Module Instructions: The Genetic Code - Transcription and Translation (Lab)

Pre-Lab: Genetic Code + Transcription and Translation

There are actually two Pre-Labs associated with this module: (1) Genetic Code and (2) Transcription and Translation. 

Before you get started I highly suggest watching the very short animation (not mine) from the Central Dogma module. It is a good review. And then watch the other animations in this module (they are all pretty short) in this week's module. I never really understood these processes until I watched these videos.

This module is pretty difficult in my experience. Take your time and focus on this one. If you need any help understanding content, feel free to reach out to me via email or Canvas Inbox.

Lab: Transcription, RNA Processing and Translation

This lab can be a challenge. Below are points for the most commonly made mistakes in the lab.

In Figure 1, be sure to label the boxes template and coding. The greyed out strand is the mRNA connecting to the opened DNA.

In Figure 3, you are to identify the exposed strand as either the template or coding strand. This will be the strand that connects directly to the incoming ribonucleotides.

In Figure 6, question 7 asks for the structure. The answer is not RNA Polymerase. That is the larger structure. What is the structure inside the RNA Polymerase that splits the two strands of the DNA?

For question 14, you will need to go back and read about the termination of transcription in prokaryotes and compare that with the reading for the termination of eukaryotes to answer the question.

In Figure 9, use the following labels: exons, intron and snRNPs.

For problem 15, calculate the number of possibilities for the triplet code. A singlet code is 4¹ = 4. That would give us 4 possible combinations. A doublet code would be 4² = 16, or 16 possible combinations. What would a triplet code be?

For problem 16, choose from the three abovementioned properties of the code: redundant, unambiguous or universal.

In Figure 12, use the reading to place the following labels in the figure: protein, primary transcript, mature mRNA and ribosome.

Figure 13 is a reading exercise. Look for the figure call outs in the reading and use the information to guide you in labelling. I have shown you an example for Figure 13a.

Figure 13e is a little tricky. To identify the amino acid, we use the amino acid table, which is based off of the mRNA. So in this case the mRNA, AAG, would yield the amino acid LYS. As for the anticodon, it would be the complement of AAG based on the rules of complementary base pairing.

For problem 19, you will predict the amino acid sequence based off of the DNA. Remember the template strand creates the mRNA strand based on complementary base pairing (not the coding strand). The first three bases of the template strand (TAC) would yield the mRNA codon (AUG). Using the amino acid chart, we see that the amino acid would be MET (the start codon). Repeat that for the remainder of the template strand listing out the sequence of amino acids, known as the polypeptide chain.

Post-Lab: Genetic Code + Transcription and Translation

Submit the Quiz in Canvas by Friday at 5pm on the date given in the calendar.