Notes:Bio 99 Homework 1

 

 

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel.  You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene.  The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR: 

DNA, RNA, and protein
Nonsense and missense mutations affect the DNA which also affects the RNA and protein.

 

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals.  You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients.  You will determine this using hybridization techniques with samples from healthy and CF patients.  Which of the following would allow you to accomplish this?

Using an RNA probe complementary to the region removed by the truncation; using an DNA probe complementary to the region removed by the truncation

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled.  Which of the following samples must you test?

The genome of a healthy individual who does not have CFTR.
The genome of a CFTR patient known to have the specific truncation you are trying to identify.
You need a way to be able to identify and compare the gene of a normal person with the mutated gene.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe.  Which would be ideal to use and why?  Select the best answer.

A DNA probe because it is more stable than RNA.

RNA is less stable than DNA because of its 2' OH group

Which of the following will lower the Tm of a given DNA strand? Select all that apply.

Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.

The cation can bind to the backbone of DNA shielding it, increasing the Tm. If you decrease the concentration of the salt, then the DNA is less shielded and the Tm decreases.

One step of the Hershey/Chase experiment involved blending the virus/cell mixtrue before centrifugation and probing the pellet for radioactivity.  Why was the blending step necessary?

To separate the bacteria from the bacteriophages.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment.  Would radioactivity still have been found in the pellet? 

Yes, because DNA and RNA nucleotides are similar.

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information.  Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?  Select all that apply.

Heat killed non-virulent bacteria is added to a live virulent bacteria strain.

The non-virulent is not harmful to bacteria, but now its dead cause we killed it with heat. The virulent is harmful and is still alive. Adding "Heat killed non-virulent bacteria is added to a live virulent bacteria strain." will kill the mouse because the virulent is still alive.

The human genome consists mostly of non-coding DNA.  Which of the following are benefits of this? 

Random DNA mutations generally won't affect RNA and protein function.

The existence of introns can lead to multiple variations of proteins encoded by a single gene. 

Since there are more non-coding regions of DNA, if a random point mutation happens, it is more likely to happen in a non-coding region and thus won't affect the RNA and protein.