Guided Practice: Dimensions of Bacterial Chromosome Microdomains

Q1. If a microdomain in a bacterial chromosome contains about 10,000 base pairs (bp), and each base pair is 0.34 nanometers (nm) in length, how long (in micrometers) would the microdomain be if stretched out linearly?

Background

Topic: Chromosome Structure in Bacteria

This question tests your understanding of the physical dimensions of DNA and how to convert between units to determine the length of a DNA segment based on the number of base pairs it contains.

Key Terms and Formulas

• Base pair (bp): A pair of complementary nucleotides in a double-stranded DNA molecule.

• Length per base pair: $0.34$ nm per bp

• Conversion: $1$nm$= 0.001$micrometers ($\mu$m)

• Formula: $\text{Total length} = \text{number of bp} \times \text{length per bp}$

Step-by-Step Guidance

1. Start by identifying the number of base pairs in the microdomain: $10,000$ bp.

2. Recall that each base pair is $0.34$ nm in length.

3. Multiply the number of base pairs by the length per base pair to get the total length in nanometers: $10,000 \times 0.34$ nm.

4. Convert the result from nanometers to micrometers by dividing by $1,000$ (since $1 \mu$m = $1,000$ nm).

Try solving on your own before revealing the answer!

Final Answer: 3.4 micrometers ($\mu$m)

$10,000 \times 0.34 = 3,400$ nm. $3,400$ nm $\div 1,000 = 3.4$ $\mu$m.

This means that if the DNA in a microdomain were stretched out linearly, it would be 3.4 micrometers long.

Q2. If a bacterial microdomain was circular, what would be its diameter? (Use the formula: Circumference = $\pi D$, where $D$ is the diameter of the circle.)

Background

Topic: DNA Organization and Geometry

This question asks you to apply geometric reasoning to DNA structure, specifically relating the linear length of DNA to the diameter of a circular loop.

Key Terms and Formulas

• Circumference ($C$): The total length around a circle.

• Diameter ($D$): The distance across the circle through its center.

• Formula: $C = \pi D$

• Known value: The circumference is the linear length of the DNA, which you calculated in the previous part.

Step-by-Step Guidance

1. Recall the total linear length of the microdomain from the previous part (in micrometers).

2. Set this length equal to the circumference of the circle: $C = \pi D$.

3. Rearrange the formula to solve for the diameter: $D = \frac{C}{\pi}$.

4. Plug in the value for $C$ (the length you found in part A) and use $\pi \approx 3.14$.

Try solving on your own before revealing the answer!

Final Answer: Approximately 1.08 micrometers ($\mu$m)

$D = \frac{3.4}{3.14} \approx 1.08$ $\mu$m.

This is the diameter of the circular microdomain if the DNA is arranged in a perfect circle.

Q3. Is the diameter of the circular microdomain calculated in part B small enough to fit inside a bacterium with a cytoplasm about 0.5 $\mu$m wide and 1.0 $\mu$m long?

Background

Topic: DNA Packaging in Bacteria

This question tests your ability to compare calculated DNA dimensions to the physical dimensions of a bacterial cell, to assess whether the DNA can fit inside the cell in the given configuration.

Key Terms and Formulas

• Bacterial cell dimensions: Width = 0.5 $\mu$m, Length = 1.0 $\mu$m

• Microdomain diameter: Value calculated in part B

Step-by-Step Guidance

1. Recall the diameter of the circular microdomain from part B.

2. Compare this diameter to the width and length of the bacterial cytoplasm.

3. Consider whether a circle of this diameter could fit within the cell's dimensions.

4. Think about the implications for DNA packaging if the diameter is larger than the cell's width or length.

Try solving on your own before revealing the answer!

Final Answer: The diameter (1.08 $\mu$m) is larger than the width (0.5 $\mu$m) and about the same as the length (1.0 $\mu$m) of the bacterium.

This means that a circular microdomain of this size would not fit comfortably inside the bacterial cell, especially across its width. This illustrates why DNA must be highly compacted and organized within the cell.