Describe the differences between the types of stereoisomers found in biomolecules

Calculate standard free energy change of a reaction

Enthalpy ( H ) is the heat exchanged between the system and the surroundings at constant pressure

Problem :

• Given:

  • $\mathrm{Glucose}+\mathrm{P}{\phantom{A}}_{\mathrm{i}}⟶\mathrm{glucose}6\text{-}\mathrm{phosphate}$$\ce{Glucose + P_i -> glucose 6-phosphate}$ ; ΔG°= 13.8 kJ/mol
  • $\mathrm{ATP}⟶\mathrm{ADP}+\mathrm{P}{\phantom{A}}_{\mathrm{i}}$$\ce{ATP -> ADP + P_i}$ ; ΔG°= -30.5 kJ/mol

• What is ΔG° for :

  • $\mathrm{Glucose}+\mathrm{ATP}⟶\mathrm{glucose}6\text{-}\mathrm{phosphate}+\mathrm{ADP}$$\ce{Glucose + ATP -> glucose 6-phosphate + ADP}$

Describe the importance of hydrogen bonds in water and biomolecules such as DNA and proteins

• important for stabilization , protein folding , clathrate cage formation

Describe the factors that determine hydrogen bond strength

• Geometry of the bond

  • linear = stronger
  • bent = weaker

Describe the function of a buffer in biochemical systems

• resists changes to pH

  • helps keep things in equilibrium
  • provides protection in acidic environments

Define and calculate pH and $p{K}_a$$pK_a$

• pH = available hydronium ion in solution

  • $pH=-log([H^{+}])$$pH = -log\left(\ [H^+] \ \right)$

• small $p{K}_a$$pK_a$ = strong acid

• large $K_a$$K_a$ = strong acid

Problem :

• 1 liter of a phosphate buffer ( $p{K}_a=6.86$$pK_a = 6.86$ ) is made using $0.042M$$0.042\ M$ of $\mathrm{NaH}{\phantom{A}}_2\mathrm{PO}{\phantom{A}}_4$$\ce{NaH2PO4}$ and $0.058M$$0.058\ M$ $\mathrm{Na}{\phantom{A}}_2\mathrm{HPO}{\phantom{A}}_4$$\ce{Na_2HPO4}$
• How much would the pH of the buffer change if $1.0mL$$1.0\ mL$ of $10.0M$$10.0\ M$ $NaOH$$NaOH$ is added?

Problem :

• Does a strong acid have a greater or lesser tendency to lose its proton than a week acid?

  • greater

• Does the strong acid have a higher or lower $K_a$$K_a$ than the weak acid?

  • higher

• Does the strong acid have a higher or lower $p{K}_a$$pK_a$ than the weak acid?

  • lower

Describe the general structure of amino acids

• Peptide bond = condensation reaction

• Planar = because of thermodynamics ( steric hindrance )

• Rigid = double bonds ( pi bonding )

  • Because of Resonance structure , 40% of it is in a double bond

• Because peptides are planar and rigid , it gives us our “torsional angles” aka $\psi$$\psi$ and $\varphi$$\phi$

• $\varphi$$\phi$ = bond between the Nitrogen and the alpha carbon

• $\psi$$\psi$ = bond between alpha carbon and the carbonyl carbon

• Both can rotate freely from -180 to +180

  • In reality they don’t get this full range because of steric hindrance

$pI$$pI$ = When the net charge = 0

Calculate pI for an amino acids and small peptides

• 1. organize all $p{K}_a$$pK_a$ values

     • 1 = $N$$N$-terminal amino group
     • 2 = C-terminal carboxylic group
     • 3 ... = Any ionizable side chains on any of the amino acids

  2. sort all of the $p{K}_a$$pK_a$ values from lowest to highest

  3. make an educated guess for where it "might" be neutral

     • default , start at pH = 7
     • try , just above or just below each of your $p{K}_a$$pK_a$ values

  4. write out and solve each $pH<p{K}_a$$pH < pK_a$ $?$$?$ test

     • Protonated Amine = +1
     • Deprotonated Amine = 0
     • Protonated Carboxylic Acid = 0
     • Deprotonated Carboxylic Acid = -1

     xxxxxxxxxx
     if ph < pKa:
         protonated
     if pH > pKa:
         deprotonated
     

  5. adjust your "test" pH value up or down depending on the last net charge you calculated

  6. once you find where the peptide is neutral ,

     • take the first $p{K}_a$$pK_a$ value that is below the neutral pH

     • take the first $p{K}_a$$pK_a$ value that is above the neutral pH

     • take their average

       $$pI=\frac{p{K}_{a_{\text{below neutral}}}+p{K}_{a_{\text{above neutral}}}}{2}$$

Describe the condensation reaction that forms peptides

• hydroxyl on C-terminus carboxylic acid combines with hydrogen on N-terminus amine to release a water molecule

Polar , uncharged R side chains can form hydrogen bonds

• Serine , Threonine , Asparagine , Glutamine

Cysteine and Methionine forms disulfide bonds

Problem :

• Calculate the pI of the peptide : Glu-His-Trp-Ser-Gly-Leu-Arg-Pro-Gly

• What is the charge of the molecule at pH 3 , 8 , and 11?

  • pH = 3 :

    • 2.0

  • pH = 8 :

    • 0.0

  • pH = 11 :

    • -1

Describe how proteins are separated and quantified by chromatography

Evaluate the best type of protein separation for a given research scenario

Problem :

• A researcher has produced a protein of interest using a peptide synthesizer. The product material is a mixture of the desired protein and smaller fragments ( incomplete synthesis ). For the experiment the researcher needs a large quantity of the protein with the smaller fragments removed.

  • size-exclusion chromatography with beads with small pores

Identify the importance of determining amino acid sequences of proteins

• drug design , general understanding

Identify the different methods of determining the primary structure of proteins

Determine the amino acid sequence of a small polypeptide

Problem :

• Sequencing of an unknown polypeptide has yielded the following information. What is the sequence of the polypeptide?

  1. Gly, Leu, Phe, and Tyr are in a 2:1:1:1 molar ratio

  2. Treatment with 1-fluoro-2,4-dinitrobenzene ( FDNB ) yielded complete hydrolysis and 2,4-dinitrophenyl tyrosine and no free tyrosine

  3. Digestion with chymotrypsin yielded free tyrosine and leucine , and a tripeptide of Phe and Gly

     • Chymotrypsin = Phe , Trp , Tyr ( C )

  • First bullet point = tells us molar ratio only

    • Gly , Gly , Leu , Phe , Tyr

  • 2nd bullet point = tells us peptide was labeled on N-terminus , shot it apart with an acid , tells us that the N-terminus was tyrosine

  • 3rd bullet point = the only way chymotrypsin could be applied and it cleaves a peptide length of 3 , is for phenylalanine to be on the end

    • Gly-Gly-Phe

  • We know leucine and tyrosine are “free”

  • Write down all combinations for chymotrypsin cleavage

  • Via trial and error , we know where chymotrypsin cleaved

    • Tyr || Gly-Gly-Phe || Leu

  • Structure must be :

    1	2	3	4	5
    Tyr	Gly	Gly	Phe	Leu

Diagram secondary structure features of proteins

Predict the type of secondary structure of a protein segment

Problem :

• Which of the following peptides would be more likely to form an $\alpha$$\alpha$-helix?

  • LKAENDEAARAMSEA

    • there are no glycine or prolines here

  • CRAGGFPWDQPGTSN

• How many turns would the helix have?

  $$\frac{15}{3.6}=4.167$$

Describe the structural differences between fiberous and globular proteins

• Fiberous = scaffolding , and give structure

  • Collagen , keratin
  • Make up the majority of the “volume”
  • woven together polymers of either helixes or beta sheets

• Globular = cytosolic protein

  • large proteins with domains / motifs
  • Enzymes
  • We have more globular proteins , majority of the “population”

Evaluate the structural features of a protein

• different domains / motifs

  • combinations / orientations of them

Problem :

• characterize 3-D quaternary structure

Describe how factors such as heat contribute to protein denaturation

• breaks hydrogen bonds , eventually breaks peptide bonds

Diagram how chaperones / chaperonins contribute to protein folding

• they can provide a scaffold for where different functional groups can undergo chemical changes

Problem :

• Why does the Gro EL/ES system only function in one direction ?

  • the inside cavity provides a hydrophobic environment

  • the ATP hydrolysis used to power , creates an energy barrier almost impossible to reverse

    • Once you hydrolyze ATP , the Delta G to reattach phosphate group is almost impossible

Describe oxygen binding to myoglobin using both graphical and mathematical approaches

Describe cooperative ligand binding

• shifting the partial pressure to the right is not enough

  • We need some conformational change

• It starts in the low affinity state

  • Once it reaches a certain partial pressure of oxygen , it changes from low affinity state to high affinity

    • Becomes saturated more quickly now ( cooperativity ) ( positive homotrophic allosteric effector )

• When first oxygen binds , it moves heme to be in plane with the iron

  • Normally , there is an amino acid that causes the Tense / Strain state

  • That causes the proximal histidine to move with the iron

    • The proximal histidine is attached to the F-helix

      • This moves the F-helix

        • causes 15-degree rotation at interface between alpha and beta subunits

• Deoxystate = 4 oxygens are closed off

  • Once 15 degree rotation happens , it moves to high affinity state

    • Oxygen binding sites are more open and accessible

Proximal Histidine = works with $F{e}^{2+}$$Fe^{2+}$ to stabilize oxygen binding

Distal Histidine = further stabilizes oxygen binding

Problem :

• Could myoglobin transport $O_2$$O_2$ ?

  • No , myoglobin has to high of a binding affinity to oxygen for it to work as a transport protein

Problem :

• A protein binds a ligand with associtation rate of $8.9\ast {10}^3{M}^{-1}{s}^{-1}$$8.9 * 10^{3}\ M^{-1}s^{-1}$ and an overall dissociation constant of $10nM$$10\ nM$. What is $kd$$kd$ ?

  $$K_d=\frac{k_d}{K_a}$$
  $$K_d=k_d\ast K_a$$
  $$K_d=8.9\ast {10}^3\ast (10\ast {10}^{-9})=8.9\ast {10}^{-5}{s}^{-1}$$

Describe the following and how they affect $O_2$$O_2$ transport :

• Bohr effect

• 2,3-BPG

  • In the R-state , the positive charges can no longer interact with 2,3-BPG , causing 2,3-BPG to leave

  • 2,3-BPG = negative allosteric effector , causes right shift , increases $\mathrm{\Delta }Y-O_2$$\Delta Y-O_2$

  • Adult hemoglobin has positively charged histidine

    • Binds 2,3-BPG
  • Oxygen = positive homotrophic effector
  • PH = negative heterotrophic
  • 2,3-BPG = negative heterotrophic

• Fetal hemoglobin

  • Fetal hemoglobin has the histidine swapped out for serine

    • Does not bind 2,3-BPG

      • Left Shift

• Sickle cell anemia

  • When hemoglobin is in R state , the glutamate is not exposed

Conformational change from the T state to the R state involves breaking ion pairs between the ${\alpha }_1-{\beta }_2$$\alpha_1-\beta_2$ interface

Acidic environment stabilizes the $T$$T$ state

• leads to the release of $O_2$$O_2$ ( in the tissues )

Basic environment stabilizes the $R$$R$ state

$CO$$CO$ has a filled lone electron pair that can be donated to vacant d-orbitals on the $F{e}^{2+}$$Fe^{2+}$

Problem :

• Hemoglobin S homozygotes who are severely anemic often have elevated levels of BPG in their red blood cells. Is this good or bad?

  • Bad , because even though the 2,3-BPG promotes more oxygen release in the tissues ,

    • it promotes an even bigger problem of polymerization of the erythrocytes