Minimum Codon Length for 20 Amino Acids with 6 Nitrogenous Bases

[toothpaste666]

Homework Statement

Suppose we discovered a form of life like our own which utilized 20 amino acids, but had 6 nitrogeneous bases instead of the 4 our life uses. What would be the minimum length of a codon which could represent all 20 currently used amino acids?

The Attempt at a Solution

I am a little confused about what I am being asked here. Would this mean that the length of a codon must be divisible by 6 making 24 the minimum?

 

[jedishrfu]

What is the definition of a codon?

How many possible codes can each digit of the codon be?

How many digits in a codon?

Now think about your problem and the answer should come to you.

 

[toothpaste666]

Can you explain what a digit of a codon is? I have not heard that term. This is not even for a bio class we are briefly going over this stuff in an astronomy class and I have never heard most of it before

 

[toothpaste666]

I know the codon is the three part "word" that codes for an amino acid. does this mean each digit is a letter in that word? so since each codon can code for one amino acid and there are 20 amino acids does that mean it would take 6x20 or 120 codons to code them all?

 

[toothpaste666]

actually i think i confused myself. in the problem there are 6 bases so each digit can be 1 of 6 different codes. there are 3 digits in the codon so each codon so that means there are 18 possible codons right?

 

[jedishrfu]

Sorry you are right each codon has three letters and each letter has 4 values so the maximum is 4x4x4 = 64 but since nature is redundant it reduces down to 20.

So in your case you have 6 bases instead of 4 bases so how many possible choices are there?

 

[toothpaste666]

6x6x6 = 216 possible codons. but in order to represent all 20 amino acids i wouldn't need all of these like you said right? how do i figure out what the minimum is? you said the 64 reduces to 20 but how come 20?

 

[jedishrfu]

It not that it reduces its that nature chose to be redundant mapping multiple codon codes to one of the 20 choices.

So now you know it can't exceed 216 choices for 6 bases. I guess if you compare them you might argue that 20/64 ratio would hold and come up with a practical number. However I'd still make it clear about where the 216 came from in case that's what the prof wanted.

 

[Borek]

You can always find the minimum number using logarithms, but it is not necessary, just trial and error will give answer much faster.

Imagine 2 letter codons (using 6 bases) - are there enough different codons to cover all necessary 20 aminoacids?

 

[toothpaste666]

it would be 6x6 = 36 so there would be enough i think

 

[toothpaste666]

but you are saying it would be 20/216 to find the minimum number?

 

[toothpaste666]

or wait i think you mean there is a ratio of 20/64 and we have 216 so the answer would be 216(20/64) = 67.5

 

[Borek]

it would be 6x6 = 36 so there would be enough i think

And that's the correct answer.

 

[toothpaste666]

so the minimum of the codon is 2 letters?

 

[Borek]

Yes. With 6 letters 1 letter codon will be able to code up to 6 aminoacids. 2 letter codon would be capable of coding up to 36 aminoacids and so on. 20 is more than 6, so it needs more than 1 letter, but less than 36, so it doesn't need more than 2.

 

[toothpaste666]

thank you !