409/CH1/EX1.9/Example1_9.sce


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clear;
clc;

// Example 1.9
printf('Example 1.9\n\n');
//Page no. 29
// Solution

um = 3 ;//[kb]
kb = 1000 ;//[bp]
bs_prs = (3*um*kb)/(1*1);
printf('The number of base pairs are %i bp. \n',bs_prs);