Problem 19 - Biochemistry: A Short Course

Question

Around we go. Assuming that 145 base pairs of DNA wrap around the histone octamer 1\(^{3 / 4}\) times, estimate the radius of the histone octamer. Assume 3.4 Ä per base pair, and simplify the calculation by assuming that the wrapping is in two rather than three dimensions and neglecting the thickness of the DNA.

Step-by-Step Solution

Answer

The estimated radius of the histone octamer is about 44.9 Å.

Step 1: Understanding the Problem

We have 145 base pairs of DNA wrapping around the histone octamer 1.75 times. Each base pair contributes 3.4 Å to the circumference. We're tasked with determining the radius of this octamer.

Step 2: Calculate Total Circumference

First, calculate the total length of DNA that wraps around the octamer. Multiply the number of base pairs (145) by the length per base pair (3.4 Å): \(145 \times 3.4 \text{ Å} = 493 \text{ Å}\).

Step 3: Determine the Total Circumference Per Wrap

Since the DNA wraps around the octamer 1.75 times, we need to divide the total DNA length (493 Å) by the number of wraps to find the circumference of the circle formed by one complete wrap: \( \frac{493}{1.75} \text{ Å} = 281.71 \text{ Å} \approx 282 \text{ Å}\).

Step 4: Use Circumference to Find Radius

The relationship between the circumference \(C\) of a circle and its radius \(r\) is given by \(C = 2\pi r\). Set \(C = 282 \text{ Å} \) and solve for \(r\): \[ r = \frac{C}{2 \pi} = \frac{282}{2 \times 3.1416} \approx \frac{282}{6.2832} \approx 44.9 \text{ Å} \].

Step 5: Concluding the Calculation

Thus, the estimated radius of the histone octamer is approximately 44.9 Å.

Key Concepts

Histone Octamer

The histone octamer is an important protein complex in the structure of chromatin. Inside the cell nucleus, DNA is not just floating around loosely. Instead, it is packaged neatly to ensure that it fits within the confines of the microscopic nucleus. Here’s where histone octamers come into play.

• The histone octamer is made up of eight protein molecules.
• These proteins are called histones and they play a crucial role in DNA packaging.
• Histones help to condense DNA by allowing it to wrap around them. This creates a nucleosome, the fundamental unit of DNA packaging in eukaryotic cells.

So, understanding the formation of the histone octamer allows us to grasp how DNA is condensed and organized effectively in the cell nucleus.

Base Pairs

Base pairs are fundamental units of the DNA structure. DNA is composed of two strands that are twisted together to form a double helix. Base pairs hold these strands together. Here’s what you need to know:

• Base pairs are formed by the pairing of nitrogenous bases.
• The main bases are adenine (A), thymine (T), cytosine (C), and guanine (G).
• A pairs with T, and C pairs with G, creating a specific complementary mechanism.
• Each base pair contributes a specific distance to the DNA length, which in this case is 3.4 Å per base pair.

Base pairs are essential for the genetic encoding and transmission of genetic information. Their specific pairing ensures the fidelity of DNA replication.

Nucleosome

The nucleosome is the basic structural unit of chromatin and plays a pivotal role in DNA packaging. In the context of histone octamers, nucleosomes are significant as they establish the first level of DNA compaction:

• DNA wraps around the histone octamer approximately 1.75 times to form a nucleosome.
• This structure resembles a bead on a string when viewed under a microscope.
• The spacing and organization of nucleosomes allow efficient packaging and accessibility of DNA for transcription and replication processes.

Recognizing the formation of nucleosomes helps in understanding how genetic material is condensed yet available for cellular processes.

Circumference Calculation

Circumference calculation is a key mathematical concept applied in figuring out the spatial configuration of DNA around histones. Here’s how it relates to the problem at hand:

• To estimate how DNA wraps around the histone, we calculate the total length of DNA using the number of base pairs and the distance each base pair contributes.
• Once the total length is known, dividing this by the number of times the DNA wraps around the histone gives the circumference of a single wrap.
• The circumference formula, \( C = 2\pi r \), then allows us to solve for the radius \( r \).
• In this exercise, with \( C = 282 \text{ Å} \), calculations show the radius to be approximately 44.9 Å.

Understanding circumference calculations in this context helps in visualizing the tight packing of DNA around histones, providing insight into the structural organization within the nucleus.