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- Two proteins bind to the same ligand with the following Kd's. Protein 1: 10 μΜProtein 2: 100 nM Which protein would be bound to more ligand at a ligand concentration of 10 μM? (Assume a limiting concentration of each protein)Two proteins, A and B, bind to the same ligand, L, with the binding curvesshown below. What is the dissociation constant, Kd , for each protein? Which protein (A or B) has a greater affinity for ligand L?How many copies of a protein need to be present in a cell in order for it to be visible as a band on an SDS gel? Assume that you can load 100 µg of cell extract onto a gel and that you can detect 10 ng in a single band by sil ver staining the gel. The concentration of protein in cells is about 200 mg/mL, and a typical mammalian cell has a volume of about 1000 μm³ and a typical bacterium a vol ume of about 1 µm³. Given these parameters, calculate the number of copies of a 120-kd protein that would need to be present in a mammalian cell and in a bacterium in order to give a detectable band on a gel. You might try an order-of-magnitude guess before you make the calcula tions.
- What are the four components for the translocation of protein across endoplasmic reticulum? and briefly explain ORF?Which factor has NOT been shown to play a role in determining the specificity of protein kinases? a. protein tertiary structure b. protein quaternary structure c. primary sequence at phosphorylation site d. disulfide bonds near the phosphorylation site e. residues near the phosphorylation siteFind a method that uses some form of HPLC for the analysis of proteins. What was the stationary phase used? How does this kind of stationary phase separate the proteins? What kind of mobile phase was used? Was the method isocratic or was a gradient used? How were the proteins detected?
- Which of the following statements are true? Electrostatic interactions are the dominant forces in protein molecular recognition. When two proteins form a complex there is an unfavorable loss of rotational-translational entropy. Protein-protein interfaces are most often dry. The exclusion of water results in an unfavorable loss in rotational-translational entropy. The free energy change associated with the formation of an enzyme-substrate complex almost always results in an unfavorable reduction in conformational entropy of the proteins. Burial of an uncompensated positive charge inside proteins is usually unfavorable. So-called van der Waals’ interactions are essentially electrostatic in origin. Steric complementarity of the two partners forming a complex is essential to achieve optimal free energy of binding. Structural models of proteins obtained from low temperature crystallography are excellent descriptions of all biochemically relevant aspects of their function.1. Channel proteins used for the bulk passage of water molecules across the cell membrane?2. The tonicity of the extracellular fluid that maintains the turgid appearance of plant cells?3. The tonicity of the extracellular environment causing plasmolysis of plant cells?4. Plant cells are said to be flaccid in this type of cellular environment or solution?Using the the enzyme acid hydrolase in the lysosome: What is the final destination in which the protein will function? Which features will the protein receive during its manufacture? What is the primary structure (general)? Where is the primary structure made? Where are the secondary and tertiary structures made? Will the protein travel through any organelles during its manufacture? Which ones? What would be the overall result if some part of the manufacture process went wrong, such that the protein ended up as nonfunctional?
- An intermediate folding stage seen in protein denaturation or renaturation is called : a) domain b) motif c) subunit d) molten globule Proteins which do not renature spontaneously when denaturation conditions are removed may need the assistance of: a) a prosthetic group b) a higher salt concentration c) a lower temperature d) a chaperone protein The information needed for correct protein folding is encoded in: a) the surrounding molecules b) the protein’s amino acid sequence c) the pH of the aqueous medium d) the electrolyte composition of the aqueous solutionHurray! The bacteria did their job and expressed your protein. Now, you need to isolate biochemisfunase from the other bacterial proteins. To do this, you first need to break open the bacterial cells, a process called cell lysis. The lysis buffer you are going to use requires 225 mM potassium chloride (KCl), which is available as a crystalline solid. The lysis buffer also requires 22.5% (v/v) glycerol, which is available as a 100% stock solution. Describe how you would make 250 ml of lysis buffer and show all calculations.Protein A binds to Ligand X with a Kd of 1 μM. Protein B binds to Ligand Y with a Kd of 100 nM. Which of the following statements is TRUE? A. Protein A binds Ligand X ten times tighter than Protein B binds Ligand Y B. At a sufficiently high concentration of ligand, the binding becomes irreversible C. When the concentration of Ligand X is 1 μM, 50% of Protein A is bound to ligand D. Protein B binds to Ligand Y with a Kd of 1 × 10-8 mol/L. E. When both binding reactions are at equilibrium, Protein A has more Ligand X bound than Protein B has ligand Y bound The answer is C, could you show how to get the answer? Thanks