According to Kleiber’s law, for the vast majority of animals, an animal’s metabolic rate is proportional to m 0.75, where m is the animal’s mass. When we plot the graph of the common logarithm of the mass on the horizontal axis and the common logarithm of the metabolic rate on the vertical axis, we get a straight line. Explain why the graph is a straight line. What is the slope of the line?

Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places

In mathematical terms, what characteristic of a graphed line is a measure of enzymereaction rate?

How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s

The Nutrition Facts label on most food products shows % daily values based on a food energy requirement of 8,360 kJ (2,000 kcal) at normal resting state. a) Assuming that the efficiency of converting food energy to ATP is 50%, calculate the mass of ATP (in kg) that is harvested daily by the human body from 8,360 kJ of food energy. A total of 30.5 kJ of energy is needed to synthesize one mole of ATP under standard conditions. The molar mass of ATP is 505 g/mol. b) For an average 68-kg (150 lbs.) human adult, calculate the % mass of extracted ATP (from your answer in 2a) relative to body weight.

Could you explain step by step how to solve this? I'm not sure how to get all the numbers that are shown in the solved answer.

The following question focuses on how the parameters regulating enzyme function might change, and how these might appear graphically on a Michaelis-Menten plot and a Lineweaver-Burke plot. Carbonic anhydrase is an enzyme that will convert CO2 and water into HCO3. CO2 + H20 > H+ + HCO3 There are many different isoforms of this enzyme. (see for instance http://en.wikipedia.org/wiki/Carbonic_anhydrase 1 Assume that one variant has a Km of 10 µM and a different variant has a Km of 100 µM. Draw on the same graph a typical Michaelis-Menton plot showing the alteration in the rate of carbonic anhydrase as the CO2 level is varied for the two different variants of enzyme, assuming the concentration of the enzyme (10 mM) in the test tube is kept constant. Assume that you have equal amounts of the two different variants of carbonic anhydrase in a number of test tubes and that the Vmax for both enzymes are the same. Be sure to label the axes. For the same conditions as above, draw a…

Aconitase catalyzes the conversion of Citrate to Isocitrate in the TCA cycle. The standard free energy change (AG°) for this reaction is +6.7 kJ/mol. The observed free energy change (AG) for the reaction in pig heart mitochondria is +0.8 kJ/mol. The ratio of [isocitrate]/[citrate] in these mitochondria is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 10 b. 1 100 0.01 e f 0.33 0.10 h 0.25 i 2

The enzyme, fumarate, has the following kinetic constants: k 1 k 2 k -1 where k 1 = 10 9 M -1 s -1 k -1 =4.4 x 10 4 s -1 k 2 = 10 3 s -1 a. What is the value of the Michaelis constant for this enzyme? b. At an enzyme concentration of 10 -6 M, what will be the initial rate of product"

Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitor competitive or noncompetitive? [S] (M) 10,000 1 x 104 2.000-5 x 104 666673 1.5 x 10-3 400- 2.5 x 10-³ 2005x103 V, No Inhibitor (Arbitrary units) 0.026 38,461 0.092 10.87 37100 0.136 7.35 Wak 0.150 -67 0.165 6.06 L slope: 0033 yint=5057 полесте um.x=0.198 km 6.5x 10-4 0.010 0.040 25 10.086 11.63 0.120 8.33 0.142 7.04 2500 V, Inhibitor Present (Arbitrary Units) 100 ↓ $op.= .009 yint: 5.21 900x12 km: 1.73 y/o conf. V10-3 321000