Question

Calculate the number of clones required to obtain with a probability of 0.99 a specific 5-kb fragment from C. elegans (Table 3-3).

Short answer

The number of clones is$8.91\times {10}^4$.

Step-by-step solution

Introduction

The production of multiple identical organisms from a single ancestor is known as cloning. The term "clone" refers to a group of cells that have been infected with a vector that contains the identical desired DNA.

Number of clones to obtain the probability

Shotgun cloning is a technique for reproducing genomic DNA that is governed by probability principles. The probability (P) that a collection of N number clones has a fragment that makes up a fraction (f) of an organism's genome can be calculated by the following equation:

$\mathrm{N}=\frac{\log (1-\mathrm{P})}{\log (1-\mathrm{f})}$

Given,

$$\begin{gathered} \mathrm{Probability}\mathrm{of}\mathrm{clones}=0.99 \\ \mathrm{Size}\mathrm{of}\mathrm{the}\mathrm{fragments}=5\mathrm{kb} \end{gathered}$$

The genome size of C.elegans is 97,000 kb.

Therefore, fraction can be calculated as:

f=size of the fragment/ genome size

f=5/97000

f=5.2x10^-5

On substituting the values in the equation,

$\mathrm{N}=\frac{\log (1‐0.99)}{\log (1‐5.2\times {10}^{-5})}$

Therefore,

$\mathrm{N}=8.91\times {10}^4$

Thus, 8.91×10⁴ isthe required number of clones produced by using the shotgun cloning method that subjects the laws of probability.