Respiration in Plants — NEET UG practice

1. Locate glycolysis first. Glycolysis is the very first step of cellular respiration and it takes place entirely in the cytoplasm of the cell, outside the mitochondria.

This gives us:

$$A \, (\text{Glycolysis}) \rightarrow III \, (\text{Cytoplasm})$$

2. Locate the Electron Transport System (ETS). The protein complexes of the ETS are embedded within the inner mitochondrial membrane, since this is where electrons are passed along the chain.

So:

$$B \, (\text{ETS}) \rightarrow I \, (\text{Inner mitochondrial membrane})$$

3. Locate proton accumulation. As electrons move through the ETS, protons ($H^+$) are actively pumped out of the matrix into the intermembrane space, building up a concentration there.

Hence:

$$C \, (\text{Accumulation of protons}) \rightarrow IV \, (\text{Intermembrane space})$$

4. Locate Krebs' cycle. The enzymes required for the Krebs' cycle (citric acid cycle) are found dissolved in the mitochondrial matrix, where the cycle's reactions occur.

Therefore:

$$D \, (\text{Krebs' cycle}) \rightarrow II \, (\text{Mitochondrial matrix})$$

5. Combine all matches. Putting it together: A-III, B-I, C-IV, D-II.

Hence, the answer is Option C: A-III, B-I, C-IV, D-II.

1. Recall what glycolysis produces. Glycolysis is the pathway that breaks down one molecule of glucose (a 6-carbon sugar) into two molecules of pyruvic acid (a 3-carbon compound).

So, for every 1 glucose molecule:

$$1 \text{ Glucose} \rightarrow 2 \text{ Pyruvic acid}$$

2. Set up the ratio for the given amount of glucose. Since the question gives us 206 molecules of glucose, we simply scale the above relationship.

Given that each glucose molecule yields 2 pyruvic acid molecules:

$$206 \times 2 = 412$$

3. State the conclusion. Therefore, 206 molecules of glucose will yield 412 molecules of pyruvic acid at the end of glycolysis.

Hence, the answer is Option A: 412.

1. Define the Respiratory Quotient (RQ). RQ tells us what type of respiratory substrate (carbohydrate, fat, or protein) is being used by an organism. It is defined as the ratio of the volume of $CO_2$ released to the volume of $O_2$ consumed during respiration.

So, the defining formula is:

$$RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}}$$

2. Identify the relevant coefficients from the balanced equation. The reaction shows that for every 145 molecules of $O_2$ consumed, 102 molecules of $CO_2$ are released.

Therefore, substituting these values into the RQ formula:

$$RQ = \frac{102}{145}$$

3. Calculate the numerical value. Dividing 102 by 145 gives:

$$RQ \approx 0.7$$

4. Interpret the result. A fat molecule being respired typically gives an RQ less than 1 because fats contain relatively less oxygen compared to carbon and hydrogen, so more external $O_2$ is needed to fully oxidize them compared to $CO_2$ released. An RQ of 0.7 fits within the range for fats, which generally falls between 0.5 and 0.95.

Hence, the answer is Option A: Between 0.5 and 0.95.