BMB 7500 - Exam 3

1. What is the pH of the following buffer mixture : $100mL$$100\ mL$ of $1M$$1\ M$ acetic acid and $100mL$$100\ mL$ of $0.5M$$0.5\ M$ sodium acetate $(p{K}_a=4.76)$$\left(\ pK_a = 4.76 \ \right)$

   $$pH=p{K}_a+lo{g}_{10}\left(\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}\right)$$
   $$100mL\text{Acetic Acid}\ast \frac{1L}{1000mL}\ast \frac{1mol\text{Acetic Acid}}{1L}=0.1\text{moles of Acetic Acid}$$
   $$100mL\text{Sodium Acetate}\ast \frac{1L}{1000mL}\ast \frac{0.5mol\text{Sodium Acetate}}{1L}=0.05\text{moles of Sodium Acetate}$$
   $$pH=4.76+lo{g}_{10}\left(\frac{0.05M}{0.1M}\right)=4.4589700043360185$$

The process of folding a protein $\left(\text{unfolded}\stackrel{-\rightharpoonup }{\leftharpoondown -}\text{folded}\right)$$\left(\ \ce{\text{unfolded} <=> \text{folded}} \ \right)$ has a $\mathrm{\Delta }{H}^{\circ }$$\Delta H^{\circ}$ of $-280kJ/mol$$-280\ kJ/mol$ and a $\mathrm{\Delta }{S}^{\circ }$$\Delta S^{\circ}$ of $\frac{-790J}{mol\cdot K}$$\frac{-790\ J}{mol \cdot K}$

2. What is the $\mathrm{\Delta }{G}^{\circ }$$\Delta G^{\circ}$ of the folding of this protein at $25{}^{\circ }C$$25\ ^{\circ}C$ ?

   $$\mathrm{\Delta }G=\mathrm{\Delta }H-(T\ast \mathrm{\Delta }S)$$
   $$\mathrm{\Delta }G=\frac{280kJ}{1mol}-((25{}^{\circ }C+273.15)\ast \left(\frac{-790J}{mol\cdot K}\right)\ast \frac{1kJ}{1000J})=-44.54kJ$$

    

3. At what temperature would the unfolding of this protein become favorable ?

   $$0=\frac{-280kJ}{1mol}-(T\ast \frac{-0.79kJ}{mol\cdot K})$$
   $$T\ast \frac{-0.79kJ}{mol\cdot K}=\frac{-280kJ}{1mol}$$
   $$T=\frac{\frac{-280kJ}{1mol}}{\frac{-0.79kJ}{mol\cdot K}}=\frac{-280kJ}{1mol}\ast \frac{mol\cdot K}{-0.79kJ}=354.4303797468354{}^{\circ }K$$

    

4. For the peptide : $\text{Gly-His-Ala-Trp-Ala-Lys}$$\text{Gly-His-Ala-Trp-Ala-Lys}$ , calculate the $pI$$pI$

   • N-terminus = Gly = 9.6

   • C-terminus = Lys = 2.2

   • Side-Chains :

     • 6.0 = amine
     • 10.0 = amine

   • 2.2 , 6.0 , 9.6 , 10.0

   • @ pH = 9.7 :

     • 9.7 < 2.2 ? = False = Deprotonated Carboxylic Acid = -1
     • 9.7 < 6.0 ? = False = Deprotonated Amine = 0
     • 9.7 < 9.6 ? = False = Deprotonated Amine = 0
     • 9.7 < 10.0 ? = True = Protonated Amine = +1

   • We know its neutral at pH 9.7

   $$pI=\frac{9.6+10.0}{2}=9.8$$

    

5. What is the overall charge on this peptide : $\text{Gly-His-Ala-Trp-Ala-Lys}$$\text{Gly-His-Ala-Trp-Ala-Lys}$ , at pH of 4 , 8 , and 10 ?

   • @ pH = 4 :

     • 4 < 2.2 = False = Deprotonated Carboxylic Acid = -1
     • 4 < 6.0 = True = Protonated Amine = +1
     • 4 < 9.6 = True = Protonated Amine = +1
     • 4 < 10.0 = True = Protonated Amine = +1
     • Net-Charge = $-1+1+1+1=2$$-1 + 1 + 1 + 1 = 2$

   • @ pH = 8 :

     • 8 < 2.2 = False = Deprotonated Carboxylic Acid = -1
     • 8 < 6.0 = False = Deprotonated Amine = 0
     • 8 < 9.6 = True = Protonated Amine = +1
     • 8 < 10.0 = True = Protonated Amine = +1
     • Net-Charge = $-1+0+1+1=1$$-1 + 0 + 1 + 1 = 1$

   • @ pH = 10 :

     • 10 < 2.2 = False = Deprotonated Carboxylic Acid = -1
     • 10 < 6.0 = False = Deprotonated Amine = 0
     • 10 < 9.6 = False = Deprotonated Amine = 0
     • 10 = 10.0 = Half-Deprotonated Amine and Half-Protonated Amine = $\frac{(+1)}{2}=+0.5$$\frac{\left(\ +1 \ \right)}{2} = +0.5$
     • Net-Charge = $-1+0+0+0.5=-0.5$$-1 + 0 + 0 + 0.5 = -0.5$

    

6. In three to four sentences explain the importance of the hydrophobic effect on protein folding and structure

   • The hydrophobic effect kick starts protein folding

   • There are enumerable conformations a protein could fold into

   • Immediately , the proteins non-polar side chains seek to minimize their interactions with water.

     • So they fold inward in response to their polar cytosolic environment

    

7. In three to four sentences explain how oxygen affects ( at the molecular level ) the process of hemoglobin transitioning from the $T$$T$ to the $R$$R$ state

   • Hemoglobin and oxygen binding has "cooperativity"

   • After the first oxygen binds to the T-state , it opens up the pore of hemoglobin , causing it to transition into the R-state

     • Because the first oxygen has already opened up the pore slightly , the second oxygen has an easier time accessing its binding site.
     • The second oxygen binds and further opens up the pore for the 3rd oxygen.
     • This continues , and the 4th oxygen has the easiest time binding because there is no steric hinderance

   • Each time oxygen binds , it further promotes stabilization of the R-state , and makes it increasingly likely that more oxygen will bind

    

8. What is the sequence of the polypeptide ?

   Sequencing of an unknown polypeptide has yielded the following information :

   1. Vl , Ile , Trp , and Phe are in a 2:1:1:1 kolar ratio
   2. Treatment with 1-flouro-2,4-dinitrobenzene ( FDNB ) yielded complete hydrolysis and 2,4-dinitrophenyl tryptophan , and no free tryptophan
   3. Digestion with chymotrypsin yielded free tryptophan , a dipeptide containing only Val , and a dipeptide of Ile and Phe

   • Try || IIe-Phe || Val-Val
   • Sequence = $\text{Trp-Ile-Phe-Val-Val}$$\text{Trp-Ile-Phe-Val-Val}$

    

9. A child in the midst of a hissy fit takes several rapid deep breaths ( hyperventilates ) right before holding his breath to the point of passing out to prove his point. Did the hyperventilating increase $O_2$$O_2$ content delivered to tissue in his body? Why or why not ?

   • They child expelled a lot of $C{O}_2$$CO_2$ ( an acid )
   • This caused a shift in the oxygen saturation / heme binding curve
   • Basic environments stabilizes the $R$$R$-state
   • NO , it causes less oxygen to be released to tissues

    

10. Assume you need to separate the $A$$A$ and $B$$B$ chains of insulin ( see image below ) after the disulfide bonds have been reduced. Size exclusion isn't an option due to the resolution of the method not being high enough for this particular separation. What remaining method would you use to separate the two chains and why? Assume it has been maintained at physiological conditions.

    

    • You could use affiinity chromatography if you could make a receptor that bound specifically for either the alpha or beta chain only
    • You can't use polarity , they both seem to have an equal distribution of polar and non-polar amino acids
    • Bot most likely they will have different net charges , so you could separate them based on Ion-Exclusion Chromatography

     

11. A protein bind a ligand with association rate of $8.9\ast {10}^3{M}^{-1}\cdot s^{-1}$$8.9 * 10^{3}\ M^{-1} \cdot s^{-1}$ and an overall dissociation constant of $10nM$$10\ nM$. What is $k_d$$k_d$ ?