DNA: replication, transcription and translation

If you are in Year 4 at a liceo scientifico (the Italian science-focused high school), in October science usually moves on to molecular biology: DNA, how it is copied, how it becomes protein. Often the topic has just started or is about to start, and the test that includes it falls in roughly six to nine weeks, between mid-November and December. There is time to build a solid framework, and it is worth doing it now, because gene regulation and mutations, which come afterwards, all rest on this part.

The thread of logic is a single one: DNA → RNA → protein. Replication copies the DNA, transcription produces a messenger RNA from it, translation uses the mRNA to build the protein. For cell division, where the DNA is copied before being distributed, I wrote mitosis and meiosis with exercises; for the cell and biomolecules of the first two years there is the post on the cell and biomolecules; for the admission-test angle, cell and molecular biology for Medicine. Here I concentrate on the mechanism, step by step.

The structure of DNA, in brief

DNA is a double helix of two antiparallel strands: one runs in the 5’→3’ direction, the other in 3’→5’. Each nucleotide has a sugar (deoxyribose), a phosphate group and a nitrogenous base. The bases always pair in the same way:

Adenine and thymine form two hydrogen bonds, guanine and cytosine three. A useful consequence for exercises: in double-stranded DNA $A=T$ and $G=C$, so $A+G=T+C$.

Replication: semi-conservative

Before the cell divides, the DNA has to be copied. The process, in order:

1. helicase opens the double helix, forming the replication fork;
2. primase builds a short RNA primer;
3. DNA polymerase adds complementary nucleotides, only in the 5’→3’ direction;
4. DNA ligase joins the fragments.

Since the strands are antiparallel and the polymerase works in only one direction, one strand (the leading, or fast, strand) is synthesised in a single piece, following the fork. The other (the lagging, or slow, strand) is synthesised in stretches, the Okazaki fragments, later joined by the ligase.

The result is semi-conservative: each daughter molecule has one old strand and one new strand. After $n$ cycles starting from one molecule you have $2^n$ molecules, of which always and only 2 contain an original strand.

Transcription: from DNA to mRNA

RNA polymerase binds to a region of the DNA (the promoter), opens the helix and builds a strand of mRNA by reading only one strand of DNA, the template, in the 3’→5’ direction; the mRNA therefore grows in the 5’→3’ direction. Two differences from DNA to keep in mind:

• in RNA the sugar is ribose;
• in place of thymine there is uracil ($A\leftrightarrow U$).

The mRNA therefore has the same sequence as the coding strand of the DNA (the one that is not read), with $U$ in place of $T$. In eukaryotes the mRNA is processed before leaving the nucleus (introns cut out, exons spliced together), but for basic exercises you work on the mRNA that is already ready.

The genetic code and translation

The genetic code is made of triplets: each group of three bases of the mRNA, a codon, specifies an amino acid or a signal. A few codons to know:

The code is degenerate (several codons for the same amino acid) but not ambiguous (each codon indicates only one). In the test the full table is usually provided; what is assessed is whether you can read it.

Translation takes place on the ribosome: the ribosome attaches to the mRNA, finds the AUG and reads one codon at a time. Each tRNA carries an amino acid and has an anticodon complementary to the codon; the chain of amino acids grows until a stop codon arrives, and then the protein is released.

Typical mistakes

• T and U mixed up. In mRNA there is never a T; in DNA there is never a U.
• Template and coding strand swapped. The mRNA is complementary to the template, so it is the same as the coding strand (with U). Whoever copies it from the template without complementing gets everything wrong.
• Antiparallel ignored. If you read a template written 3’→5’, the mRNA comes out 5’→3’. Always write the ends and, when you have to report the sequence, rewrite it 5’→3’.
• The stop codon counted as an amino acid. It is not: there is no corresponding tRNA.
• Replication and transcription mixed up. Replication: all the DNA, both strands, DNA polymerase. Transcription: one gene, one strand only, RNA polymerase.

Exercise 1: from the DNA sequence to the protein

A stretch of template strand is, written 3’→5’:

Find the mRNA and the sequence of amino acids.

Solution. The mRNA is complementary to the template, with U in place of T, and is written 5’→3’. Complementing base by base (T→A, A→U, C→G, G→C):

I split it into codons: AUG · UUU · GGC · GAA · AAA · UGG · GCU · CUG · UCU · UAA. I translate: Met – Phe – Gly – Glu – Lys – Trp – Ala – Leu – Ser – stop. The protein has 9 amino acids (30 nucleotides = 10 codons, of which the last is the stop).

The coding strand is therefore 5’-ATGTTTGGCGAAAAATGGGCTCTGTCTTAA-3’: the same sequence as the mRNA, with T.

Exercise 2: a mutation

In the same gene, the G of the codon GAA (the fourth) is replaced by a U in the mRNA. What changes?

Solution. The codon becomes UAA, that is, a stop. Translation stops at the fourth codon: the protein is Met – Phe – Gly, 3 amino acids instead of 9. It is a nonsense mutation and usually produces a non-functioning protein. In the coding DNA the change is GAA → TAA.

Exercise 3: calculations on DNA and replication

A fragment of double-stranded DNA has 400 nucleotides in total and A is 30 %. How many bases of each type?

Solution. $A=T$, so $A+T=60\%$ and $G+C=40\%$, that is, $G=C=20\%$. Out of 400 nucleotides: $A=T=120$ and $G=C=80$. Check: $120+120+80+80=400$.

If this fragment is replicated 3 times in a row, how many molecules are obtained and how many contain an original strand?

Solution. After 3 cycles: $2^3=8$ molecules. The two original strands are still present, one in each of 2 molecules: 2 molecules are mixed, the other 6 are made only of new strands.

Finally: a gene codes for a protein of 29 amino acids. How many nucleotides does the coding region have, excluding the stop, and how many with the stop? Each amino acid is a codon of 3 nucleotides: $29\times 3=87$. Adding the stop codon (3 nucleotides) the gene has 90, that is 30 codons, as in the FAQ below.

Test yourself

If you want to practise on science tests and the other high school subjects, there is the practice test on high school tests: try it after redoing the three exercises above on your own, without looking at the solution. If the step from template to mRNA still confuses you, in my one-to-one online lessons I always start it from a hand-written sequence, step by step.