Protein synthesis, mutation and cystic fibrosis

What you’ll learn

• Trace information from a DNA template to a polypeptide.
• Explain triplet, non-overlapping and degenerate coding.
• Link different mutations to possible protein and whole-organism effects.

Understand the idea

A protein-coding gene contains information for a polypeptide sequence. During transcription, RNA polymerase uses the template, or antisense, DNA strand to produce complementary RNA. In a eukaryotic cell, mature messenger RNA reaches ribosomes in the cytoplasm. During translation, transfer RNAs bring amino acids; their anticodons pair with mRNA codons. The ribosome joins amino acids by peptide bonds. A start codon establishes the reading frame and a stop codon ends translation without specifying an amino acid.

The code is triplet because three bases form a codon, non-overlapping because each base is read in one codon within a reading frame, and degenerate because more than one codon can specify the same amino acid. A base substitution may be silent, change one amino acid, or create a stop codon. An insertion or deletion not divisible by three can shift the reading frame. Some mutations have no detectable effect, whereas others disrupt important proteins or the regulation of cell division, contributing to a genetic disorder or cancer.

Cystic fibrosis results from particular variants of the CFTR gene, affecting a membrane protein involved in chloride transport. Altered ion movement changes water movement and leads to dehydrated, sticky mucus in several organs. Airways can become obstructed and infection-prone; blocked pancreatic ducts reduce delivery of digestive enzymes to the intestine; reproductive effects include obstruction or abnormal development of ducts and altered cervical mucus. The explanation should connect gene, protein, transport and tissue effects instead of saying that the gene directly makes mucus.

Key formulae

$$\text{DNA}\xrightarrow{\text{transcription}}\text{mRNA}\xrightarrow{\text{translation}}\text{polypeptide}$$

$$\text{coding bases}=3\times\text{number of amino-acid codons}$$

Worked example

A DNA template segment is written $3'\text{-TAC GGA-}5'$. Write the complementary mRNA segment, giving its direction. The supplied codon table states AUG = methionine and CCU = proline.

1. RNA is complementary and antiparallel to its DNA template: T pairs with A, A with U, C with G and G with C.
2. The RNA sequence is $5'\text{-AUG CCU-}3'$.
3. Reading the supplied codons in that direction gives methionine followed by proline.

Answer: $5'\text{-AUG CCU-}3'$; methionine–proline.