AOL 2 : Protein Structure & Translation

Discuss the four levels of protein structure and relate them to the function of a protein.

• Primary structure is the linear sequence of amino acids joined by peptide bonds.

  • By convention it is written from the amino end (N-terminus) to the carboxyl end (C-terminus), so Thr-Ile-Pro-Ser is not the same polypeptide as Ser-Pro-Ile-Thr.

  • It is the only level encoded directly by the gene, and it determines every higher level.

• Secondary structure consists of local folds stabilized by hydrogen bonds in the polypeptide backbone, between a carbonyl group (C=O) and an amide group (N–H).

  • The α helix is a right-handed coil with 3.6 amino acids per turn. Each carbonyl bonds to the amide four amino acids ahead, and the R groups point outward.

  • The β sheet is a pleated sheet in which strands lie side by side and hydrogen bond to each other. Strands can be antiparallel (more stable) or parallel.

  • Secondary structure is relatively independent of the R groups.

• Tertiary structure is the overall three-dimensional shape of a single polypeptide.

  • It is determined mainly by interactions between R groups: hydrophobic effect, van der Waals forces, ionic bonds, hydrogen bonds, and disulfide bonds (covalent S–S bridges between cysteines).

  • Hydrophobic R groups tend to be buried in the interior, and charged or polar R groups tend to be on the surface.

  • Amino acids far apart in the sequence can end up close together in space.

• Quaternary structure is the assembly of two or more polypeptide subunits into one functional protein.

  • Subunits can be identical (an HIV enzyme with two identical subunits) or different (hemoglobin, with two α and two β subunits).

  • Subunits can influence one another. When one hemoglobin subunit binds oxygen, the others bind it more easily.

• Relationship to function:

  • Function depends on the three-dimensional shape. Folded proteins can have pockets that bind small molecules (enzymes), surfaces that bind other proteins or DNA, rigid rods for structural support, or hydrophobic surfaces that anchor them in membranes.

  • The chain of dependence is primary sequence → folding → shape → function.

  • Anfinsen’s ribonuclease A experiment showed this. After denaturing with urea and 2-mercaptoethanol and then removing the denaturants, the enzyme refolded and regained activity, so the primary sequence alone dictates folding.

Define “denatured” as it relates to protein structure.

• Denaturation is the process by which a protein unfolds and loses its three-dimensional structure.

• It disrupts the weak interactions that hold secondary, tertiary, and quaternary structure together, mainly hydrogen bonds and ionic bonds.

  • Common causes are high temperature and chemical treatment (for example, urea breaks hydrogen bonds and the hydrophobic core, and 2-mercaptoethanol breaks disulfide bonds).

• The primary structure is not destroyed, because the covalent peptide bonds remain intact.

• Consequence: because function depends on shape, a denatured protein usually loses its activity.

• Denatured proteins expose their hydrophobic groups, which can cause inappropriate aggregation. Cells use chaperones to shield slow-folding or denatured proteins.

• Some denaturation is reversible if the primary sequence can refold, as with ribonuclease A.

Describe the structure and function of the ribosome.

• Structure:

  • The ribosome is made of ribosomal RNA (rRNA) and ribosomal proteins.

  • It has a small subunit and a large subunit, each containing 1 to 3 types of rRNA and 20 to 50 types of ribosomal protein.

  • Eukaryotic ribosomes are larger than prokaryotic ribosomes.

  • The large subunit has three tRNA-binding sites:

    • A (aminoacyl) site: where a new charged tRNA enters.

    • P (peptidyl) site: holds the tRNA attached to the growing polypeptide.

    • E (exit) site: where an uncharged tRNA leaves.

• Function:

  • The ribosome is the site of translation. It binds mRNA and tRNAs.

  • It establishes the correct reading frame, reading the mRNA in successive, nonoverlapping groups of three nucleotides (codons).

  • An RNA in the large subunit catalyzes peptide bond formation, so the ribosome acts partly as a ribozyme.

  • It moves along the mRNA one codon at a time.

• Location: in prokaryotes, ribosomes bind mRNA while it is still being transcribed. In eukaryotes, ribosomes are in the cytoplasm, and rRNA is made in the nucleolus.

Explain the major events that occur during initiation, elongation, and termination of translation.

• Initiation: the start codon is found and the ribosome is assembled.

  1. Initiation factors bind the mRNA. In eukaryotes they bind the 5′ cap.

  2. Initiation factors recruit the small ribosomal subunit and a tRNA charged with methionine (tRNA^Met).

  3. The complex scans along the mRNA until it reaches the first AUG (start codon). This AUG sets the reading frame.

  4. The large subunit joins and the initiation factors are released. tRNA^Met sits in the P site.

• Elongation: amino acids are added one at a time. The cycle repeats.

  1. A charged tRNA whose anticodon pairs with the next codon enters the A site.

  2. A peptide bond forms. The growing chain is transferred from the P-site tRNA to the amino group of the amino acid on the A-site tRNA. This is catalyzed by RNA in the large subunit.

  3. The ribosome moves (translocates) one codon toward the 3′ end. The uncharged tRNA moves to the E site and is released, the peptide-bearing tRNA moves to the P site, and the A site is empty for the next tRNA.

  • Elongation factors bound to GTP provide energy by breaking GTP’s high-energy bonds.

  • The chain grows from its amino end to its carboxyl end.

• Termination: synthesis ends and the polypeptide is released.

  1. The ribosome reaches a stop codon (UAA, UAG, or UGA). No tRNA matches these codons.

  2. A protein release factor binds the A site.

  3. The release factor causes the bond between the polypeptide and the final tRNA to break, completing the carboxyl end.

  4. The polypeptide is released and the ribosomal subunits dissociate from the mRNA and from each other.

Define the terms codon, anticodon, and reading frame as they relate to translation.

• Codon:

  • A nonoverlapping group of three adjacent nucleotides in mRNA that specifies one amino acid or a stop signal.

  • There are 64 codons: 61 specify the 20 amino acids and 3 are stop codons.

  • Example: AUG codes for methionine and is also the start codon.

• Anticodon:

  • The three nucleotides in the anticodon loop of a tRNA that base pair with the mRNA codon.

  • Pairing is antiparallel, so the codon’s 5′ base pairs with the anticodon’s 3′ base.

  • Example: codon 5′-AUG-3′ pairs with anticodon 3′-UAC-5′.

• Reading frame:

  • One of the three possible ways of dividing an mRNA sequence into codons, depending on which nucleotide is the first of the first codon.

  • Analogy: THEBIGBOYSAWTHEMANRUN makes sense only when read as THE BIG BOY SAW THE MAN RUN.

  • The start codon (AUG) establishes the reading frame. A protein is correct only if translated in the proper frame.

  • Khorana’s repeating-sequence mRNAs showed this, because one repeating sequence (UCA) could be read in three frames giving three different polypeptides.

Discuss the role of tRNAs in the process of translation.

• Structure:

  • tRNAs are small RNAs of 70 to 90 nucleotides. They fold by self-pairing into a cloverleaf in two dimensions and an L-like shape in three dimensions.

  • The anticodon loop has the anticodon, which pairs with the mRNA codon.

  • The 3′ end has the sequence CCA. The hydroxyl of the terminal A is the amino acid attachment site.

• Role as the “adaptor” that translates the code:

  • Each tRNA links a specific anticodon to the specific amino acid that the genetic code assigns to that codon.

  • Aminoacyl tRNA synthetases attach the correct amino acid to the correct tRNA. Most organisms have one synthetase per amino acid.

  • A tRNA with its amino acid attached is charged, and one without is uncharged.

  • These enzymes are very accurate, with fewer than 1 error in 10,000.

  • They are directly responsible for translating codons into amino acids.

• Action on the ribosome:

  1. A charged tRNA enters the A site if its anticodon base pairs with the codon.

  2. Its amino acid is joined to the growing chain in the P site, and the tRNA moves to the P site.

  3. It then moves to the E site and is released uncharged to be recharged.

• Initiator tRNA: tRNA^Met pairs with AUG to start translation.

• Stop codons have no matching tRNA, which is why translation terminates.

Compare and contrast translation in prokaryotes and eukaryotes.

• Similarities:

  • Ribosomes have a small and a large subunit made of rRNA and protein.

  • The genetic code, tRNAs, and aminoacyl tRNA synthetases are used.

  • Elongation and termination are very similar, with the A, P, and E sites, elongation factors using GTP, and release factors at stop codons.

  • Translation starts with AUG and methionine.

• Differences:

  • Location and timing:

    • Prokaryotes have no nucleus, so transcription and translation are coupled in space and time. Ribosomes begin translating the mRNA while the 3′ end is still being transcribed.

    • Eukaryotes transcribe in the nucleus and translate in the cytoplasm. This separation allows RNA processing (5′ cap, poly(A) tail, splicing) and extra gene regulation.

  • mRNA processing:

    • Prokaryotic primary transcript = mRNA, with no cap or tail.

    • Eukaryotic mRNA has a 5′ cap and poly(A) tail and has had its introns removed.

  • Initiation:

    • Eukaryotes: initiation factors bind the 5′ cap, and the complex scans for the first AUG.

    • Prokaryotes: no cap. The ribosome binds an internal Shine-Dalgarno sequence (5′-AGGAGGU-3′ in E. coli), followed by an AUG about eight nucleotides downstream.

  • Number of proteins per mRNA:

    • Prokaryotes often make polycistronic mRNA from an operon, because each Shine-Dalgarno sequence is an independent start site.

    • Eukaryotic mRNAs usually encode a single polypeptide.

  • Ribosome size: eukaryotic ribosomes are larger.

Evaluate how changes in particular amino acids of a protein may affect protein structure and function.

• General principle: a change in one amino acid changes the primary structure, which can alter folding, shape, and function. The effect depends on which amino acid changes and where.

• Changes in R-group properties:

  • Hydrophobic ↔ hydrophilic or charged: a hydrophobic residue buried in the core replaced by a charged one disrupts the hydrophobic effect and can cause misfolding. A surface charged residue replaced by a hydrophobic one can cause aggregation.

  • Loss or change of charge: this can break ionic bonds that stabilize tertiary or quaternary structure, or disrupt binding to DNA or other molecules.

  • Polar → nonpolar: this removes the ability to form hydrogen bonds.

• Special amino acids:

  • Cysteine: losing a cysteine removes a disulfide bond (a covalent cross-bridge), weakening stability.

  • Proline: adding or removing it adds or removes a kink and restricts backbone rotation. This can break an α helix.

  • Glycine: replacing it with a bulky residue reduces flexibility and may cause steric clashes.

• Location matters:

  • A change in the active site or binding pocket can destroy function even if overall folding is normal.

  • A conservative change (for example, one hydrophobic residue for another of similar size) on the surface may have little effect.

• Possible outcomes: no effect, reduced activity, complete loss of function, or misfolding and aggregation, which is linked to diseases such as Alzheimer’s disease and Parkinson’s disease.

• Example: sickle cell disease results from replacing a polar glutamic acid with a hydrophobic valine in hemoglobin’s β subunit, which causes the proteins to aggregate.

Define the term “open reading frame” as it relates to the translation of genetic information.

• An open reading frame (ORF) is a stretch of nucleotides that can be translated into a polypeptide.

  • It begins at a start codon (AUG).

  • It continues in the same reading frame, in unbroken codons that each specify an amino acid.

  • It ends at a stop codon (UAA, UAG, or UGA) with no earlier in-frame stop codon.

  • “Open” means no premature stop codon interrupts it.

• Why it matters:

  • The AUG sets the reading frame. A different frame would give a different amino acid sequence and often a stop codon soon after.

  • The ORF is the protein-coding portion of an mRNA. Regions before the start codon (such as the leader region and 5′ cap) and after the stop codon are not translated.

• Prokaryotes: a polycistronic mRNA contains several ORFs, each with its own Shine-Dalgarno sequence and AUG.

• Eukaryotes: a mature mRNA usually contains one ORF, which is formed after introns are spliced out so the exons join in frame.

Predict the anticodon given a mRNA sequence or the amino acid associated with the anticodon.  

• Rules:

  • The anticodon is complementary and antiparallel to the codon.

  • Pair A–U and G–C.

  • Because the strands are antiparallel, the codon’s 5′ base pairs with the anticodon’s 3′ base.

• Steps: mRNA codon → anticodon

  1. Write the codon 5′→3′.

  2. Write the complement beneath it, running 3′→5′.

  3. Flip it to read 5′→3′ if needed.

  • Example: codon 5′-GCA-3′ → anticodon 3′-CGU-5′, which is 5′-UGC-3′ (the amino acid is Ala).

  • Example: codon 5′-AUG-3′ → anticodon 3′-UAC-5′ (the tRNA carries Met).

• Steps: anticodon → amino acid

  1. Write the anticodon 5′→3′ and reverse-complement it to get the codon.

  2. Look up the codon in the genetic code table.

  • Example: anticodon 5′-GAA-3′ → codon 5′-UUC-3′ → Phe.

• Notes:

  • Stop codons (UAA, UAG, UGA) have no anticodon, because no tRNA matches them.

  • Because the code is redundant, several codons for one amino acid can be read by tRNAs with different anticodons.

Predict how changing tRNAs or other components of the translation machinery could alter the process and products of translation.

• Change in a tRNA’s anticodon (but not its attached amino acid): the tRNA pairs with a different codon but still delivers its original amino acid, so wrong amino acids are inserted at that codon.

• Wrong amino acid attached to a tRNA (a defective aminoacyl tRNA synthetase): the correct anticodon-codon pairing occurs, but the wrong amino acid is added. Synthetases are normally very accurate, so a faulty one causes widespread errors.

• A tRNA that matches a stop codon: it could read through a stop codon and make a longer protein.

• Missing or nonfunctional tRNA: the ribosome stalls at the codons that tRNA reads and translation is slowed or incomplete.

• Initiation components:

  • Loss of the 5′ cap (eukaryotes) prevents ribosome binding and translation.

  • Loss of initiation factors prevents initiation.

  • Altering the start codon or Shine-Dalgarno sequence changes or eliminates the start site.

• Ribosome: defects in rRNA affect peptide bond formation, because the large subunit’s RNA is the catalyst.

• Elongation factors: if they cannot use GTP, ribosome movement and peptide bond formation stall.

• Release factors: defective ones cause failure to terminate, so the ribosome continues past the stop codon and produces an abnormally long protein.

• Overall: changes can lead to no protein, truncated protein, altered sequence, or longer protein.

Use a codon table to translate a nucleotide sequence into an amino acid sequence.

• Steps:

  1. Write the mRNA sequence 5′→3′. If given DNA, transcribe it first (see the last card).

  2. Find the first AUG to set the reading frame.

  3. Divide the sequence into nonoverlapping groups of three.

  4. For each codon, use the table: the first base (5′) selects the row, the second base selects the column, and the third base selects the specific entry.

  5. Write the amino acids in order, starting from the amino end (Met).

  6. Stop at UAA, UAG, or UGA. Stop codons do not code for an amino acid.

• Example: 5′-AUG GCA UUC GAA UAG-3′

  • AUG = Met

  • GCA = Ala

  • UUC = Phe

  • GAA = Glu

  • UAG = Stop

  • Result: Met-Ala-Phe-Glu (amino end → carboxyl end)

• Reminders:

  • The code is redundant: UUU and UUC both code for Phe, and the variation is mostly in the third position.

  • Use the amino acid abbreviations, for example Met (M) and Phe (F).

Relate the length of a gene to its primary and mature mRNA.

• Primary transcript:

  • It is a copy of the entire transcribed region of the gene from the promoter’s start site to the terminator.

  • It is complementary to the template strand and contains exons and introns.

  • Its length is therefore about the same as the transcribed length of the gene.

• Mature mRNA in prokaryotes: it is the same as the primary transcript, because there is no processing, so length ≈ transcribed gene length.

• Mature mRNA in eukaryotes: it is shorter than the gene’s transcribed region because processing changes it.

  • Splicing removes introns, which are often thousands of nucleotides long. Most human genes contain at least one intron, and some have dozens. The spliceosome joins the exons.

  • A 5′ cap (7-methylguanosine) and a poly(A) tail (about 250 As) are added. These add a small amount of length but are not encoded in the gene.

  • Alternative splicing lets one gene give several mature mRNAs of different lengths and exon combinations, making different proteins.

• Takeaway: gene length, as transcribed, ≥ primary transcript length ≥ spliced mRNA length (in eukaryotes, before the poly(A) tail is added), and the mature mRNA is often much shorter than the DNA region that encodes it.

• Note: the mature mRNA also contains untranslated sequence, so it is longer than its ORF.

Relate the number of nucleotides in an open reading frame to the the number of amino acids in a protein produced from it.

• Rule: each codon (3 nucleotides) specifies one amino acid, so the number of amino acids is based on nucleotides ÷ 3.

• Counting the stop codon:

  • If the ORF count includes the stop codon:

    • Number of amino acids = (nucleotides ÷ 3) − 1

    • Nucleotides = 3 × (amino acids + 1)

  • If the count includes only the amino-acid-coding codons: amino acids = nucleotides ÷ 3.

• Example: an ORF of 300 nucleotides (including the stop codon) has 100 codons, so it codes for 99 amino acids plus the stop signal.

• Example: a protein of 150 amino acids requires an ORF of 3 × (150 + 1) = 453 nucleotides.

• Notes:

  • The initiator Met counts as an amino acid, although it may be cleaved off after synthesis.

  • An ORF length that is not a multiple of 3 means a frameshift or a counting error.

  • Because the code uses 3 nucleotides per amino acid, the mRNA is always longer than the protein’s amino acid count.

Transcribe and translate a gene based on a given DNA sequence.

• Step 1: Identify the template strand.

  • The template strand is read by RNA polymerase. The nontemplate (coding) strand has the same sequence as the mRNA, with T in place of U.

  • If given only one strand, determine whether it is the template or coding strand from the question’s wording.

• Step 2: Transcribe.

  • RNA polymerase reads the template 3′→5′ and builds RNA 5′→3′.

  • Pair A–U, T–A, G–C, C–G (the template’s A pairs with U).

• Step 3: Translate.

  1. Find the first AUG in the mRNA.

  2. Read codons in frame, 5′→3′.

  3. Look up each codon in the table.

  4. Stop at UAA, UAG, or UGA.

• Worked example:

  • Template strand: 3′-TACCGTAAGCTTATC-5′

  • Coding strand: 5′-ATGGCATTCGAATAG-3′

  • mRNA: 5′-AUGGCAUUCGAAUAG-3′

  • Codons: AUG | GCA | UUC | GAA | UAG

  • Protein: Met-Ala-Phe-Glu (UAG is the stop codon)

• Tips:

  • Always label 5′ and 3′ ends. If you are given a 5′→3′ template, reverse and complement it carefully.

  • Eukaryotes: splice out introns before translating the mature mRNA.

  • The mRNA is complementary to the template, but the same as the coding strand (T→U).