Chalk−1

Biology · Introductory biology · Worked example

Find the length of a protein from its mRNA

The human β-globin mRNA is 628 bases long, not counting its poly-A tail. Its coding sequence runs from base 51, the A of the start codon, to base 494, the last base of the stop codon. How many codons does it hold, and how many amino acids are in the chain? How long does a ribosome take to make the chain at 6 amino acids per second?

Count the coding bases

The coding sequence includes both of its end bases, so add 1 to the difference of the positions.

494−51+1=444

Count the codons

Each codon is three bases, read in frame from AUG.

4443=148

Take away the stop codon

Why take one away? No tRNA pairs with a stop codon, so the last codon adds no amino acid.

148−1=147

Account for the rest of the mRNA

The other bases are the 50 before the start codon and the 134 after the stop. They are part of the mRNA, and the ribosome never reads them as codons.

628−444=184

Find the time

At 6 amino acids per second, the rate measured in mouse stem cells, the chain takes about 25 seconds. This counts only elongation and leaves out the time to start.

1476=24.5

Check against the protein

The protein database lists 147 amino acids for β-globin, which matches. The cell later removes the first methionine, so the chain in hemoglobin has 146, and the sickle-cell change is usually called position 6.

Result

The 444-base coding sequence holds 148 codons, which give a chain of 147 amino acids. The other 184 bases of the mRNA are never read as codons. At 6 amino acids per second a ribosome makes the chain in about 24.5 seconds.

Your turn

1. One like the example

Human preproinsulin, the chain the insulin gene codes for, has 110 amino acids. How many bases long is its coding sequence, stop codon included?

Show the answer and explanation

333 bases.

110 amino acids need 110 codons, and the stop codon makes 111. Each codon is three bases, 3×111=333, which is the length the gene record gives.

110+1=1113⁢(111)=333

2. A variation

An E. coli ribosome adds about 20 amino acids per second. How long does it take to make a 300-amino-acid protein, and how many bases of mRNA does it read, stop codon included?

Show the answer and explanation

About 15 seconds, reading 903 bases.

300 amino acids at 20 per second take 300⁢/20=15 seconds. The ribosome reads 300 codons and the stop, 3⁢(300+1)=903 bases.

30020=153⁢(300+1)=903

3. Somewhere new

Instead of the sickle-cell substitution, suppose the A at base 20 of the β-globin coding sequence were deleted. What would happen to the chain?

Show the answer and explanation

It would be a frameshift. The first six amino acids stay the same, the rest change, and a stop codon comes after 18 amino acids instead of 147.

Removing one base moves every later codon boundary by one. From codon 7 on, the ribosome reads new triplets, Gly, Arg, Ser and so on, until the new frame meets UGA at codon 19. A substitution at the same base changes only one amino acid.

Keep exploring

β-globin opens in Sequence & protein with the sickle-cell change at base 20, GAG to GTG. Then the counts open in Math, starting from 494−51+1=444, and each row you add is checked.

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