# Question 2dd67

Sep 23, 2016

You could synthesize 4.096 × 10^(15) polypeptides.

#### Explanation:

You have 20 choices for the first amino acid in the sequence.

For each of these, you have 20 choices for the second position, another 20 for the third, and so on.

Hence the number of possible choices is

underbrace(20 × 20 × 20 × … × 20 × 20)_color(red)("12 times") = 20^(12) = 4.096 × 10^(15)

Thus, you could synthesize 4.096 × 10^(15)# polypeptides.