Answer:
Total ATP molecules produced = 66 molecules of ATP
Explanation:
A 10-carbon fatty acid when it has undergone complete oxidation will yield 5 acetyl-CoA molecules and 4 FADH₂ and 4 NADH molecules each. Each of the 5 acetyl-CoA molecules enters into the citric acid cycle and is completely oxidized to yield further ATP and FADH₂ and NADH molecules.
The total yield of ATP in the various enzymatic step is calculated below:
Acyl-CoA dehydrodenase = 4 FADH₂
β-Hydroxyacyl-CoA dehydrogenase = 4 NADH
Isocitrate dehydrogenase = 5 NADH
α-Ketoglutarate dehydrogenase = 5 NADH
Succinyl-CoA synthase = 5 ATP (from substrate-level phosphorylation of GDP)
Succinate dehydrogenase = 5 FADH₂
Malate dehydrogenase = 5 NADH
Total ATP from FADH₂ molecoles = 9 * 1.5 = 13.5
Total NADH molecules = 19 * 2.5 = 47.5
Total ATP molecules produced = 13.5 + 47.5 + 5
Total ATP molecules produced = 66 molecules of ATP
The particles are moving fastest in Region D) 5.
As temperature increases, so does the average kinetic energy (speed) of the molecules.
Answer is: a mixture.
Crude oil is mixture (solution) consist of several different hydrocarbons (alkanes, alkenes...). This hydroalkanes can be separated from mixture (crude oil) using distilation (usually fractional distillation), because they have different boiling points.
Proportion of hydrocarbons is diiferent in varios parts of the world.
Solution is homogeneous mixture composed of two or more substances.
Answer:
A)glyceraldehyde 3-phosphate dehydrogenase reaction,
B)Thehexose bisphosphate that accumulates is fructose 1,6-bisphosphate
C)glyceraldehyde 3-phosphate dehydrogenase reaction to yield an acyl arsenate
Explanation:
The fermentation of ethanol in yeast has the following overall equation Glucose 2ADP 2Pi88n2 ethanol 2CO22ATP 2H2O which makes it clear that phosphate is required for the continued operation of glycolysis and formation of ethanol . In extracts to which glucose is added, fermentation proceeds until ADP and Pi(present in the extracts) are exhausted.(a)Phosphate is required in the glyceraldehyde 3-phosphate dehydrogenase reaction, and glycolysis will stop at this step when Piis exhausted. Because glucose remains, it will be phosphorylated by ATP, but Piwill not be released.(b)Fermentation in yeast cells produces ethanol and CO2rather than lactate . Without these reactions (in the absence of oxygen), NADH would accumu-late and no new NADwould be available for further glycolysis ). Thehexose bisphosphate that accumulates is fructose 1,6-bisphosphate; in terms of energet-ics, this intermediate lies at a “low point” or valley in the pathway, between the energy-input reactions that precede it and the energy-payoff reactions that follow.(c)Arsenate replaces Piin the glyceraldehyde 3-phosphate dehydrogenase reaction to yieldan acyl arsenate, which spontaneously hydrolyzes. This prevents formation of fructose1,6-bisphosphate and ATP but allows formation of 3-phosphoglycerate, which continuesthrough the pathway.
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) =
Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
![[ H_{3} O^+] = M* \alpha](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%20M%2A%20%5Calpha%20)
![[ H_{3} O^+] = 0.010* 2.23*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%200.010%2A%202.23%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 0.0223*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%200.0223%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%202.23%2A10%5E%7B-5%7D%20%5C%3Amol%2FL)
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
![[ H_{3} O^+] = 2.23*10^{-5}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%202.23%2A10%5E%7B-5%7D)
Formula:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)
Solving:
Note:. The pH <7, then we have an acidic solution.
