CHEM 350 Exam IIFridayMarch 22, 20191Name (print): _______________________Signature: _______________________( 1 pt )05101501243___ (mM)_____ (mM/min)52. Below is a Michaelis-Menten plot of an enzyme catalyzed reaction. Label the axes and label Vmaxand Kmon the graph. At a substrate concentration of 40 mM, what will the v0of the reaction be? Write the equation used to solve this problem, then solve it. Finally, draw the new curve that would be observed if a competitive inhibitor reduced activity to ½ its current activity. What would the new Km value be? (16 pts)201. Assuming that life may still be possible without enzymes, what consequences might we expect to observe? What effect do enzymes have on the equilibrium of a reaction? (3 pts)• The time-scale of life and biological reactions would be extremely slow. • The complexity of life would be very low• The conditions for life might be more extreme (not physiological) to speed up rxns• Reaction equilibrium is unchanged. +2+1Any of these 1stthree, or answers based on similar concepts+1v0=Vmax [ S ]Km + [ S ]=(5 mM/min) x (40 mM)10 mM + 40 mMv0=VmaxKmv0[ S ]=4 mM/min200 mM2/min50 mM+1+1+1+1+3+2 +1+2+1+1+1New Km (20 mM)New curve+1New curve should indicate that Vmax will eventually still be reachedNew Km should be double the initial Km20page123456pointsscore_____________________________20191322.51918Total Points 111.524. A modified nucleobase called X (below) was discovered. It is a modified form of adenine. Circle or label the modification on the base. Give it a name that matches the common nomenclature for modified bases. Based on these structures, what other nucleobase would you predict that X can base-pair with? How many H-bonds would be possible? Are there any other properties of the base that could be affected? (6 pts) 5. A novel small noncoding RNA has been identified that functions as an antibiotic. Like other antibiotics, it selectively disrupts proper protein synthesis by the bacterial ribosome. Researchers observe that this small RNA binds to the A site of the ribosome and slows protein production. If concentrations of this mystery RNA are high enough then translation stops completely. What step(s) of protein synthesis by the ribosome are being affected by this antibiotic? Propose how and why it might be working this way. Be specific. (8 pts) E PApolypeptidemRNA3. If you add high concentrations of the molecule shown below to a solution of DNA, would anything happen to the DNA structure? If so, explain. How would you experimentally measure the changes that you are predicting? (5 pts) Yes, the molecule is a hydrophobic planar structure. The DNA bases would prefer to interact with it via pi-stacking and hydrophobic packing rather than stacking with each other. Thus, DNA structure would be destabilized or denatured due to loss of base-stacking.The effect could be measured by absorbance. The Tmshould decrease and absorbance should increase.+0.5+0.5+0.5+1+1+1+0.5+12-thioadenine (or 2-sulfhydryl adenine)It could pair with guanosine (G) to make 3 H-bonds[now has a Watson-Crick pairing face similar to cytidine (C)] Because double-bond character is changed in the ring, absorbance spectra or wavelength (λmax) might change+1Partial credit for names that include the 2 position or the thiol / sulfurPartial credit for Cytidine (C)+1+2+1It probably mimics an aminoacyl-tRNA and competes for binding. This would slow translation or block it at high enough concentrations. Thus, it would block the first step of the elongation microcycle by preventing the next aminoacyl-tRNA from binding to the A site.In this mechanism, it might interact with the ribosomal RNA or mRNA through base-pairing. It may also have a shape that is similar to tRNA.+1.5+1.5+1.5+1.5+1+1Partial credit for answers that express similar concepts1938. Give the full name of the structure shown below. What is the specific class of cofactor for each of these molecules. Circle the parts that are important for their function and use by enzymes. What class of enzyme uses each of these molecules? (6 pts)Enzyme classes: Cofactor type: Full name: 6. Match the nucleic acid structure parameters shown below to the structures shown by drawing arrows or circling. Name the structures. ( 3 pts )Helix WidthBase OrientationBase Spacing/StackingBase RotationHelix Configurationanti32.6°2.6 Åright-handednarrow & deep20 ÅPyr - antiPur - syn3.7 Å-60°/ 2 nucleotidesleft-handedflat16 - 17 ÅMajor Groove7. (A) What are the two primary driving forces that establish double-helical structure of polynucleotides. (B) What percentage does each contribute to stability and what property of thermodynamics is involved? (4 pts)A.B.+1Z-DNAB-DNA+1+0.5+0.5Hydrophobic stacking of bases and hydrogen bonding of base-pairs+1.5Base stacking = ~80%, hydrogen bonding = ~20%+0.5+1.5+0.5Values do not have to be exactly %, but close approximation is fine.Partial credit for saying “entropy” cosubstratecosubstrate½ credit for “coenyzyme”S-adenosyl methionine Flavin mononucleotide½ credit for “SAM”½ credit for “FMN”transferase oxidoreductase+1+1+1+1+0.5+0.5+0.5+0.5134substrate9. The substrate and inhibitor of an enzyme reaction are shown below. (A) What type of inhibitor is it? Explain how you know. On the plots shown below, draw a new line indicating the effect of the inhibitor. Label the axes, the intercepts, and the new line as (+)I. Be sure to include relevant enzyme kinetic parameters and what changed. (12 pts) (-)I(-)IA.B.C.10. Complete the expanded version of a chemical equation shown below for an enzyme catalyzed reaction. Be sure to label all constants. Circle the steps and constants that are usually omitted for simplicity. Indicate with a box which step is usually considered reversible. (5.5 pts)11. (A) For the enzyme catalyzed chemical equation you wrote above, what does the k2rate constant represent in terms of enzyme property? (B) What does the ratio of k-1/k1 represent and what is its common name? (C) If the Vmaxof the reaction is 200 mM/sec and the total enzyme concentration is 50 mM, what is the value of the rate constant named in part B? (5 pts)inhibitorVmaxKmnew Vmax1/Vmaxnew1/Vmax-1/Km(+)I(+)IVmaxgoes down, Kmis unaffectedv0[S]1/v01/[S]+0.5+1+1+0.5+1Non-competitive inhibitionbecause the structures do not have any significant similaritiesin shape or functional groups+2+1+0.5+1+0.5+0.5+1+0.5+1k-1k1E + S _____ _____ _________ESEP E +