SOLUTION: The protein-coding strand of the average human gene consists of 1,350 nucleotides. Assuming that each nucleotide can take any of four values (A, T, C, or G), how many different gen

Question 1197117: The protein-coding strand of the average human gene consists of 1,350 nucleotides. Assuming that each nucleotide can take any of four values (A, T, C, or G), how many different genes with exactly 1,350 nucleotides are possible?

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

There are 4 choices for slot 1, then 4 choices for slot 2, then 4 choices for slot 3, and so on. We continue this pattern having 4 choices until reaching slot 1350

We'll have 1350 copies of "4" multiplied together
4*4*4*...*4 = 4^1350
Exponents are very useful to represent repeated multiplication.

My calculator says "overflow" when trying to calculate 4^1350
Your calculator may have the same result.

When using WolframAlpha, it says
4^1350 = 6.039*10^812
which is approximate.
This helps show how massive this number really is
The 812 exponent means "move the decimal point in 6.039 exactly 812 spaces to the right". This large number will have 812-3 = 809 zeros after the 6039 string
I.e. it's roughly 6039 followed by 809 zeros.
Something like 1 billion = 1,000,000,000 = 1 followed by 9 zeros to try to get a sense of scale.

In any event, I would write the answer as 4^1350 as it's easier to think about.
Ask your teacher for further clarification.

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