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- QUESTION NO. 1Targeting a protein to be degraded within proteasomes usually requires ubiquitin. In the function of ubiquitin all of the following are true except: A. ATP is required for activation of ubiquicin. B. a peptide bond forms between the carboxyl terminal of ubiquitin and an ε-amino group of a lysine . C. linkage of a protein to ubiquitin does not always mark it for degradation. D. the N-terminal amino acid is one determinant of selection for degradation. E. ATP is required by the enzyme that transfers the ubiquitin to the protein to be degraded QUESTION NO. 2Much of procollagen formation occurs in the endoplasmic reticulum and Golgi apparatus which requires signal peptide. All of the following statements about targeting a protein for the ER are true except. A. signal peptide usually has a positively charged N-terminus and a stretch of hydrophobic amino acids. B. signal peptide emerging from a free ribosome binds signal recognition…QUESTION 2An isocitrate dehydrogenase assay was performed on the enzyme sample and found to give an absorbance change at 340nm of 0.5 absorbance units perminute. Given that the molar absorption coefficient (E) is 6220 M-1 cm-1 and the pathlength is 1cm, what is the rate of the enzyme catalysed reaction in umol perminute per mL?Question 1: When the CAC is run in reverse by microorganisms that use it to fix carbon, the citrate synthase reaction is different and is catalyzed by an enzyme called ATP-citrate lyase. Write the reaction catalyzed by ATP-citrate lyase, then briefly (in one sentence) explain why the use of different chemistry (different from the ‘normal’ direction) makes sense here.
- QUESTION NO. 1L-Carnitine is synthesized primarily in the liver but also in the kidneys and then transported to other tissues. It is most concentrated in tissues that use fatty acids as their primary fuel, such as skeletal and cardiac muscle. In this regard, L-carnitine plays an important role in energy production by conjugating to fatty acids for transport from the cytosol into the mitochondria. L-carnitine shuttle is an example of A. ion driven active transport B. facilitated diffusion C. simple diffusion D. ATP driven active transportE. symport F. antiportQUESTION NO.2 Statements: (1) Glucose is both a hexose and a aldose. (2) There can never be more than three enantiomers for a molecule. (3) All common disaccharides have beta-one-four linkages. Which statements are true?Question 11. // Hint: Isoelectric focusing separates proteins based on their pI values, and can separate proteins that only differ by a net charge of ±1.±1. Recall that an amino acid residue with a negatively charged R group has a relatively low isoelectric point (pI) where it has zero net charge. Likewise, an amino acid residue with a positively charged R group has a relatively high isoelectric point (pI) where it has zero net charge. Order from Low pH to High pHQuestion: A decapeptide composed of ser, ala, IIe. his, trp, phe was treated with 1-flouro- 2,4-dinitrobenzene. It gave a DNP-his on the N terminal and free trp when treated with carboxypeptidase. Upon partial hydrolysis of the peptide, the following fragments were obtained. a. his-lle-phe-ala c. his-ala-phe e. ser-lle-his b. ala-phe-trp d. phe-ala-ser Give the amino acid sequence of the above decapeptide.
- Question: Fatty acid degradation (breakdown) and synthesis have many similarities but are uniquely different.Explain two ways that these pathways are similar and two ways that they differ. Please give four seperate factors in point form. Two similarities and two differences 4 overall.QUESTION 26 During gluconeogenesis, whereby liver cells convert pyruvate to glucose, Fructose-6-phosphate (F6P) is converted to Glucose-6-phosphate (G6P). If the standard equilibrium concentrations are: [F6P] = 0.52 M and [G6P] = 1.48 M, then Keq’ is ______ and the reaction is ________. Fructose-6-P ó Glucose-6-P > 1; exergonic > 1; endergonic < 1; exergonic < 1; endergonicQuestion:- The enzyme aromatase is found in the cytoplasm of some cells and converts testosterone to estrogen. You decide to test aromatase from a particular cell, and oops, your lab partner admits he drastically increased the pH in all the test tubes. Which of the following is a likely result? a. The enzyme will be denatured and the substrate will not bind to the active site. b. The enzyme will convert testosterone to estrogen at a faster rate. c. The mistake will have no effect on the experiment, because enzymes are not sensitive to pH. d. The free energy will be lowered and the reaction will not proceed spontaneously.
- Question: What is the isoelectric point of Cysteine and Glutamate, Illustrate structures and net charges (Determine the isoelectric point based on the pka) please give clear handwritten answer!Question: A. To explore the consequences of coupling ATP hydrolysis under physiological conditions to a thermodynamically unfavorable biochemical reaction, consider the hypothetical transformation X⟶Y, for which Δ?′°=20.0 kJ/mol. What is the ratio of [Y]/[X][Y]/[X] at equilibrium? B. Suppose XX and YY participate in a sequence of reactions during which ATP is hydrolyzed to ADP and Pi. The overall reaction is X+ATP+H2O⟶Y+ADP+Pi Calculate [Y]/[X] for this reaction at equilibrium. Assume that the temperature is 25.0 °C and the equilibrium concentrations of ATP, ADP, and Pi are 1.00 M each. C. We know that [ATP], [ADP], and [Pi] are not 1.00 M under physiological conditions. Calculate [Y]/[X] for the ATP‑coupled reaction when the values of [ATP], [ADP], and [Pi] are those found in rat myocytes. Metabolite Concentration in rat myocytes (M) ATP 8.05x10-3 ADP 0.93x10-3 Pi 8.05x10-3Question 1Predicting Secondary Structure Which of the following peptides is more likely to take up an -helical structure, and why? (a) LKAENDEAARAMSEA (b) CRAGGFPWDQPGTSN