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Complex II

CO2 : NADPH2 ...

Question

$C O_{2} : N A D P H_{2} : A T P$ ratio for $C_{4}$ plants is

A

1 : 2 : 3

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B

2 : 3 : 4

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C

1 : 2 : 5

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D

1 : 5 : 10

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Solution

The correct option is A $1 : 2 : 5$
In C $_{4}$ pathway for production of one molecule of glucose we require 12 NADPH $_{2}$ , 30 ATP and 6CO $_{2}$ are required.

CO $_{2}$ : NADPH $_{2}$ : ATP = 6 : 12 : 30

CO $_{2}$ : NADPH $_{2}$ : ATP = 1 : 2 : 5

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Similar questions

c_{4}

plants, number of ATP and

N A D P H_{2}

molecules required to fix 2 molecules of

C O_{2}

are respectively.

Find the mode from the given data.

1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2 , 3, 4, 1, 2, 3, 4, 1 ,2 , 3, 4, 5, 5, 1, 5, 5, 5, 3, 5, 1, 2, 4, 5, 6, 1, 2, 4, 2, 1

(1) 2² + 3²

(2) 3² − 2²

(3) 2² × 3³

(4) (−2)² × (1)²

(5) 1² + 5²

(6) 1² + 2⁰

(7) 7⁰ + 1²

(8) 6¹ − 4⁰

(9) 10⁰ − 2⁰

(10) 5¹× 3¹

Q. Name the property under multiplication used in each of the following:
(i)

\frac{2}{3} \times \frac{4}{5} = \frac{4}{5} \times \frac{2}{3}

(\frac{- 3}{4}) \times 1 = \frac{- 3}{4} = 1 \times (\frac{- 3}{4})

(\frac{- 17}{28}) \times (\frac{- 28}{17}) = 1

\frac{1}{5} \times (\frac{7}{8} \times \frac{4}{3}) = (\frac{1}{5} \times \frac{7}{8}) \times \frac{4}{3}

\frac{2}{7} \times (\frac{9}{10} + \frac{2}{5}) = \frac{2}{7} \times \frac{9}{10} + \frac{2}{7} \times \frac{2}{5}

Solve

(a) (b)

(c) (d)

(e) (f)

(g) (h)

(i) (j)

(k) (l)

(m) (n)

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