Kinesiology 2230A/B Lecture Notes - Lecture 3: Enolase, Activation Energy, Glyceraldehyde

24 views7 pages

yellowcod194

15 Jan 2016

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

The point in a metabolic pathway where ultimate control (slowest reaction) is exerted. In contrast, average free energy of a molecules remains same in uncatalyzed versus catalyzed rxns (conducted at same temperature) Binding to enzymes and facilitating the match between the enzyme and the other reactant to give you the desired product. Example: metals: iron, magnesium, cobalt, copper, zinc. Make up a part of the active site. Lipoamide (fad) flavin adenine dinucleotide (nad+) nicotinamide adenine. Dinucleotide (coa) coenzyme a (mg2+) metal ion. Cycling"s version of the atp-pcr and anaerobic energy systems. Combination of pcr system and atp system that is required for the output of the. Atp is high energy compound used to fuel exercise. Cells contain storage form of high energy known as phosphocreatine (pcr) or creatine phosphate (cp) Discus or any event that is explosive that takes less than a. Breaking down atp, no replenishment happening because too second and a half short a duration.

Related Questions

A hypothetical metabolic pathway can be diagrammed as A-->B--->C--->D,, and x, y z are the enzymes that catalyze the reactions The free energy changes for the reactions are

x= -0.2kJ/mol

y= -12.3kJ/mol

z= -1.2 kJ/mol

a) Which reaction is likely to be a major regulatory point for the pathyway? b) If your answer to a was in fact the case, in the presence of an inhibitor that blocks the activity of enzyme "Z", would the concentrations of A,B,C,D increase, decrease or not be affected? Explain your answer.

The following two reactions proceed along the nucleophilic substitution mechanistic pathway whose rate determining step is unimolecular. For the alcohol, water acts as the leaving group in the presence of a hydrogen halide as demonstrated in lecture. For the alkyl halide, the halogen atom is the leaving group. In the space below the reaction, draw a proper elementary step by elementary step mechanism for each reaction indicating how the reactants progress to their final product.

Draw only the structure of the product for the second reaction above, where the starting alcohol is 3-methyl-2-butanol instead of 2-methyl-2-butanol.

To understand how to use integrated rate laws to solve for concentration.

A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours?

This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145.

55 mi/hrÃ2 hr=110 miles traveled

milemarker 145â110 miles=milemarker 35

If we were to write a formula for this calculation, we might express it as follows:

milemarker=milemarker0â(speedÃtime)

where milemarker is the current milemarker and milemarker0 is the initial milemarker.

Similarly, the integrated rate law for a zero-order reaction is expressed as follows:

[A]=[A]0ârateÃtime

[A]=[A]0âkt

since

rate=k[A]0=k

A zero-order reaction (Figure 1) proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates (Figure 2) do change over time as the reactant concentration changes.

Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction.

The integrated rate law for a first-order reaction is expressed as follows:

[A]=[A]0eâkt

where k is the rate constant for this reaction.

The integrated rate law for a second-order reaction is expressed as follows:

1[A]=kt+1[A0]

where k is the rate constant for this reaction.

PART A

The rate constant for a certain reaction is k = 3.50Ã10^â3 sâ1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 5.00 minutes?

Express your answer with the appropriate units.

PART B

A zero-order reaction has a constant rate of 5.00Ã10^â4M/s. If after 75.0 seconds the concentration has dropped to 2.50Ã10^â2M, what was the initial concentration?

Express your answer with the appropriate units.

coffeefly836

Kinesiology 2230A/B Lecture Notes - Lecture 3: Enolase, Activation Energy, Glyceraldehyde

Document Summary

Get access

Related Documents

Kinesiology 2230A/B Lecture Notes - Lecture 3: Enzyme Inhibitor, Metabolic Pathway, Phosphofructokinase

Kinesiology 2230A/B Lecture Notes - Lecture 3: Endergonic Reaction, Exergonic Process, Phosphate

Kinesiology 2230A/B Lecture Notes - Lecture 3: Acetyl-Coa, Citric Acid Cycle, Isocitric Acid

Related Questions