Bioinformatics Translation Exercise

Instructor’s Guide

[Additional information in italics]

Purpose:

The following activity is an introduction to the Biology Workbench, a site that allows users to make use of the growing number of tools for bioinformatics analysis. It is also a review of translation, mutation, and restriction analysis.

The activity should take students 1.5 to 2 hours to complete.

Background:

Options:

Give the students an overview of translation using overheads from text.
Have the students perform a short translation by hand.
Use an animation of translation.
Use a Chime tutorial of translation.

Discuss start and stop codons and that there are 6 reading frames. (3 forward and 3 reverse.)

Procedure:

Go to the Biology Workbench page ( http://workbench.sdsc.edu/ ) and choose Session Tools

Select START NEW SESSION and click RUN. Name the session sickle cell (or something similar).
Go to Nucleic Tools and choose ADD NUCLEIC SEQUENCES from the window.
Copy and paste the human beta globin cDNA sequence below into the file (You can only browse for a file if the document contains nothing but the sequence) Select save.

[Teachers: you may want this document loaded on computers students are using so that they can cut and paste. Or you can create a document with just the sequences below in Word, Notepad, or other word processing programs.]

Label:

>normal B-hemoglobin

Sequence:

Repeat the process with the sickle-cell hemoglobin.

Label:

>sickle-cell B-hemoglobin

Sequence:

Check the box beside the human beta globin cDNA and select VIEW.

How large is it (how many base pairs)? 626 bp

How large a protein could it encode (how many amino acids)?

626/3 = 208 a.a

TRANSLATE THE cDNA

Click on RETURN to go back to Nucleic Tools and select the box next to human beta globin cDNA and then choose the application SIXFRAME from the window of tools (you will probably scroll down to find it). Click on the RUN button.
Leave the default settings as they are and hit SUBMIT
Scroll down to analyze the results. Blue M’s represent possible start codons (methionine) and red * represent stop codons (code for no amino acid).

Which reading frame had the longest protein (scroll to the bottom of the screen and look for the longest ORF or Open Reading Frame)?

Frame 3

Frame 4 is fairly long, but is not considered the longest protein. Why not?

No Methionine at the beginning.

How long are the shorter proteins (look at a few sections from start codon to stop codon and estimate the average)?

10-30 a.a

Why would you predict these proteins to be short if they were not the correct reading frames?

There are 3 stop codons out of a total of 64 codons. Thus you would predict a stop codon every 20 codons.

Why should we choose the longest open reading frame as the correct translation?

It does not have any internal stop codons and has a start codon and therefore probably encodes the complete protein.

The longest protein is not 208 a.a. long (as predicted in question 4b). Why not? The start codon is not at the very beginning of the sequence and the stop codon is not at the very end of the sequence.

Normal B-hemoglobin gene sequence showing start and stop codon locations

Import the longest open reading frame by checking the box in front of it and clicking the IMPORT SEQUENCES button. It will be imported into your Protein Tools section. If your Protein Tools is empty you forgot to check the box before you imported.
Repeat this procedure (step 6, 7, and 9) with the sickle cell DNA.

You may want students to go to a Protein Explorer tutorial of hemoglobin showing its 3-D structure http://www.umass.edu/microbio/chime/hemoglob/2frmcont.htm.

This Website requires downloading a plugin called Chime for your browser.

ALIGN PROTEINS FOR COMPARISION

Go to Protein Tools and select both the normal and sickle cell cDNA by clicking on the boxes in front of their names.
Select the application CLUSTALW from the tools window to align the two protein sequences.
Leave the default preferences as they are and hit the SUBMIT button.
Scroll down and look for the mutated amino acid(s). Describe the mutation.

The base T has been substituted for the base A

Can you find the mutation in the DNA that caused the mutation in the amino acids (suggestion: you can run CLUSTALW in nucleic tools as well as protein tools)? What kind of mutation is this? A substitution

In the Nucleic Tools, select the normal human beta globin sequence and select Edit Nucleic Sequences. Add or delete a base somewhere near the middle of the sequence. It will be saved with “edited” after the name.
Select the newly edited file and translate it using SIXFRAME. What impact did this have on frame 3 of the possible reading frames? Is it still 147 a.a? (you may have to look at the sequences and find the start and stop codons)

answers vary, but it is probably shorter

What kind of mutation was introduced?

. A frameshift

What impact might this have on the function of the protein?

It would probably not be functional since every codon after the mutation would be altered.

RESTRICTION ANALYSIS

Restriction enzymes are like spell checking enzymes, they look for specific “words” in the DNA and then cut them. If the “word” is misspelled, then the enzyme will not cut the DNA. A model of a restriction enzyme can be found at: http://www.worthpublishers.com/lehninger3d/index.html

We will use the enzyme DdeI. It cuts the normal beta globin cDNA at CTGAG, but it will not cut the sickle cell cDNA at CTGTG.
Go to Nucleic Tools, select the normal beta globin and then select the application TACG from the tools window. This program will allow you to digest the DNA with different enzymes.
Scroll down to User Specified Enzymes: and type in DdeI. This enzyme will cut the normal cDNA at the sequence CTNAG (N can be any nucleotide and it will still be cut) but not the sickle cell mutant. Scroll down to Smallest Fragment Cutoff Size for Simulated Gel Map: and change it to 10 (our smallest segments will be smaller than the default of 100 base pairs). Select SUBMIT.
What was the output of Fragment Sizes by Enzyme? (how many fragments? What lengths?)

Seven Fragments: 37bp 50bp 68 84 89 139 159

Repeat steps 17 and 18 with the sickle cell sequence and compare the two.
What was the output?

Six Fragments: 37bp 50bp 84 89 139 227

Notice that the lengths of the two missing fragments add up the the length of the new fragment. Why is that?

The restriction enzyme didn’t cut at the site of the mutation, leaving one long fragment instead of two shorter ones.

Use the outputs from steps 19 and 21 to label the band sizes on the gel at the end of this exercise.

From banding patterns on the gels and the pedigree chart, indicate whether the person on the pedigree has normal hemoglobin (2 normal genes or homozygous dominant), has sickle-cell anemia (one normal, one sickle-cell gene or heterozygous), or sickle-cell disease (two sickle-cell genes or homozygous recessive).

Parents

1. Ss Sickle-cell anemia (has 68 and 159 bp bands from normal and 227 bp band from mutated DNA.

2. Ss Sickle-cell anemia

Offspring

3. ss Sickle-cell disease (missing 68 and 159 bp bands from normal)

4. Ss

5. Ss

6. SS Normal (missing 227 bp band from mutated DNA)