examine the diagram of aerobic cellular respiration below and use it to help you answer the following question:
question: all glucose molecules are made of six carbon (c) atoms, twelve hydrogen (h) atoms, and six oxygen (o) atoms. when glucose is metabolized in aerobic respiration, the molecule is destroyed and the atoms are used to create new molecules. match the atoms to the molecules where they will be found.
note: you will only use three of the options!

options:
- becomes the h in water (h₂o)
- becomes the o in carbon dioxide (co₂)
- becomes the c in carbon dioxide (co₂)
- becomes the o in water (h₂o)

atoms:
- oxygen (o)
- hydrogen (h)
- carbon (c)

Question

examine the diagram of aerobic cellular respiration below and use it to help you answer the following question:
question: all glucose molecules are made of six carbon (c) atoms, twelve hydrogen (h) atoms, and six oxygen (o) atoms. when glucose is metabolized in aerobic respiration, the molecule is destroyed and the atoms are used to create new molecules. match the atoms to the molecules where they will be found.
note: you will only use three of the options!

options:
- becomes the h in water (h₂o)
- becomes the o in carbon dioxide (co₂)
- becomes the c in carbon dioxide (co₂)
- becomes the o in water (h₂o)

atoms:
- oxygen (o)
- hydrogen (h)
- carbon (c)

Accepted Answer

# Brief Explanations:
- **Carbon (C)**: Glucose has 6 C atoms. During aerobic respiration, the carbon in glucose is released as carbon dioxide ($\ce{CO_{2}}$) during the citric acid cycle (Krebs cycle) and the breakdown of pyruvate. So Carbon (C) matches "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)".
- **Hydrogen (H)**: Glucose has 12 H atoms. The hydrogen atoms from glucose are used in the electron transport chain, and ultimately combine with oxygen to form water ($\ce{H_{2}O}$). So Hydrogen (H) matches "Becomes the H in water ($\ce{H_{2}O}$)".
- **Oxygen (O)**: The oxygen we inhale (from the air) is used in the electron transport chain to accept electrons and hydrogen ions to form water. But also, some of the oxygen in glucose? Wait, no—wait, the glucose has 6 O atoms. Wait, the options: the oxygen from glucose? Wait, the problem says "when glucose is metabolized in aerobic respiration". Wait, the oxygen atoms in glucose: during the breakdown, some oxygen in glucose may end up in $\ce{CO_{2}}$? Wait no, let's re - think. Wait, the three options for the bottom are: "Becomes the O in carbon dioxide ($\ce{CO_{2}}$)", "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)" (but C is already matched), and "Becomes the O in water ($\ce{H_{2}O}$)". Wait, the oxygen from glucose: in glucose, the formula is $\ce{C_{6}H_{12}O_{6}}$. When glucose is broken down, the oxygen in glucose—some of it goes to $\ce{CO_{2}}$? Wait, no, the correct matching for Oxygen (O) from glucose: Wait, the other option for O: "Becomes the O in water" is for the inhaled oxygen. Wait, no, the problem is about the atoms in glucose. Wait, maybe I made a mistake. Wait, the three bottom boxes: let's list the top atoms (C, H, O) and the bottom descriptions.

Top atoms:
1. Carbon (C)
2. Hydrogen (H)
3. Oxygen (O)

Bottom descriptions (to match):
- Becomes the C in carbon dioxide ($\ce{CO_{2}}$)
- Becomes the H in water ($\ce{H_{2}O}$)
- Becomes the O in water ($\ce{H_{2}O}$) → Wait, no, the bottom boxes are: "Becomes the O in carbon dioxide ($\ce{CO_{2}}$)", "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)", "Becomes the H in water ($\ce{H_{2}O}$)", "Becomes the O in water ($\ce{H_{2}O}$)". Wait, the note says "you will only use three of the options". So:

- Carbon (C) → Becomes the C in carbon dioxide ($\ce{CO_{2}}$) (because C from glucose is in $\ce{CO_{2}}$)
- Hydrogen (H) → Becomes the H in water ($\ce{H_{2}O}$) (H from glucose combines with O to form $\ce{H_{2}O}$)
- Oxygen (O) → Becomes the O in water ($\ce{H_{2}O}$)? No, wait, the oxygen in glucose: in $\ce{C_{6}H_{12}O_{6}}$, the O atoms—when glucose is broken down, some of the O in glucose goes to $\ce{CO_{2}}$? Wait, no, the correct matching:

Wait, the three matches are:

- Carbon (C) → Becomes the C in carbon dioxide ($\ce{CO_{2}}$)
- Hydrogen (H) → Becomes the H in water ($\ce{H_{2}O}$)
- Oxygen (O) → Becomes the O in water ($\ce{H_{2}O}$)? No, that can't be. Wait, maybe the oxygen from the air (the $\ce{O_{2}}$) is used to form water, but the oxygen in glucose: let's look at the chemical equations.

The overall equation for aerobic respiration is $\ce{C_{6}H_{12}O_{6} + 6O_{2} -> 6CO_{2} + 6H_{2}O + ATP}$.

From this equation:

- The 6 C atoms in $\ce{C_{6}H_{12}O_{6}}$ become the 6 C atoms in $6\ce{CO_{2}}$. So Carbon (C) → Becomes the C in carbon dioxide ($\ce{CO_{2}}$)
- The 12 H atoms in $\ce{C_{6}H_{12}O_{6}}$ become the 12 H atoms in $6\ce{H_{2}O}$ (since each $\ce{H_{2}O}$ has 2 H, 6*2 = 12). So Hydrogen (H) → Becomes the H in water ($\ce{H_{2}O}$)
- The 6 O atoms in $\ce{C_{6}H_{12}O_{6}}$: in the products, $6\ce{CO_{2}}$ has 12 O atoms (6*2) and $6\ce{H_{2}O}$ has 6 O atoms. The $6\ce{O_{2}}$ (from the air) has 12 O atoms. So total O in products: 12 (from $\ce{CO_{2}}$) + 6 (from $\ce{H_{2}O}$) = 18. O from glucose: 6, O from air: 12. So the O in glucose: some of it goes to $\ce{CO_{2}}$? Wait, but the option for O: the available bottom option that's left (after C and H are matched) is "Becomes the O in water ($\ce{H_{2}O}$)"? No, that doesn't fit. Wait, maybe the problem is simplifying. Let's go back to the options. The three bottom boxes (the dashed ones) are:

1. Becomes the O in carbon dioxide ($\ce{CO_{2}}$)
2. Becomes the C in carbon dioxide ($\ce{CO_{2}}$)
3. Becomes the H in water ($\ce{H_{2}O}$)
4. Becomes the O in water ($\ce{H_{2}O}$)

But the note says "you will only use three of the options". So we have three top atoms (C, H, O) and three bottom descriptions. So:

- C: Becomes the C in $\ce{CO_{2}}$ (since C from glucose is in $\ce{CO_{2}}$)
- H: Becomes the H in $\ce{H_{2}O}$ (since H from glucose is in $\ce{H_{2}O}$)
- O: Becomes the O in $\ce{CO_{2}}$? Wait, from the equation, the $\ce{CO_{2}}$ has O from glucose? Wait, in $\ce{C_{6}H_{12}O_{6}}$, the O is 6, and in $6\ce{CO_{2}}$, the O is 12. So the extra O in $\ce{CO_{2}}$ comes from $\ce{O_{2}}$. But the O in glucose: let's see, the glucose's O: in the breakdown of glucose, the pyruvate is $\ce{C_{3}H_{4}O_{3}}$, so each glucose gives 2 pyruvates. Then, when pyruvate is broken down to acetyl - CoA, $\ce{CO_{2}}$ is released. So the O in pyruvate (which comes from glucose) is part of the $\ce{CO_{2}}$ released. So the O from glucose ends up in $\ce{CO_{2}}$. So Oxygen (O) → Becomes the O in carbon dioxide ($\ce{CO_{2}}$)

Wait, I think I made a mistake earlier. Let's re - assign:

- Carbon (C): Becomes the C in $\ce{CO_{2}}$ (correct, as C from glucose is in $\ce{CO_{2}}$)
- Hydrogen (H): Becomes the H in $\ce{H_{2}O}$ (correct, as H from glucose is in $\ce{H_{2}O}$)
- Oxygen (O): Becomes the O in $\ce{CO_{2}}$ (because the O in glucose is released as $\ce{CO_{2}}$ during the breakdown, along with O from $\ce{O_{2}}$)

But the bottom options: the three options are "Becomes the O in carbon dioxide ($\ce{CO_{2}}$)", "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)", "Becomes the H in water ($\ce{H_{2}O}$)", "Becomes the O in water ($\ce{H_{2}O}$)". Wait, the "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)" is for Carbon (C), "Becomes the H in water ($\ce{H_{2}O}$)" is for Hydrogen (H), and "Becomes the O in carbon dioxide ($\ce{CO_{2}}$)" is for Oxygen (O). The "Becomes the O in water ($\ce{H_{2}O}$)" is for the oxygen from the air (the $\ce{O_{2}}$ we breathe), but the problem is about the atoms in glucose. So the correct matches are:

- Carbon (C) → Becomes the C in carbon dioxide ($\ce{CO_{2}}$)
- Hydrogen (H) → Becomes the H in water ($\ce{H_{2}O}$)
- Oxygen (O) → Becomes the O in carbon dioxide ($\ce{CO_{2}}$)

# Answer:
- Carbon (C) is matched with "Becomes the C in carbon dioxide ($\ce{CO_{2}}$)"
- Hydrogen (H) is matched with "Becomes the H in water ($\ce{H_{2}O}$)"
- Oxygen (O) is matched with "Becomes the O in carbon dioxide ($\ce{CO_{2}}$)"

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