Question

Assume that a viral DNA molecule is a 11-μm-long strand with a uniform 20-Å diameter. Assuming complete flexibility of the molecule, will it fit inside a viral head that is a 8×10−2-μm-diameter sphere?

Part A)What is the volume of the viral head in cubic angstrom (Å3)? Hint: Recall that the volume of a sphere can be calculated using this formula: Volume= (4/3) x pi x r2, where r stands for radius of the sphere. Also note that 1Å=0.1 nm. Express your answer to three significant digits.

Part B)

What is the volume of the DNA molecule in cubic angstrom (Å3)?

Hint: Think of the DNA molecule as a cylindrical shape. The volume of a cylinder can be calculated using this formula: Volume= pi x r² x h, where r stands for radius and h stands for height.

Express your answer to three significant digits.

Answer 1

To determine whether the viral DNA will fit inside the viral head, the volume occupied by the viral DNA as well as the viral head needs to be calculated.

PART A

Volume of the viral head

Since the volume asked here is in cubic angstrom (ų), all the units need to be converted to Å

1 Å = 10^-10 m = 10^-4 μm,

therefore 1μm = 10⁴ Å

diameter of viral head = 8 x 10^-2 μm = 8 x 10^-2 x 10⁴ Å = 800 Å

radius of viral head (r) = 800 Å = 400 Å

volume of the viral head (sphere) = (4/3)πr³ = (4/3) x 3.14 x (400Å)³

= 267.947 x 10⁶ ų

PART B

Volume of the viral DNA molecule

Since the volume asked here is in cubic angstrom (ų), all the units need to be converted to Å

1 Å = 10^-10 m = 10^-4 μm

therefore 1μm = 10⁴ Å

diameter of viral DNA = 20 Å

radius of viral DNA (r) = 20 Å / 2 = 10 Å

length (height h) of viral DNA = 11 μm = 11x 10⁴ Å

volume of viral DNA (cylinder) = πr²h = 3.14 x (10 Å)² x (11 x 10⁴ Å)

= 34.540 x 10⁶  Å³

From the calculations we see that the viral DNA is much larger compared to the viral DNA volume. Since the volume of viral DNA is many folds lower than the volume of the viral head, we can say that the viral DNA can easily fit inside the viral head.