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tatuchka [14]
3 years ago
7

For each molecule of glucose processed during glycolysis, the net yield is ____. 1. two molecules of NADH, two of ATP, and two o

f pyruvate2. two molecules of NAD , two of ATP, and two of pyruvate 3. two molecules of NAD , four of ATP, and two of pyruvate 4. two molecules of NADH, four of ATP, and two of pyruvate
Chemistry
2 answers:
Lana71 [14]3 years ago
5 0

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.

Natali5045456 [20]3 years ago
5 0

Answer:

1. two molecules of NADH, two of ATP, and two of pyruvate.

Explanation:

Glycolysis is defined as the breakdown of sugar which takes place in the cytoplasm. The various processes and enzymes involved are summarised in the Kreb's cycle.

Glycolysis is present in nearly all living organisms. Glucose is the source of almost all energy used by cells. Overall, glycolysis produces two pyruvate molecules, a net gain of two ATP molecules, and two NADH molecules.

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2.17

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The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho
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Explanation:

(a)   It is known that relation between energy and mass is as follows.

            E = 2 \times mc^{2}

where,    E = energy

              m = mass

              c = speed of light = 3 \times 10^{8} m/s

As it is given that mass is 9.109 \times 10^{-31} kg. So, putting the given values into the above formula as follows.

             E = 2 \times mc^{2}

                       = 2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s

                       = 1.638 \times 10^{-13} J

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is 1.638 \times 10^{-13} J.

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

                    E = h \times \nu   ..... (1)

where,     h = plank's constant = 6.626 \times 10^{-34} J.s

              \nu = frequency

Also,     E = 2 \times mc^{2} ........ (2)

Hence, equating equations (1) and (2) as follows.

                    h \times \nu = 2 \times mc^{2}        

So,    

6.626 \times 10^{-34} Js \times \nu = 1.638 \times 10^{-13} J

                           \nu = 1.236 \times 10^{20} Hz

Thus, we can conclude that the frequency is 1.236 \times 10^{20} Hz.

5 0
3 years ago
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