Determine the number of ATP equivalents needed to form palmitic acid from acetyl-CoA. (Assume for this calculation that each NADPH is worth 3.5 ATPs.

Short Answer

Expert verified
64 ATP equivalents are needed to form palmitic acid from acetyl-CoA.

Step by step solution

01

Identify the number of cycles

First, identify the number of cycles needed to form palmitic acid. Palmitic acid is a 16-carbon fatty acid, and two carbon atoms are added in each cycle of fatty acid synthesis. Therefore, \( \frac{16}{2} = 8 \) cycles are needed.
02

Number of NADPH used per cycle

In each round of the cycle, 1 acetyl-CoA is converted to malonyl-CoA using 1 ATP. Additionally, 2 NADPH are used in the cycle for reductions. Therefore, for 8 cycles, \( 8 \times 2 = 16 \) NADPH are used.
03

Convert NADPH to ATP equivalents

Each NADPH is equivalent to 3.5 ATPs. So, for the 16 NADPH, \( 16 \times 3.5 = 56 \) ATP equivalents are used.
04

Compute ATP used for acetyl-CoA conversion

Apart from the NADPH used in the cycle, 1 ATP is used per cycle for the acetyl-CoA to malonyl-CoA conversion. Therefore in 8 cycles, \( 8 \times 1 = 8 \) ATP are used for the conversion.
05

Sum total ATP equivalents

Summing up, the total ATP equivalents used are the ATP equivalents used for the NADPH and the ATP used for the conversion. So , \( 56 + 8 = 64 \) ATP equivalents are used to form palmitic acid from acetyl-CoA.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free