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Capítulo 3
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1. In the case of chymotrypsin, a drop in pH might suppress the loss of a proton from the serine, whereas the hydroxyl ions resulting from a rise in pH might interfere with histidine's ability to influence the charge on the serine.

2. Inhibition of the first enzyme (or more accurately, the first reaction in a pathway that is thermodynamically irreversible) shuts down the entire pathway, allowing the substrate to be used for other purposes. If one of the later enzymes were inhibited, energy would be wasted and unusable metabolites might accumulate.

3. #a, b, e are true.

4. The substrate concentration is saturating at all three concentrations and thus the enzyme is operating at approximately V max .

5. 0.5

6. One. Zero.

7. That the ratio of ATP/ADP in the cell is much greater than that at standard conditions, which would be 1.0.

8. No effect since it would not change the ratio of reactants to products.

9. -4.2 kcal/mol; +4.2 kcal/mol; 10 2.85 .

10. greater; greater.

11. Yes, because the ΔG for ATP hydrolysis in the cell is approximately -12 kcal/mol

12. (3) less than one-tenth.

13. X + ATP -> XP + ADP; XP + Y -> Z + P

14. Probably not, because its formation is too endergonic (+11.8 kcal/mol), which is the reason that it can be used to form ATP by substrate-level phosphorylation (i.e., its phosphate transfer potential is much higher than that of ATP).

15. -12.9 kcal/mol; the ratio of ADP/ATP would be approximately 10 6 ; -7.3 kcal/mol

16. ΔG = ΔG o + RT ln [P]/[R]
At equilibrium, ΔG = 0
-ΔG o ' = 2.3 RT log 10 K' eq
-3 kcal/mol = 1.4 kcal/mol• log10 K' eq
log 10 K' eq = -3/1.4 = -2.2
K' eq = 10 -2.2 = 6.5 X 10 -3
No. If the concentration of reactants are large enough, the reaction will go from left to right. This will occur when the term
_[glucose 6-P]_
[glucose] [P i ]
is less than 6.5 X 10 -3 .


17.

18. ΔG = ΔG o ' + RT ln _[glutamine]_
[glutamic acid] [NH 3 ]
At equilibrium, ΔG o ' = - 2.3 RT log 10 K' eq
3.4 kcal/mol = -1.4 kcal/mol•log 10 K' eq
-2.4 = log 10 K' eq
K' eq = 3.5 X 10 -3
At equilibrium, K' eq = 3.5 X 10 -3 = ______[glutamine]_____
[glutamic acid] [10 X 10 -3 M]
35 X 10 -6 = __[glutamine]__ = 1/3.5 X 10 -5
[glutamic acid]
To proceed spontaneously from left to right, the concentration of glutamic acid must be greater than 2.9 X 10 4 times that of glutamine

19.

20. There would be no effect of increasing substrate concentration. The V max would be reduced by a noncompetitive inhibitor. The K M would be unaffected by a noncompetitive inhibitor.

21. Enzymes have such a high catalytic efficiency that others were working with very low concentrations of enzyme that were below the level of detection at the time. Sumner, on the other hand, was working with pure protein crystals.







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