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All digestible forms of carbohydrates are eventually transformed into glucose; which is then able to provide energy in the form of ATP The enzyme hexokinase catalyz...

Question

All digestible forms of carbohydrates are eventually transformed into glucose; which is then able to provide energy in the form of ATP The enzyme hexokinase catalyzes the following reaction in the glycolytic pathway in muscle cells: Glucose ATP == Glucose-6-phosphate ADPThe AG" for the reaction is 16.7kJlmol, whereas the AG in the cellis 33.5 kJlmol:It canbe shown that the ratio of products to reactants under intracellular conditions, Keq 0.0015.(Remember: AG = AG"' RT In Keq R= 8

All digestible forms of carbohydrates are eventually transformed into glucose; which is then able to provide energy in the form of ATP The enzyme hexokinase catalyzes the following reaction in the glycolytic pathway in muscle cells: Glucose ATP == Glucose-6-phosphate ADP The AG" for the reaction is 16.7kJlmol, whereas the AG in the cellis 33.5 kJlmol: It canbe shown that the ratio of products to reactants under intracellular conditions, Keq 0.0015.(Remember: AG = AG"' RT In Keq R= 8.314 J mol" K-l , assuming T = 310K) Ifyou run a sprint; your ATP levels will sharply decrease. Based on the calculated Keq" ~ above; does this make glycolysis more or less favorable? Briefly explain your answer:



Answers

Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as
$$\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \operatorname{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$
where ADP represents adenosine diphosphate. For this reaction, $\Delta G^{\circ}=-30.5 \mathrm{kJ} / \mathrm{mol} .$
a. Calculate $K$ at $25^{\circ} \mathrm{C}$ .
b. If all the free energy from the metabolism of glucose
$$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$
goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?

We know Standard gives free energy at normal STP conditions. That is 298 kelvin temperature. One more look concentration and one bar pressure is mathematically represented by question. Negative. Artie, Ellen, Ellen is natural law. Kippy gate is equally brim pressure Here are is universal gas constant temperature. KP equally broom constant on delta G is standard gives free energy considering the reaction for auction e, we have standard Gibbs Free and ADI Change equals 1.7 kilo jewels, which is a positive value here. If Delta G outstanding temperature pressure condition is greater than zero, the reaction is non spontaneous. Option B. The expression that requires for calculating gives free energy is Delta G equals delta G et standard condition Bless our D natural law. KP substituting in the values we have Delta G equals negative two point. Oh, hello, Jules. Form over this case, Delta ji is less than they do this mean above reaction proceed in forward direction, favoring the formation off the products. This reaction is known spontaneous

So here we're continuing on with some some a dynamic work on what we have is a situation a 298 kelvin on. We have the following equation to consider Delta G not is equal to negative rt l on k. We're looking for the K value. So lnk is equal to Delta Jeannot divided by negative rt. So we plug in the values that we got we get Ln k is equal to 12.309 k is equal to the exponent off 12.309 K is equal to 2.22 times 10 to the five. Yeah, so in the second part now we have the metabolism of glucose that is shown any following equation. We have C six h six, age 12 06 My mistake age 12 06 at 602 breaks down to six c 02 at six h, 20 So we're looking at the value of the Gibbs free energy of the reaction. So I doubt that, you know, value is negative. 2875000 jewels per mole. And now we're looking to the time the number of ATP molecules formed. So that is equal to the delta. G metabolism of Bleeker's Divided by Dr G No hydraulic sis off ATP. So on to the next page, we have negative to eight 75 multiplied by 10, 23 jewels. Permal all divided by a negative 30.5 times 10 to the power three Jules Park mall. So you can see our units cancel nicely there. And so what we get? There's units, councils, We have a unit Lis Valley. We can already tell. We get 94.26 and we can just call that 94 by rounding down. Therefore, we just have 94 units of ATP molecule for every molecule arc, Lucas.

Question Number 62 is a Hess's law question using standard Delta Jeez of Chemical are of two chemical reactions. One in particular is the like. Collis is of glucose going to glucose six phosphate in the presence of ATP, where ATP also produces a D. P. And then the other one is the Delta G for at peace simply going to a D. P. Both of these are negative. If, however, we want to determine what the delta G is for just glucose going to glucose six phosphate, then we need to reverse this first this second reaction so that the A. T P s will cancel and the day a DPS will cancel and when we reverse it, we need to change the sign on Delta G over here. Then we can sum everything up and we'll get the glucose going to glucose six phosphate. And while some of the Delta G s and will get positive 13 killer jewels. So why does ATP need to be present in order to drive the process? Well, the process by itself has a positive delta G, suggesting it is non spontaneous. It only becomes spontaneous with the assistance of a teepee going to a d. P. Giving us a net negative 17 killer jewels, which is spontaneous, which then will occur in the presence of ATP.

So I just answered the question about 80 p The previous question. Number 34. Using some approximate values for, uh, the Delta Cheese Delta G zero standard free energy changes, Um, and replacing those with more accurate values plus 13.4 and surplus 15 and thinking of 31 instead of negative 30. Then the when we can combine two reactions to give 1/3 reaction, then the sum of their delta. Jeez, is Theo or Delta G zeros is Thea is the Delta G zero for that reaction. So the reaction of the phosphor relation of glucose to produce glucose six phosphate has a negative. A Delta G zero about negative 17.6 village rules. And so, using the equation that relates free energy to the equilibrium constant just divide both sides by a negative rt and I can calculate the value of the log of K and so plugging the energy change for the reaction in and this would be in the biological system, say a person at 37 degrees C Uh, or, uh, 3 10 k. Uh, I get a log of natural log of K of 6.83 in a value for the equilibrium constant. That is roughly 1000. So it's highly favorable in the presence of 80 p for a glucose to gain that phosphate. But that phosphate, the P I hear stands for really, it's not age three p 04 It's really H p o for double minus and h two p o for minus that already at equilibrium in the biological system. But anyway, uh, that a teepee makes it favourable by about a factor of 1000 in relation to the equilibrium constant.


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