translation

Instruction on Using the In-Class Guide

  • Follow along with the slides and take notes.

Learning Objectives

  • 6.1 Translate information coded by an mRNA into an amino acid sequence using a codon table.

  • 6.2 Predict the effect of given mutations on protein structure and function.

  • 6.3 Describe the process of translation.

Pre-Class Review

  1. Transcription uses DNA as a template to make mRNA, which is then used in translation to code for a protein.

  2. The organelle responsible for protein translation is the ribosome.

  3. Ribosomes are located in:

    • a. Cytoplasm

    • b. Rough ER

    • c. Smooth ER

    • Select all that apply.

  4. In Eukaryotes, the processes that take place in the nucleus include:

    • a. Transcription

    • b. RNA Processing

    • Select all that apply.

  5. The letter that indicates the anticodon of the tRNA is B.

  6. The letter that indicates where the amino acid is bound to the tRNA is C.

Structure of Ribosomes and Translation Process

  1. Ribosomes are made of rRNA and proteins.

  2. The three binding sites of the ribosome are the:

    • a. A site

    • b. P site

    • c. E site

  3. Peptide bonds are formed between the P site and the A site.

  4. In Eukaryotes, the parts of the mRNA recognized by the ribosome to initiate translation include (Select all that apply):

    • a. The AUG start codon

    • c. The 5’ cap

  5. Initiation in Prokaryotes: The mRNA strand contains a ribosome binding site, also known as the Shine-Dalgarno sequence, which is recognized by the ribosome.

  6. Initiation in Eukaryotes: The ribosome recognizes the 5' cap and the AUG (start codon).

  7. Termination occurs when a stop codon (UAA, UAG, UGA) indicates the end of the protein. The ribosome recognizes this, and translation ends after a release factor fills the A-site of the ribosome.

Application and Case Study of Cystic Fibrosis

  1. The gene for the Cystic fibrosis transmembrane conductance regulator (CFTR) is described as follows:

    • a. The CFTR gene is located on chromosome 7.

    • b. It contains 188,703 nucleotides long.

    • c. It contains 27 exons.

    • d. It codes for a 1480 amino acid protein.

    • e. Mutations in CFTR can result in Cystic fibrosis, which is an autosomal recessive disorder characterized by thick mucus, difficulty breathing, and inability to transfer O2 from the air to the blood. Patients often die from respiratory failure.

  2. A portion of the CFTR mRNA has the sequence: 5’ CUG AGU GGA GGU CAA GCA 3’.

    • What protein sequence does this mRNA sequence code for? (Assume the start codon has already passed and this sequence is in frame).

  3. Predict the effect of each of the following mutations, keeping in mind the primary, secondary, and tertiary structure of proteins:

    • a. Silent mutation: No effect on the amino acid sequence.

    • b. Missense mutation: Results in a different amino acid.

    • c. Nonsense mutation: Results in a premature stop codon.

    • d. Frameshift mutation: Alters the reading frame of the mRNA.

  4. A portion of the middle of an exon in a mutant CFTR gene has the sequence: 3’ GAC TCA CCT CTA GTT CGT 5’.

    • What protein sequence does this code for?

    • Compare to your sequence in number 15. What amino acid is mutated? What is the original amino acid and the new amino acid?

    • What type of mutation is this? Possible categories include:

      • i. Nonsense

      • ii. Missense

      • iii. Silent

      • iv. Frameshift

Experimental Techniques

  1. Western blots allow detection of relative amounts of specific proteins, indicating levels of translation.

  2. Experiment involving engineered different cells to express various CFTR mutants included performing a western blot with antibodies specific to these mutant proteins. Conclusions drawn from this western blot include:

    • a. The 1410X mutant contains the fewest amino acids compared to other CFTR mutants.

    • b. The mutant CFTR proteins negatively affect organismal function.

    • c. Cells normally express all of these mutant CFTR proteins.

    • d. The G551D mutant was translated the most.

  3. Measurements of chloride ion flow through normal (wild type (WT)) and G551D CFTR channel proteins in rat cells provide insights into the G551D mutation's effect on CFTR function:

    • a. The mutant CFTR channel is denatured and hydrolyzed in the cytoplasm.

    • b. The mutant CFTR channel is not folded properly.

    • c. The mutant CFTR channel is not translated properly.

    • d. The mutant CFTR channel does not get properly transported to the membrane.

Genetic Terminology and Ribosome Function

  1. Match the letters to the following terms:

    • i. Anti-codon

    • ii. Peptide

    • iii. Amino acid

    • iv. Peptide bond

    • v. tRNA

    • vi. Codon

  2. The ribosome reads mRNA in:

    • a. 5’ to 3’ direction.

    • b. 3’ to 5’ direction: not applicable.

    • c. Either direction: not applicable.

  3. The protein insulin is a heterodimer composed of 2 subunits:

    • a. Interacts with a membrane receptor to signal when blood glucose is high, resulting in glucose uptake from the blood into the cells.

    • b. Disorders related to insulin can result from:

      • i. A mutation in the insulin protein.

      • ii. Destruction of the insulin-producing cells.

      • iii. Errors in the insulin receptor.

  4. Since insulin is a dimer of 2 subunits, the highest level of protein structure is:

    • a. Primary

    • b. Secondary

    • c. Tertiary

    • d. Quaternary.

  5. Translation typically starts and ends at specific sequences, including:

    • a. Start codon (AUG).

    • b. Stop codon (UAA, UAG, or UGA).

    • c. Ensure the correct reading frame is maintained.

  6. To translate a sequence:

    • From an mRNA sequence: Translation STARTS and STOPS at specific points.

    • Find the start codon – AUG. Read each amino acid in frame from this AUG.

    • Your reading frame is defined by your first AUG (sets of 3).

Insulin Translation Activities

  1. Insulin Translation: Insulin is essential for the uptake and regulation of blood sugar, interacting with a membrane receptor to signal when blood sugar is high and glucose needs to be take up. Disorders can stem from:

    • A mutation in the insulin protein itself.

    • Mutations in the insulin receptor.

  2. Normal Insulin: Below is the normal insulin mRNA sequence (5’ to 3’). The nucleotide numbers indicate the beginning of each row, with each grouping consisting of 10 nucleotides.

Normal Insulin mRNA Sequence (5’ to 3’)

  • GCUGCAUCAGAAGAGGCCAUCAAGCACAUCACUGUCCUUCUGCCAUGGCCCUGUGGAUG

  • CGCCUCCUGCCCCUGCUGGCGCUGCUGGCCCUCUGGGGACCUGACCCAGCCGCAGCCUU

  • UGUGAACCAACACCUGUGCGGCUCACACCUGGUGGAAGCUCUCUACCUAGUGUGCGGGG

  • AACGAGGCUUCUUCUACACACCCAAGACCCGCCGGGAGGCAGAGGACCUGCAGGUGGGG

  • CAGGUGGAGCUGGGCGGGGGCCCUGGUGCAGGCAGCCUGCAGCCCUUGGCCCUGGAGGG

  • GUCCCUGCAGAAGCGUGGCAUUGUGGAACAAUGCUGUACCAGCAUCUGCUCCCUCUACC

  • AGCUGGAGAACUACUGCAACUAGACGCAGCCCGCAGGCAGCCCCCCACCCGCCGCCUCC

  • UGCACCGAGAGAUGGAAUAAAGCCCUUGAACCAGC

  1. Insulin Mutants: Below are several mutated insulin mRNA sequences. For each, you should:

    • a. Find the start codon and circle it.

    • b. From the start codon, circle every set of three bases after it, this is your reading frame.

    • c. Stop at the stop codon.

    • d. Translate the amino acid featured at the mutation (highlighted in yellow) and compare it to the same amino acid in the wild-type insulin.

    • e. Identify the type of mutation:

      • i. Nonsense mutation

      • ii. Missense mutation

      • iii. Silent mutation

      • iv. Frameshift mutation

    • f. Predict the effect of the mutation on insulin function and explain your reasoning.

      • Insulin mRNA Mutant #1

    • Additional mutated sequences follow similar instructions, prompting circled start codon, identification of mutations, and function predictions.