Question

activity b (continued from previous page)

1. summarize: based on what you have seen, write a simplified formula for cellular respiration.

c₆h₁₂o₆ + __ → + __
turn on show input/output formula to check. were you correct? ________

1. challenge: to balance the inputs and outputs of cellular respiration, there should be the same number of carbon, oxygen, and hydrogen atoms on each side of the arrow.

a. is the formula balanced as written? why or why not? ________________________
________________________
b. now balance the input/output formula by adding coefficients to each molecule. write the balanced formula below, and then check your work by clicking balance.
__ + → + __

1. compare: the aerobic phase of cellular respiration in the mitochondrion produces a net of about 28 to 30 atp molecules. how does this compare to the energy released in glycolysis?

________________________
________________________
(note: some textbooks state that up to 36 atp molecules are produced in this phase of cellular respiration. in reality, some energy is lost in the process due to the cost of transporting molecules and imperfect membranes.)

1. extend your thinking: when you think of the word

espiration,\ you might think about the process of breathing, which is actually called ventilation. (the respiratory system consists of the windpipe, lungs, etc.)
how is breathing related to cellular respiration? (hint: think about both the inputs and the outputs of cellular respiration.)
________________________
________________________
________________________
________________________

Explanation:

6.A

Step1: Count atoms

Original formula: \(C_6H_{12}O_6+O_2
ightarrow CO_2 + H_2O\).
For carbon (\(C\)): Left - \(6\) (from \(C_6H_{12}O_6\)), right - \(1\) (from \(CO_2\)).
For hydrogen (\(H\)): Left - \(12\) (from \(C_6H_{12}O_6\)), right - \(2\) (from \(H_2O\)).
For oxygen (\(O\)): Left - \(6\) (from \(C_6H_{12}O_6\))+\(2\) (from \(O_2\))=\(8\), right - \(2\) (from \(CO_2\))+\(1\) (from \(H_2O\))=\(3\).

Step2: Check balance

Since the number of \(C\), \(H\), and \(O\) atoms are not equal on both sides, the formula is not balanced.

6.B

Step1: Balance carbon

Let’s start with carbon. We have \(6\) \(C\) atoms in \(C_6H_{12}O_6\). So, we put a coefficient of \(6\) in front of \(CO_2\). The formula becomes \(C_6H_{12}O_6+O_2
ightarrow6CO_2 + H_2O\).

Step2: Balance hydrogen

We have \(12\) \(H\) atoms in \(C_6H_{12}O_6\). So, we put a coefficient of \(6\) in front of \(H_2O\). The formula is \(C_6H_{12}O_6+O_2
ightarrow6CO_2 + 6H_2O\).

Step3: Balance oxygen

On the right - hand side, oxygen atoms: \(6\times2\) (from \(CO_2\))+\(6\times1\) (from \(H_2O\))=\(18\). On the left - hand side, oxygen atoms: \(6\) (from \(C_6H_{12}O_6\))+\(2x\) (from \(O_2\)). So, \(6 + 2x=18\), \(2x = 12\), \(x = 6\). The balanced formula is \(1C_6H_{12}O_6+6O_2
ightarrow6CO_2 + 6H_2O\).

8

Step1: Recall glycolysis ATP yield

Glycolysis produces a net of \(2\) ATP molecules.

Step2: Compare

The aerobic phase (in mitochondria) produces \(28 - 30\) (or up to \(36\) in some textbooks) ATP. So, the aerobic phase produces much more ATP ( \(14 - 15\) times more if \(28\div2 = 14\), \(30\div2=15\)) than glycolysis.

9

Step1: Consider inputs and outputs

Cellular respiration inputs: \(O_2\) (from breathing) and glucose. Outputs: \(CO_2\) (exhaled during breathing) and \(H_2O\). Breathing (ventilation) brings in \(O_2\) which is used as an input in cellular respiration. Also, the \(CO_2\) produced in cellular respiration is removed from the body through breathing.

Answer:

6.A

No. Because the number of carbon, hydrogen, and oxygen atoms are not the same on both sides of the arrow.

6.B

\(1C_6H_{12}O_6+6O_2
ightarrow6CO_2 + 6H_2O\)

8

The aerobic phase of cellular respiration in the mitochondrion produces much more ATP ( \(14 - 15\) times more, considering glycolysis produces \(2\) ATP) than the energy released in glycolysis.

9

Breathing brings in \(O_2\) (an input for cellular respiration) and removes \(CO_2\) (an output of cellular respiration).