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Molecular genetics · From DNA to protein

Gene expression

Having the sequence of a gene is of no use until it is used to make the protein it codes for. This use, gene expression, takes place in two stages, transcription and translation, which make up the so-called central dogma of molecular biology: DNA → RNA → protein. This section describes the molecular machinery of both stages, the code that links them and the differences between bacteria and eukaryotes.

Before you start, make sure you know: Gene expression (4th ESO)
Before you start, make sure you know: DNA and RNA (4th ESO)

Transcription: from gene to messenger RNA

In eukaryotic cells the DNA stays confined within the nuclear membrane, but proteins are made in the cytoplasm: how do the instructions of a gene travel to the cell's factory without the DNA leaving the nucleus?

Transcription in a eukaryotic cell: inside the nucleus, RNA polymerase opens the DNA double helix, runs along the template strand in the 3'→5' direction and synthesises a complementary mRNA in the 5'→3' direction; the finished mRNA leaves through a nuclear pore into the cytoplasm, where it binds to a ribosome. nucleus cytoplasm 3' 5' 5' 3' coding strand DNA: template strand (read 3'→5') promoter RNA polymerase moves along mRNA: copy (5'→3') 5' 3' finished mRNA leaves for the cytoplasm mRNA 5' ribosome nuclear pore nuclear membrane
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Which enzyme catalyses the transcription of DNA into RNA? State on which of the two strands of the gene it acts and in which direction it synthesises the RNA molecule.

Which enzyme catalyses the transcription of DNA into RNA? State on which of the two strands of the gene it acts and in which direction it synthesises the RNA molecule.

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State two structural differences between DNA and RNA, and explain which bases pair up during transcription when the template strand shows an adenine or a thymine.

State two structural differences between DNA and RNA, and explain which bases pair up during transcription when the template strand shows an adenine or a thymine.

Solve
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From the DNA template strand 3'-TACGGAACT-5', obtain the resulting mRNA sequence indicating its polarity, and state the maximum number of amino acids this fragment can…

From the DNA template strand 3'-TACGGAACT-5', obtain the resulting mRNA sequence indicating its polarity, and state the maximum number of amino acids this fragment can encode.

Solve

The genetic code

Diagram of the genetic code: the AUG codon of the mRNA, made up of the bases A, U and G, codes for the amino acid methionine and signals the start; there are 64 codons for 20 amino acids, so the code is degenerate. A U G AUG codon (start) methionine 64 codons 20 amino acids → degenerate code
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What is a codon? State how many nucleotides make it up, how many distinct codons exist, and how many different amino acids they encode in total.

What is a codon? State how many nucleotides make it up, how many distinct codons exist, and how many different amino acids they encode in total.

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Look at the diagram of the flow of genetic information. State which molecules are labelled A, B and C, which processes occur at arrows 1 and 2, and what the area…

Look at the diagram of the flow of genetic information. State which molecules are labelled A, B and C, which processes occur at arrows 1 and 2, and what the area surrounded by the dashed line represents.

A 1 B 2 C
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Consider the following fragment of the coding sequence of an mRNA that is translated into a cytoplasmic protein: $5'\text{--- AUG - UCU - UCC - UCA - UCG --- } 3'$…

Consider the following fragment of the coding sequence of an mRNA that is translated into a cytoplasmic protein:

$5'\text{--- AUG - UCU - UCC - UCA - UCG --- } 3'$

(Note: the four codons following the AUG start codon all code for the amino acid serine).

  1. a) Based on this specific case, justify why the genetic code is degenerate and demonstrate that this property is not equivalent to saying that the code is "ambiguous".
  2. b) Suppose that two independent point mutations are introduced: Mutant 1 changes the second codon from UCU to UCC; Mutant 2 changes the third codon from UCC to UAC (which codes for tyrosine). Compare the impact of both mutations on the amino acid sequence of the protein and classify each mutation according to its functional effect.
  3. c) Reason what adaptive advantage is conferred by the fact that the degeneracy of the genetic code is almost always concentrated at the third base of each triplet (wobble hypothesis) rather than at the first or second, in the face of spontaneous point mutations.
Solve

Translation: from mRNA to protein

The mRNA is written with an alphabet of only 4 nucleotides ($A, U, G, C$), whereas proteins contain 20 different amino acids: how does the cell translate a four-letter text into a functional polypeptide sequence?

Animation of translation: a tRNA brings an amino acid from the right to the codon of the mRNA, the amino acid joins the chain, the tRNA leaves, the mRNA moves on one codon and the polypeptide grows. ribosome mRNA read codon by codon chain of amino acids = protein
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What is the start codon of translation and which amino acid does it encode? Explain what function the stop codons fulfil and what happens when the ribosome reaches one…

What is the start codon of translation and which amino acid does it encode? Explain what function the stop codons fulfil and what happens when the ribosome reaches one of them.

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Briefly describe what the ribosome does during translation: how it reads the mRNA, how the amino acids arrive and what type of bond is formed between them.

Briefly describe what the ribosome does during translation: how it reads the mRNA, how the amino acids arrive and what type of bond is formed between them.

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During translation, the mRNA is read in consecutive codons, with no separators between them. Explain what the reading frame is and reason what would happen to the…

During translation, the mRNA is read in consecutive codons, with no separators between them. Explain what the reading frame is and reason what would happen to the synthesised protein if the ribosome started reading one nucleotide before the correct position.

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mRNA processing in eukaryotes

When you measure in the laboratory the length of the RNA freshly transcribed in the nucleus and compare it with that of the mRNA entering the ribosome, you find that the mature molecule is often several times shorter: where do the missing fragments go?

Diagram of splicing in eukaryotes: the primary mRNA, with three exons and interspersed introns, loses the introns and leaves a mature mRNA made up only of the joined exons. Exon 1 Intron 1 Exon 2 Intron 2 Exon 3 Primary mRNA: exons (colour) + introns (grey) splicing Exon 1 Exon 2 Exon 3 Mature mRNA: only exons joined
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Define what a gene is from the point of view of gene expression and state which two types of functional products a gene can generate when it is expressed.

Define what a gene is from the point of view of gene expression and state which two types of functional products a gene can generate when it is expressed.

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A protein is made up of $300$ amino acids. a) Calculate how many nucleotides the mature mRNA that encodes it has, at a minimum, including the stop codon. b) Calculate…

A protein is made up of $300$ amino acids. a) Calculate how many nucleotides the mature mRNA that encodes it has, at a minimum, including the stop codon. b) Calculate how many base pairs the gene would occupy in an organism without introns. c) Reason why in eukaryotic organisms the gene is usually much longer than the corresponding mature mRNA.

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In the human genome, the fibronectin gene contains numerous exons and introns. In fibroblasts, it gives rise to an insoluble protein isoform that forms part of the…

In the human genome, the fibronectin gene contains numerous exons and introns. In fibroblasts, it gives rise to an insoluble protein isoform that forms part of the extracellular matrix, whereas in hepatocytes it originates a soluble isoform that circulates in the blood plasma.

  1. a) Justify the molecular mechanism by which a single genomic DNA sequence can give rise to these two proteins with such different functions and locations.
  2. b) A patient presents a point mutation consisting of a transition (G to A) at the splicing acceptor site of intron 4 of this gene. Predict two possible alterations in the structure of the resulting mature mRNA and their consequences for the synthesized protein.
  3. c) If we clone this intact human gene (with all its introns and exons) into a plasmid and introduce it into Escherichia coli, reason whether the bacterium will synthesize functional human fibronectin.
Solve

Gene expression in bacteria and in eukaryotes

Comparison of gene expression: in the bacterium, which has no nucleus, ribosomes translate the mRNA while RNA polymerase is still transcribing it; in the eukaryotic cell, the pre-mRNA is transcribed and matured by splicing in the nucleus, leaves through a nuclear pore and is translated afterwards in the cytoplasm. Bacterium: no nucleus DNA RNA polymerase movement ribosomes mRNA 5' polypeptide cytoplasm Coupled: ribosomes translate the mRNA while RNA polymerase is still transcribing it. Eukaryote: with a nucleus nucleus pre-mRNA introns splicing mature mRNA nuclear envelope nuclear pore cytoplasm ribosomes 5' 3' polypeptides Separated: transcription and maturation in the nucleus; translation happens afterwards, in the cytoplasm.
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Streptomycin is a bactericidal antibiotic that binds to the $30\text{S}$ subunit of the prokaryotic ribosome ($70\text{S}$), causing mRNA misreading and inhibiting…

Streptomycin is a bactericidal antibiotic that binds to the $30\text{S}$ subunit of the prokaryotic ribosome ($70\text{S}$), causing mRNA misreading and inhibiting protein synthesis. Although it is considered selective against eukaryotic cytoplasmic ribosomes ($80\text{S}$), patients undergoing intensive treatments often experience nephrotoxicity and ototoxicity associated with cellular energy dysfunction.

  1. a) Based on the endosymbiotic theory, justify why this drug can damage human cells and cause an ATP deficit despite the described target selectivity.
  2. b) When analyzing a streptomycin-resistant bacterial strain, it is verified that it does not degrade the drug nor does it possess active efflux pumps, but it carries a mutation in the bacterial chromosome gene that encodes the $16\text{S}$ rRNA. Explain how this mutation confers resistance without completely impairing ribosome function.
  3. c) In a cellular experiment, cycloheximide (an $80\text{S}$ ribosome inhibitor) and chloramphenicol (a $70\text{S}$ ribosome inhibitor) are used. Predict what will happen to the assembly of the multienzyme complexes of the mitochondrial electron transport chain (whose subunits have a dual genetic origin) in the presence of each of these compounds separately.
Solve
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To translate the $61$ sense codons in the standard genetic code, most eukaryotic cells only need to synthesize between $30$ and $45$ different tRNA species, instead of…

To translate the $61$ sense codons in the standard genetic code, most eukaryotic cells only need to synthesize between $30$ and $45$ different tRNA species, instead of $61$.

  1. a) Explain the thermodynamic and structural basis of Crick's 'wobble hypothesis' and justify why this spatial flexibility is restricted to the third nucleotide of the codon ($3'$ end) and the first of the anticodon ($5'$ end).
  2. b) The amino acid valine is coded by four codons: $5'\text{-GUU-}3'$, $5'\text{-GUC-}3'$, $5'\text{-GUA-}3'$, and $5'\text{-GUG-}3'$. If a specific tRNA has inosine (I) at the $5'$ end of its anticodon (knowing that inosine can pair with U, C, and A), determine how many and which of these codons this single tRNA will be able to read, writing the sequence of said anticodon in the $5'\to 3'$ direction.
  3. c) Reason the adaptive advantage for the organism of having the degeneracy of the genetic code and wobble concentrated at the third position of the codon, compared to them occurring at the first or second base when point mutations occur.
Solve
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When observing gene expression in bacteria (Escherichia coli*) by electron microscopy, 'Christmas tree' structures can be distinguished where multiple ribosomes…

When observing gene expression in bacteria (Escherichia coli) by electron microscopy, 'Christmas tree' structures can be distinguished where multiple ribosomes translate mRNA while it is still being synthesized by RNA polymerase coupled to genomic DNA. In contrast, this is never observed in a human cell.

  1. a) Predict what immediate consequence the addition of rifampicin (an inhibitor that blocks RNA polymerase initiation but not ribosome activity) will have on bacterial protein synthesis, and compare this effect with what would occur in the short term in the cytoplasm of a eukaryotic cell.
  2. b) Reason why bacterial transcription-translation coupling is mechanically incompatible with the presence of introns and with the splicing mechanism typical of eukaryotes.
  3. c) The gene regulation mechanism by 'attenuation' in the bacterial tryptophan operon depends on the ribosome's rate of progress through a leader sequence determining the formation of termination hairpins in the nascent RNA. Justify in a reasoned manner why this regulatory system is unfeasible in eukaryotic cells.
Solve

Following a tRNA through the ribosome

A charged tRNA brings an amino acid to the A site; the P site holds the tRNA attached to the growing peptide; the E site lets the uncharged tRNA leave. The initiator tRNA is an exception: it is initially placed in P. The peptidyl transferase activity of the ribosome catalyses the peptide bond.

Ribosome reading the mRNA from 5′ to 3′, with a tRNA in each site: aminoacyl-tRNA entering at A, peptidyl-tRNA with the growing peptide at P and uncharged tRNA leaving through E. movement of the ribosome (reading 5′ → 3′) enters aminoacyl-tRNA tRNA leaves uncharged growing peptide C 5′ 3′ mRNA N E P A

Summary

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