Back to homeBiology · 5th Year OL · Change subject
Start for free
5th Year OL · Biology
Molecular genetics · From DNA to protein

Gene expression

Imagine you are in Leaving Cert Biology class in Galway and you ask how a gene in one of your cheek cells ends up as a protein. The sequence of a gene is of no use until the cell uses it 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.

Transcription: from gene to messenger RNA

In eukaryotic cells the DNA stays confined within the nuclear envelope, 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 5' 3' 3' 5' coding strand DNA: template strand (read 3'→5', right to left) promoter RNA polymerase moves left mRNA: copy (5'→3') 5' 3' finished mRNA leaves for the cytoplasm mRNA 5' ribosome nuclear pore nuclear envelope

Try the widget below: complete the complementary strand of this short DNA sequence (A pairs with T, G with C). In transcription the same pairing applies, except that RNA uses U instead of T.

Translation is the process in which the base sequence of an mRNA is used to build a polypeptide; in eukaryotic cells it takes place on ribosomes in the cytoplasm (cytosol or rough endoplasmic reticulum), whereas transcription takes place in the nucleus.

Flow of genetic information: DNA is copied into new DNA by replication, DNA is transcribed into RNA (arrow 1) inside the nucleus, and RNA is translated into protein (arrow 2) in the cytoplasm. nucleus DNA RNA protein 1 transcription 2 translation replication

Flow of genetic information in a cell: DNA is copied into new DNA by replication (the arrow looping back on DNA); DNA is copied into RNA by transcription; RNA is used to make protein by translation. In a diagram, the area inside a dashed line usually marks the compartment where a process happens, such as the nucleus, which contains the DNA and where transcription occurs.

Practice★☆☆Not started
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

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

A short extract of the genetic code (mRNA codons, read 5'→3'):

Codon Amino acid Anticodon (5'→3')
AUG Methionine (Met), start CAU
GCU Alanine (Ala) AGC
CCU Proline (Pro) AGG
GGC Glycine (Gly) GCC
UAU Tyrosine (Tyr) AUA
UAA, UAG, UGA Stop none (no tRNA)

The tRNA anticodon is complementary and antiparallel to the codon it reads.

Practice★★★Not started
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
Practice★★★Not started
In gene therapy for Leber's hereditary optic neuropathy, researchers are investigating transferring a wild-type copy of a mitochondrial gene to the cell nucleus…

In gene therapy for Leber's hereditary optic neuropathy, researchers are investigating transferring a wild-type copy of a mitochondrial gene to the cell nucleus (allotopic expression) so that the protein is translated in the cytosol and imported into the mitochondrion. However, in the human mitochondrial genetic code, the triplet UGA codes for tryptophan, whereas in the standard nuclear code it acts as a termination codon.

  1. a) Predict what would happen if the mitochondrial gene is transferred to the nucleus without modifying its sequence and is transcribed and translated on cytoplasmic ribosomes. What impact would this have on the length and functionality of the resulting polypeptide?
  2. b) Design the site-directed mutagenesis strategy that the biotechnology team must apply to the coding sequence of the transgene before integrating it into the nucleus to ensure the synthesis of the native protein.
  3. c) Justify why the mitochondrial genome (with only $37$ genes and a very small number of tRNAs) could 'afford' drift and changes in codon assignments throughout evolution, whereas in the nuclear genome this type of alteration is lethal to the cell.
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

mRNA processing in eukaryotes

When a researcher in UCD or Trinity measures the length of the RNA freshly transcribed in the nucleus and compares it with that of the mRNA entering the ribosome, they 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
Practice★☆☆Not started
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.

Solve
Practice★★★Not started
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
Practice★★★Not started
A human gene of $24\,000$ base pairs (bp) consisting of $8$ exons and $7$ introns is studied. After being completely processed, its mature mRNA in muscle cells is $1200$…

A human gene of $24\,000$ base pairs (bp) consisting of $8$ exons and $7$ introns is studied. After being completely processed, its mature mRNA in muscle cells is $1200$ nucleotides long (including the UTR regions and the poly-A tail).

  1. a) If the template strand of genomic DNA is hybridized in the laboratory with said mature muscle mRNA and the resulting heteroduplex is observed under an electron microscope, predict what morphology it will present and how many single-stranded loops will be observed at most. Justify which nucleic acid each loop will be made of.
  2. b) A student claims that, having $1200$ nucleotides, the resulting protein will be exactly $400$ amino acids long. Refute this statement with reasoning, providing at least two arguments of a molecular nature.
  3. c) In nervous tissue, a mature mRNA originating from the same gene but with a size of $1500$ nucleotides is isolated. Reason what phenomenon has occurred and predict whether the neuronal protein will necessarily be longer than the muscular one, justifying your answer.
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.
Practice★★★Not started
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
Practice★★★Not started
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
Practice★★★Not started
The following diagram represents a functional ribosome during the elongation phase of translation: a) Justify and deduce the polarity ($5'$ and $3'$ ends) of the mRNA…

The following diagram represents a functional ribosome during the elongation phase of translation:

mRNA AUG GCU UAA UAC ? stop codon
  1. a) Justify and deduce the polarity ($5'$ and $3'$ ends) of the mRNA strand and the direction in which the ribosome will move along it.
  2. b) Write the sequence of the anticodon of the tRNA marked with a question mark (?), mandatory specifying its $5'\to 3'$ polarity, and reason which amino acid it carries knowing that the GCU codon codes for alanine.
  3. c) In a mutant strain, a suppressor tRNA is produced that is capable of recognizing the UAA codon by inserting the amino acid tyrosine. Predict what will happen to translation and to the length of the resulting protein when the ribosome reaches the position marked with the stop codon compared to the wild-type cell.
Solve

Summary

Let's practice! 0 medium exercises in this section
See all the exercises in this section