Chapter 23, Problem 85A

Solution

Interpretation : The percentage of the size of the human genome that the E.coli DNA represent is to be compared.

Concept Introduction : Deoxyribonucleic acid, or DNA, is made up of two long chains of nucleotides that are spirally coiled together. The nitrogenous bases that makeup DNA are adenine (A), guanine (G), cytosine (C), and thymine (T).

The percentage of the E.coli DNA represents $0.14\%$ of the human genome.

Escherichia coli has $\text{4}\text{.2}\times {\text{10}}^6$ base pairs of DNA.

The human cell has $3\times {\text{10}}^9$ base pairs of DNA.

The percentage of E.coli in the human genome is given as:

  $\text{\% =}\frac{\text{E}\text{.coli}\text{base}\text{pair}}{\text{human}\text{cell}\text{base}\text{pair}}\text{×100}$

To calculate the percentage of the size of E.coli base pair in the human genome, substitute the values in the above formula:

  $\begin{array}{c}\text{\% =}\frac{\text{E}\text{.coli}\text{base}\text{pair}}{\text{human}\text{cell}\text{base}\text{pair}}\text{×100}\\ \text{=}\frac{\text{4}\text{.2}\times {\text{10}}^6}{3\times {\text{10}}^9}\times 100\\ \%=0.14\end{array}$