2 NH ~3.9 CH2 ~6.5 + NH3 ~10.5
(a) (b) (c)
b. Identify each of the amino acids (a) through (c).
a = Aspartic Acid b = Histidine c = Lysine
c. Based on the structures shown below and the pKa values of the corresponding groups from above, what is the most likely range of the pH of the surrounding solution?
O O O + - + - + - H3N CH C O H3N CH C O H3N CH C O CH2 CH2 CH2 C O CH2 - N O CH2 NH CH2 + NH3
(1) between 3 and 4 (2) between 7 and 8 (3) between 10 and 11
8. Shown below is a picture of a protein.
The major interactions that hold the protein in its properly folded conformation are highlighted.
f
a e c
b d
a. Briefly describe each interaction a-e
b. Which of these interactions would be most affected if the protein were placed in reducing conditions? d, disulfide bond would be broken. c. Label each of the amino acid side chains with the full name, three letter abbreviation and one letter code. Serine, Ser, S; Valine, Val, V; Leucine, Leu, L; Lysine, Lys, K; Cysteine, Cys, C;
d. If you were to subject a dilute solution of this protein to denaturing conditions, and then remove the denaturant, would you expect the protein to refold into the conformation shown above? Discuss the thermodynamic factors that contribute to your answer.
Yes, the folded protein is thermodynamically more stable. The entropic driving force is the release of ordered water from solvation shells around hydrophobic side chains (∆S positive). This offsets the entropic penalty paid by restricting the conformational freedom of the protein (∆S negative). Enthalpic contributors are the formation of electrostatic, hydrogen bonding, and Van der Waals interactions within the protein (∆H negative). Enthalpic detractors are the loss of hydrogen bonding interactions with water around the polar side chains and backbone atoms. e. If were you were to subject a concentrated solution of protein to denaturing conditions, and then remove the denaturant, would you expect the protein to refold into the conformation shown above? Discuss the thermodynamic factors that contribute to your answer.
No, protein will aggregate. Release of ordered water is still the entropic driving force. If a single molecule of protein is in a dilute solution, the only way it can release this ordered water is have its hydrophobic side chains associate with one another away from water in the interior of the protein (i.e. the hydrophobic effect drives the protein to fold). However, if another molecule of protein is nearby, hydrophobic side chains on the two different molecules of protein can interact with each other to accomplish the same thing. Therefore, if the concentration is high enough, protein molecules can achieve the same favorable ∆S by interacting with each other rather than by folding correctly. In addition, because the protein does not have to adopt a specific properly folded conformation, there is less of an entropic penalty to aggregation than there is to folding. Enthalpically, aggregation is similar to folding because all of the same types of interactions can be formed/broken. 9. Proteins A and B bind to each other as shown at physiological pH (7.0):
H H N N O 2 NH Protein A O 2 Protein B
Glutamate Arginine sidechain sidechain pka = ~ 4 pka = ~ 12
a. What types of interactions exist between the Glu and Arg side chains at pH 7?
Ionic interaction between +/- charges; 2 hydrogen bonds between Glu oxygen lone pairs and Arg protons.
b. What would you expect to happen to the strength of the interaction as the pH is slowly increased from 7 to 14?
As the pH increases from 7 to 12, an increasing fraction of the arginine side chains on protein B become deprotonated and therefore neutral. Thus, at pH >12, the ionic interaction is lost. The hydrogen bonding interactions remain, although the overall strength of interaction is now weakened.
c. What would you expect to happen to the strength of the interaction as the pH is slowly decreased from 7 to 1?
As the pH decreases from 7 to 4, an increasing fraction of the glutamate side chains on protein A become protonated and therefore neutral. At pH < 4, the ionic interaction is lost. The hydrogen bonding interactions remain, although they are weaker because neutral oxygen is less electronegative than negatively charged oxygen.