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- The extinction coefficient or absorptivity (ɛ) of protein A at 340 nm is 6440 M-1 cm-1, whereas protein B does not absorb at 340 nm. What absorbance will be observed when light at 340 nm passes through a 5 mm cuvette containing 10 µM of protein A and 10 µM of protein B? Beer-Lambert-law; A = ɛ x C x1; A = absorbance, C= concentration, 1= pathlength).H8. Protein A interacts with biomolecule B and forms a complex AB, with a dissociation constant KD = 1 µM at 25 °C, (for dissociation, AB ⇋ A + B, KD=[A][B]/[AB]). The interface contains a phenylalanine residue. A biochemist mutated the phenylalanine of protein A to a tyrosine (A’), which introduced a hydrogen bond between the hydroxyl group of the tyrosine with B without affecting any other interactions. The formation of the hydrogen bond releases a heat at 11.4 kJ/mol. What is the KD of the complex of the mutant A’ with B? Hint: Gas constant R = 8.3145 J/K/mol, Euler’s number e = 2.7183, ln(A) - ln(B) = ln(A/B).12 Determine the type of biochemical reaction that occurs in the reaction shown below. O O O O O redox isomerization group C-C bond formation/cleavage transfer hydrolysis/condensation Previous Ex-xx-xx OH (ATP) OH OH OH OH + AMP OH OPO₂² OH OH +AMP1 OH (ADP)
- 2. . questions below. Assume the following pKa values: N-terminal –NH3®, 7.0; all -COOH groups, 4.0; Arg, 12.5; Cys, 8.4; His, 6.0; Lys, 10.0; Tyr, 10.0. Calculate the overall charge (pH 7) on the following three polypeptides and answer the (A) Ser-Tyr-Ser-Met-Glu–His–Phe–Arg-Trp-Gly-Lys-Pro–Val-Gly-Lys-Lys-Arg-Arg-Pro-Val-Lys–Val-Tyr-Pro-Asp-Ala -Gly- Glu--Asp-Gln- Ser-Ala-Glu-Ala-Phe-Pro-Leu-Arg-Glu-Phe (B) Ser-Tyr-Ser-Met-Glu–His–Phe-Arg-Trp-Gly-Ala-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu–Val-Tyr-Pro-Asp- Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu–Phe-Cys-Ser-Tyr-Ser-Met–Glu–His-Phe-Asp-Trp-Gly- Asp-Pro-Val-Gly-Pro-Asp-Ala-Gly-Asp-Gln-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu-Val-Tyr-Pro-Asp-Ala (C) Gly-Ser-Val-Arg-Asp-Pro-Val-Lys-Glu-Val-Tyr-Pro-Asp- Lys-Ala-Gly-Arg-Glu-Ser-Arg-Ala (g) Which of the three peptides would migrate the furthest away from the anode in isoelectric fo- cusing (h) Which of the peptides would migrate the closest to the cathode in isoelectric focusing?…6-One method to determine proximity of two locations in a protein molecule or protein complex is FRET. This technique uses fluorescence transfer between two fluorophores where the ability of the transfer to occur is correlated to distance between the two residues. A) On the following graph using colored pencils, pens, or markers draw the emission spectra from the donor and the absorbance spectra of the acceptor (this should be a general schematic, does not need to be a FRET pair). Include a key for your graph B) As a rescarch student in my group you want to study the proximity of two residues in clamp region of a protein. Draw a schematic of your protein and label the two locations of your fluorophores. In 2-4 sentences (or figures) explain how your FRET fluorophores would give determine if these two residues were close to one another.AH = - 1.2 kcal/mol; Кр 3 16 nM OH 1 HO AH = - 6.0 kcal/mol; Kp = 76 nM OH HOll AH = - 5.5 kcal/mol; Kp = 0.5 nM OH 3 (iv) Consider the molecule 1 being derivatized to yield molecule 2. How would you expect AG for the binding process (to their target) of 2 to compare to 1? Justify your answer.
- 2. | Calculate the overall charge (pH 7) on the following three polypeptides and answer the questions below. Assume the following pKa values: N-terminal -NH3®, 7.0; all -COOH groups, 4.0; Arg, 12.5; Cys, 8.4; His, 6.0; Lys, 10.0; Tyr, 10.0. (A) Ser-Tyr-Ser-Met-Glu-His-Phe-Arg-Trp-Gly-Lys-Pro-Val-Gly-Lys-Lys-Arg-Arg-Pro-Val-Lys-Val-Tyr-Pro-Asp-Ala -Gly- Glu-Asp-Gln- Ser-Ala-Glu-Ala-Phe-Pro-Leu-Arg-Glu-Phe (B) Ser-Tyr-Ser-Met-Glu–His-Phe-Arg–Trp–Gly-Ala-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu–Val-Tyr-Pro-Asp- Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe-Cys-Ser-Tyr-Ser-Met-Glu-His-Phe-Asp-Trp-Gly- Asp-Pro-Val-Gly-Pro-Asp-Ala-Gly-Asp-Gln-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu-Val-Tyr-Pro-Asp-Ala | (C) Gly-Ser-Val-Arg-Asp-Pro-Val-Lys-Glu–Val-Tyr-Pro-Asp- Lys-Ala-Gly-Arg-Glu-Ser-Arg-Ala (a) Which of the three peptides would elute first from a gel filtration column? (b) Which of the three peptides would migrate the fastest on SDS-PAGE (c) Which of the three peptides could be…2. | Calculate the overall charge (pH 7) on the following three polypeptides and answer the questions below. Assume the following pKa values: N-terminal -NH3®, 7.0; all -COOH groups, 4.0; Arg, 12.5; Cys, 8.4; His, 6.0; Lys, 10.0; Tyr, 10.0. (A) Ser-Tyr-Ser-Met-Glu–His–Phe–Arg–Trp-Gly-Lys-Pro–Val-Gly-Lys-Lys-Arg-Arg-Pro–Val-Lys-Val-Tyr-Pro-Asp-Ala -Gly- Glu--Asp-Gln- Ser-Ala-Glu-Ala-Phe-Pro-Leu-Arg-Glu-Phe (B) Ser-Tyr-Ser-Met-Glu-His-Phe-Arg-Trp-Gly-Ala-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu-Val-Tyr-Pro-Asp- Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe-Cys-Ser-Tyr-Ser-Met-Glu-His-Phe-Asp-Trp-Gly- | Asp-Pro-Val-Gly-Pro-Asp-Ala-Gly-Asp-Gln-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu-Val-Tyr-Pro-Asp-Ala (C) Gly-Ser-Val-Arg-Asp-Pro-Val-Lys-Glu–Val-Tyr-Pro-Asp- Lys-Ala-Gly-Arg-Glu-Ser-Arg-Ala 11. - - I C'IL- - 1! (e) Which of the above peptides would elute last from a gel filtration column? (f) Which of the three peptides would migrate the slowest on SDS-PAGE (g) Which of the three…E. PROTEIN PRIMARY STRUCTURE ELUCIDATION. 1. Determine the primary structure of the protein described below. Write the final sequence using the corresponding three-letter code for each amino acid. Example: M-F-Y-R should be written as Met-Phe-Tyr-Arg Treatment with cyanogen bromide and sequencing yields the following peptide fragments: o D-M o R-A-Y-G-N o L-F-M Chymotrypsin digestion and sequencing yields the following peptide fragments: o G-N D-M-L-F o M-R-A-Y o o
- 2. | Calculate the overall charge (pH 7) on the following three polypeptides and answer the questions below. Assume the following pKa values: N-terminal –NH3®, 7.0; all -COOH groups, 4.0; Arg, 12.5; Cys, 8.4; His, 6.0; Lys, 10.0; Tyr, 10.0. (A) Ser-Tyr-Ser-Met-Glu-His-Phe-Arg-Trp-Gly-Lys–Pro–Val–Gly–Lys–Lys–Arg-Arg-Pro-Val-Lys–Val-Tyr-Pro-Asp-Ala -Gly- Glu-Asp-GIn– Ser-Ala-Glu-Ala-Phe-Pro-Leu-Arg-Glu-Phe (B) Ser-Tyr-Ser-Met-Glu-His-Phe-Arg-Trp-Gly-Ala-Pro-Val-Gly-Glu-Glu–Cys-Asp-Pro-Val-Glu–Val–Tyr-Pro-Asp- Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe-Cys-Ser-Tyr-Ser-Met-Glu–His-Phe-Asp-Trp-Gly- Asp-Pro-Val-Gly-Pro-Asp-Ala-Gly-Asp-Gln-Pro-Val–Gly–Glu-Glu-Cys-Asp-Pro–Val-Glu–Val–Tyr-Pro-Asp-Ala (C) Gly-Ser-Val-Arg-Asp-Pro-Val-Lys-Glu-Val-Tyr-Pro-Asp- Lys–Ala-Gly-Arg-Glu-Ser-Arg-Ala (d) Which of the three peptides would migrate the closest to the anode in isoelectric focusing? (e) Which of the above peptides would elute last from a gel filtration column? (f) Which of the three…Explain why it has been very difficult to obtain an X-ray crystal structure for the complete nuclear pore complex. Explain in terms of its size, number of proteins as well as its molecular properties (e.g. charge). Suppose you have two genetic variants of a large protein that differ only in that one contains a histidine (side chain pk, = 6.0) when the other has a valine (uncharged side chain). (a) Which would be better for separation: gel electrophoresis or isoelectric focusing? Why? (b) What pH would you choose for the separation?