The template strand of a short bacterial gene reads, from its 3'-end to its 5'-end: 3'-TAC AAA CCG…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
The template strand of a short bacterial gene reads, from its -end to its -end:
-TAC AAA CCG TTT CTC ACC ATT-
(The spaces only mark groups of three bases.) Use these facts: AUG is the start codon; UAA, UAG and UGA are stop codons; every other codon in this gene codes for an amino acid.
- How many nucleotides does the mRNA of this gene have, from the start codon to the end of the stop codon?
- How many amino acids does the polypeptide have?
- How many hydrogen bonds hold the two strands of this stretch of DNA together?
- A point mutation changes the middle base of the sixth group of the template (ACC) from C to T. How many amino acids does the polypeptide have now?
- Instead, three bases are inserted between the second and third groups, forming one new codon that codes for an amino acid. How many amino acids does the polypeptide have now?
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
The mRNA. RNA polymerase reads the template strand and makes a complementary RNA, with U pairing with A. Opposite -TAC AAA CCG TTT CTC ACC ATT- it makes -AUG UUU GGC AAA GAG UGG UAA-. This is the same as the coding strand, with U in place of T.
(a) Length. codons of three bases: nucleotides.
(b) Amino acids. Translation starts at AUG and goes codon by codon. AUG, UUU, GGC, AAA, GAG and UGG each add an amino acid, and UAA stops it: amino acids.
(c) Hydrogen bonds. The stretch has A-T pairs and G-C pairs: hydrogen bonds.
(d) Point mutation. The template group ACC (read to ) gives the codon UGG. With T in place of its middle C the group is ATC, which gives UAG, a stop codon. Translation now stops after the fifth amino acid: amino acids.
(e) Insertion of three bases. Three bases make one whole codon, so one amino acid is added, and the reading frame after it is unaltered: amino acids.
Answer: (a) nucleotides. (b) amino acids. (c) hydrogen bonds. (d) amino acids. (e) amino acids.
Check. Count the codons: codons, of which is a stop codon, leave amino acids. The insertion makes codons and amino acids.
Watch out. The template is written here from to . Reading it the other way, or copying the template instead of its complement, gives a wrong mRNA. Inserting one or two bases, instead of three, would change the reading frame from that point on.
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