BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
9th Edition
ISBN: 9781319425784
Author: BERG
Publisher: Macmillan Higher Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 16, Problem 54P
Interpretation Introduction
Interpretation:
Concentration of glucose when vο is equal to 90% of Vmax.
Concept introduction:
Glycolysis is the process of breakdown of one molecule of glucose into 2 molecules of pyruvate. 2 ATP and 2 NADH molecules are yielded in this reaction. Moreover, its is divided into two phases, preparatory phase and payoff phase. In the preparatory phase, 2 ATP molecules are used, and in payoff phase, 4 ATP molecules are yielded.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
PRESSURE
OF A
WHAT IS THE asmoTIC
SOLUTION
OF BOVINE insuLIN (MOLAR MASS,
5700g mol - 1 ) aT
18 °C
IF 100.0 ML
OF
THE SOLUTION
conTains
0.103 a OF
THE INSULIN?
Calculation about delta standard G, delta H, detla S. Question attached as photo below. And my answer attempted. Need my answer verified and corrected if neccesary. Please let me know where I got wrong and what key ideas I had miss. Thanks.
please show your work so i can correct figure out where im going wrong. thank you
Chapter 16 Solutions
BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
Ch. 16 - Prob. 1PCh. 16 - Prob. 2PCh. 16 - Prob. 3PCh. 16 - Prob. 4PCh. 16 - Prob. 5PCh. 16 - Prob. 6PCh. 16 - Prob. 7PCh. 16 - Prob. 8PCh. 16 - Prob. 9PCh. 16 - Prob. 10P
Ch. 16 - Prob. 11PCh. 16 - Prob. 12PCh. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Prob. 15PCh. 16 - Prob. 16PCh. 16 - Prob. 17PCh. 16 - Prob. 18PCh. 16 - Prob. 19PCh. 16 - Prob. 20PCh. 16 - Prob. 21PCh. 16 - Prob. 22PCh. 16 - Prob. 23PCh. 16 - Prob. 24PCh. 16 - Prob. 25PCh. 16 - Prob. 26PCh. 16 - Prob. 27PCh. 16 - Prob. 28PCh. 16 - Prob. 29PCh. 16 - Prob. 30PCh. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Prob. 33PCh. 16 - Prob. 34PCh. 16 - Prob. 35PCh. 16 - Prob. 36PCh. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - Prob. 43PCh. 16 - Prob. 44PCh. 16 - Prob. 45PCh. 16 - Prob. 46PCh. 16 - Prob. 47PCh. 16 - Prob. 48PCh. 16 - Prob. 49PCh. 16 - Prob. 50PCh. 16 - Prob. 51PCh. 16 - Prob. 52PCh. 16 - Prob. 53PCh. 16 - Prob. 54PCh. 16 - Prob. 55PCh. 16 - Prob. 56PCh. 16 - Prob. 57PCh. 16 - Prob. 58PCh. 16 - Prob. 59PCh. 16 - Prob. 60PCh. 16 - Prob. 61P
Knowledge Booster
Similar questions
- Effects of Changing Metabolite Concentrations on Glycolysis In an erythrocyte undergoing glycolysis what would be the effect of a sudden increase in the concentration of a. AΤP? b. AMP? c. fructose-1.6-bisphosphate? d. fructose-2, 6-bisphosphate? e. citrate? f. glucose-6-phospthate?arrow_forwardEnergetics of the Hexokinase Reaction The standard-state free energy change. Gfor the hexokinase reaction, is — 1 6.7 kJ/mol. Use the values in Table I to calculate the value of Gfor this reaction in the erythrocyte at 37°C.arrow_forwardptidase varying given below. of ATP ge 8 of 18 b) Match each item of Column A with the most appropriate item of column B. Rewrite the correct pairs by writing the number and corresponding alphabet in your answer sheet Column A 1. Syncytial 2. Velamen 3. NAD and FAD 4. GABA 5. Acetylcholine 6. Osteoclasts 7. Fibrous cartilage 8. Light reaction 9. ETC 10. Kreb cycle Column B a) Excitatory neurotransmitter b) Inhibitory neurotransmitter c) Bone phagocytes d) Resist compression e) Electron acceptor f) Multinucleated animal cell g) Substrate level photophosphorylation h) Photophosphorylation i) Oxidative photophosphorylation j) Multinucleated plant cell. k) Proton tunnel 1) Ion channel m) Modified epidermis to absorb water. c) Fill in the blanks with appropriate word/words. i. The source of electron in photosynthesis [5] Sun glucose ii. The thick acidic mixture of gastric juice and semi-digested food that passes from stomach to small intestine is BCS/MIDTERM/BIOLOGY/2024 Page 9 of 18arrow_forward
- Do not give handwriting solutionarrow_forward13. 0.9% (m/v) NaCI solution and 5% (m/v) glucose solution are both isotonic to red blood cells. SHOW your work and watch sig figs & units. c. convert the concentration from M to % (m/v) for a 0.342 M NaC solution. (HINT: convert to g/ml and then multiple by 100%)arrow_forwardnot true about the Michaelis-Menten equation? The equation that gives the rate, v, of an the substrate concentration [S] is the Michaelis-Menten equation = Vmax[S]/(Km + [S]), where V, enzyme-catalyzed reaction for all values of max and Km are constants. Which of the following is a) for [S] << Km, V = Vmax applies to most enzymes, but allosteric enzymes have different kinetics when [S] = Km, then v = Vmax/2 gives the rate when the enzyme concentration, temperature, pH, and ionic strength are constant for very high values of [S], v approaches Vmax e) Which is correct about the constant Km in the Michaelis-Menten equation? also called the catalytic constant or turnover number equal to the number of product molecules produced per unit time when the enzyme is saturated with substrate it is the constant in the first order rate equation v = k[A] it is the constant in the second order rate equation v = equal to the substrate concentration at which the velocity or rate of a reaction is ½ the…arrow_forward
- Distinguishing the Mechanisms of Class I and Class I Aldolases Fructose bisphosphate aldolase in animal muscle is a class 1 aldolase, which forms a Schiff base intermediate between substrate (for example. fructose-1, 6-bisphosphate or dihydroxyacetone phosphate) and a lysine at the active site (see Figure I8.12). The chemical evidence for this intermediate conies from studies with aldolase and the reducing agent sodium borohydride, NaBH4. Incubation of the enzyme with dihydroxyacetone phosphate and NaBH4 inactivates the enzyme. Interestingly, no inactivation is observed if NabH4 is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.arrow_forwardThis was glossed over super quick in class. Extra help would be greatarrow_forwardCarbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydration of CO₂. H₂O + CO₂ 2 H₂CO3 This is an important process in the transport of CO₂ from the tissues to the lungs. If 5 µg of pure carbonic anhydrase catalyzes the hydration of 0.35 g of CO₂ in 1 min at 37 °C at Vmax, what is the turnover number (kcat) of carbonic anhydrase? Kcat 2.0466 ×107 Incorrect min-1arrow_forward
- 6-25 substrate-band enzyme concentrations. The the turnover number is equal to umax- b) V=Umax •57(Km+S) anstont For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, V (as a percentage of Vmax), observed at the following values? a) [S] = KM C) d) e) [S] = 0.5KM [S] = = 0.1KM [S] = 2KM [S] = 10KM w reactores -maximumrate of reaction boteles conc. Would you expect the structure of a competitive inhibitor of a given enzyme to be similar to that of its substrate?arrow_forwardTABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…arrow_forwardIt has been shown recently that the African naked GLUT5-.O Fructose (b) mole-rat can withstand up to 18 min of complete anoxia (ab- sence of O2) in the presence of high levels of CO2 without ap- parent brain damage. The basis of its capacity to tolerate such unusual conditions is that the brain, heart, and liver in the mole- rat switch to fructose-driven glycolysis. In contrast to normal mice, these tissues in the naked mole-rat have high levels of GLUT5 specific for cellular uptake of fructose. The available pathways for breakdown of fruct ose for metabolic energy production are shown on the right. KHK stands for ketohexosekinase, an enzyme that is markedly upregulated in the mole-rat compared to normal mice. Conse- quently mice must metabolize fructose largely via hexokinase (HK). In the diagram ALDOB, ALDOA, etc. represent aldolase isozymes. KHK HK Fructose-6-P PFK Fructose-1-P Fructose-1,6-BP ALDOA ALDOC ALDOB ALDOC Glyceraldehyde DHAP GAЗP Glyc-3-P 3-PGA Lactate Pyruvate mouse…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biochemistry
Biochemistry
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)
Biology
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Cengage Learning