The genome of Elradicus hypotheticus consists of 3 x 105 bp of B-DNA.
a. How many turns are present in the double-helix?
b. What is the length of the DNA in μm?
To calculate the number of turns in the double helix, we need to consider the helical pitch of B-DNA. The helical pitch is the distance covered by one complete turn of the DNA helix. In B-DNA, the helical pitch is approximately 10.4 base pairs per turn (bp/turn).
a. To calculate the number of turns in the double helix, we divide the total number of base pairs by the number of base pairs per turn:
Number of turns = Total number of base pairs / Base pairs per turn
In this case, the total number of base pairs is 3 x 10^5 bp:
Number of turns = 3 x 10^5 bp / 10.4 bp/turn
Calculating this equation will give us the number of turns in the double helix of Elradicus hypotheticus.
b. To calculate the length of the DNA in μm (micrometers), we need to know the length of one complete turn of the helix. The length of one complete turn is equal to the product of the number of base pairs per turn and the distance between adjacent base pairs, which is approximately 0.34 nm.
Length of one complete turn = Base pairs per turn x Distance between adjacent base pairs
We can then convert this length from nanometers (nm) to micrometers (μm) by dividing by 1000.
Length in μm = Length in nm / 1000
To calculate the length of the DNA, we multiply the number of turns by the length of one complete turn and then convert it to micrometers:
Length of DNA = Number of turns x Length of one complete turn (in nm) / 1000
Now that we have explained how to calculate the answers, let's calculate them.
Let's plug in the values in the equations and solve them:
a. Number of turns = 3 x 10^5 bp / 10.4 bp/turn
b. Length of DNA = Number of turns x Length of one complete turn (in nm) / 1000
By solving these equations, we will obtain the answers to the questions.