Question

Question #2dd67

Answer 1

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.