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Gemiola [76]
1 year ago
5

The net reaction of the calvin cycle is the conversion of co2 into the three-carbon sugar g3p. Along the way, reactions rearrang

e carbon atoms among intermediate compounds and use the atp and nadph produced by the light reactions. In this exercise, you will track carbon atoms through the calvin cycle as required for the net production of one molecule of g3p.
Physics
1 answer:
WINSTONCH [101]1 year ago
7 0

The conversion of CO2 into the three-carbon sugar G3p is the Calvin cycle's net reaction.

<h3 /><h3>What is Calvin Cycle?</h3>

Calvin cycle uses ATP and NADPH, which are created during the light reaction, to decrease atmospheric carbon dioxide in 3C sugar; Three steps can be distinguished in the Calvin Cycle reaction:

Carboxylation: A reaction that is catalyzed by RUBISCO, where RUBIP is the CO2 acceptor, results in the formation of 6 molecules of PGA (phosphoglycerate).

Here, there are 6 molecules with 18 carbons in PGA and 3 molecules with 3 carbons in CO2.

Reduction: One molecule of the output, glyceraldehyde 3 phosphate, is used to create sugar while the remaining five molecules go through regeneration.

Here, one glyceraldehyde-3-phosphate molecule with three carbon atoms is present.

Regeneration: From glyceraldehyde 3 phosphate, CO2 acceptor (RUBIP) is generated.

Here, 3 molecules of RUBIP and 5 molecules of dihydroxy acetone phosphate each have 15 carbons.

Know more about the Calvin cycle here:

brainly.com/question/15205817

#SPJ4

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Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

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            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

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and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

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Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

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              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

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Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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