Answer: 1. two molecules of NADH, two of ATP, and two of pyruvate
Explanation:
During glycolysis 4 molecules of ATP are synthesized by two substrate level phosphorylation (first in step 6 conversion of 1,3-Bisphospho glycerate to 3-Bisphospho glycerate by 1,3-Bisphospho glycerate kinase produces 2 molecules of ATP . Secondly 2 molecules of ATP is produced in step 9 by pyruvate kinase). But 2 molecules of ATP are used in step 1 and 3 by hexokinase and phospho fructo kinase respectively, hence a net yield of only two ATP.
2 molecules of NADH is produced in step 5 by Glyceraldehyde-3-phosphate dehydrogenase.
The end product of glycolysis is 2 molecules of pyruvate.
Answer:
0.034%
Explanation:
First, let's calculated the average density (da) of the two experimental, which the sum of them divided by 2:
da = 1.090x10⁻³ g/cm³
The relative error is given by the difference of the correct value and the avarege, divided by the correct value:
e = 3.39x10⁻⁴ x 100%
e = 0.034%
Answer:
5.00 µM
Explanation:
Given data
- Initial concentration (C₁): 250.0 µM
- Initial volume (V₁): 1.00 mL
- Final concentration (C₂): ?
- Final volume (V₂): 50.0 mL
We can find the final concentration using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 250.0 µM × 1.00 mL / 50.0 mL
C₂ = 5.00 µM
The concentration of the diluted solution is 5.00 µM.
Answer:
The answer is "-50.424° C".
Explanation:
Given:
Time = 1 hour
TB=?
Formula:
time (T) = 1 hour
= 3600 second
Calculating T_B:
