Essay about Practice 4A 1

Introduction to Biochemistry

(CHEM 501)

Final Exam

12/17/2007

1 . The anaerobic change of 1 mol of sugar to a couple of mol of lactate by fermentation is accompanied by a net gain of:

a)

b)

c)

d)

e)

you mol of ATP.

you mol of NADH.

two mol of ATP.

two mol of NADH.

none of the above.

2 . The conversion of 1 mol of fructose 1, 6-bisphosphate to 2 mol of pyruvate by glycolytic path results in a net creation of:

a)

b)

c)

d)

e)

1 mol of NAD+ and 2 mol of ATP.

you mol of NADH and 1 mol of ATP.

2 mol of NAD+ and four mol of ATP.

2 mol of NADH and 2 mol of ATP.

2 mol of NADH and four mol of ATP.

several. Which with the following statements is incorrect concerning glycolysis in anaerobic muscle? a)

b)

c)

d)

e)

Fructose 1, 6-bisphosphatase is among the enzymes from the pathway. Costly endergonic procedure.

It leads to net activity of ATP.

It leads to synthesis of NADH.

The rate is usually slowed with a high [ATP]/[ADP] ratio.

some. When a muscle tissue is induced to deal aerobically, much less lactic acid solution is formed than when it contracts anaerobically since:

a)

b)

c)

d)

glycolysis will not occur to significant extent below aerobic conditions. muscle is definitely metabolically much less active underneath aerobic than anaerobic circumstances. the lactic acid made is swiftly incorporated in to lipids below aerobic circumstances. under cardio conditions in muscle, the major energy-yielding pathway is the pentose phosphate path, which will not produce lactate.

e) beneath aerobic circumstances most of the pyruvate generated as a result of glycolysis can be oxidized by the citric acidity cycle instead of reduced to lactate.

1

5. Glycolysis in the erythrocyte produces pyruvate that is even more metabolized to: a)

b)

c)

d)

e)

LASER.

ethanol.

blood sugar.

hemoglobin.

lactate.

6. In glycolysis, fructose 1, 6-bisphosphate is converted to two products with a common freeenergy modify (О”G'В°) of 23. almost eight kJ/mol. Under what conditions (encountered in a normal cell) will the free-energy change (О”G') be adverse, enabling the response to proceed to the right? a) If the concentrations of the two products are high in accordance with that of fructose 1, 6bisphosphate. b) The response will not go to the right spontaneously under any conditions as the О”G'В° is usually positive.

c) Under standard conditions, enough energy can be released to operate a vehicle the reaction towards the right. d) When there exists a high focus of fructose 1, 6-bisphosphate relative to the concentration of products.

e) When ever there is a high concentration of goods relative to the concentration of fructose 1, 6-bisphosphate.

7. The steps of glycolysis among glyceraldehyde 3-phosphate and 3-phosphoglycerate involve all of the following other than:

a)

b)

c)

d)

e)

ATP synthesis.

catalysis by phosphoglycerate kinase.

oxidation of NADH to NAD+.

the formation of 1, 3-bisphosphoglycerate.

usage of Pi.

8. Which with the following is a cofactor in the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase?

a)

b)

c)

d)

e)

ATP

2+

Cu

heme

&

NAD

&

NADP

two

9. Glycogen is converted to monosaccharide devices by:

a)

b)

c)

d)

e)

glucokinase.

glucose-6-phosphatase

glycogen phosphorylase.

glycogen synthase.

glycogenase.

15. Which of the following transactions is inappropriate?

a) Aerobically, oxidative decarboxylation of pyruvate forms acetate that enters the citric acid routine.

b) In anaerobic muscles, pyruvate is definitely converted to lactate.

c) In yeast growing anaerobically, pyruvate is transformed into ethanol. d) Reduction of pyruvate to lactate regenerates a cofactor essential for glycolysis. e) Beneath anaerobic circumstances pyruvate would not form since glycolysis would not occur. eleven. Which one with the following claims about gluconeogenesis is fake? a)

b)

c)

d)

e)

For starting supplies, it can make use of carbon skeletons derived from particular amino acids. That consists entirely of the reactions of glycolysis, operating in the reverse path. It utilizes the enzyme glucose...



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