BIO230H1 Lecture Notes - Lecture 3: Multicellular Organism, Cyclic Adenosine Monophosphate, Enzyme
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QUESTION 1: The rabies virus primarily affects the nervous system. The specificity that the rabies virus has for neuronal host cells is primarily dictated by __________________________.
A. the helical shape of its viral capsid |
B. the type of nucleic acid used for its viral genome (single-stranded, antisense RNA) |
C. the spikes that protrude from its viral envelope |
D. the segmented nature of its viral genome |
QUESTION 2: Which of the following genome types has been observed in viruses? To be marked correct, you'll need to select all true statements, as there may be more than one correct answer.
A. Single-stranded RNA |
B. Single-stranded DNA |
C. Double-stranded RNA |
D. Protein-based |
QUESTION 3: Choose the correct statement about viral evolution.
A. RNA and DNA viral genomes evolve at equal rates. | ||||||||||||
B. Cellular genomes mutate at a faster rate than viral genomes due to their large size and increased chance of replicative mistakes. | ||||||||||||
C. DNA viruses mutate faster than RNA viruses because thymine is more susceptible to mutation than uracil. | ||||||||||||
D. RNA viruses mutate faster than DNA viruses due to a lack of proofreading replicative enzymes. QUESTION 4: Your elderly patient is affected by shingles. After careful observation, you note that the virus responsible for the infection has an icosahedral capsid, is enveloped, and has double-stranded linear DNA as its genetic material. Based on this information, in which of the following viral families would you group this viral pathogen?
QUESTION 5: Which of the following is a key difference between lytic and lysogenic bacteriophage replication cycles?
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Question 1
1 pts
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 | |
mRNA | |
protein | |
tRNA |
Question 2
1 pts
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 will allow you to accomplish this?
Using an RNA probe complementary to the region not removed by the truncation. | |
Using an RNA probe complementary to the region removed by the truncation. | |
Using an DNA probe complementary to the region not removed by the truncation. | |
Using an DNA probe complementary to the region removed by the truncation. |
Question 3
1 pts
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 genomic DNA of a healthy individual who does not have cystic fibrosis. | |
The genome of a CFTR patient known to have the specific truncation you are trying to identify. | |
The genome of a CFTR patient with a missense mutation but full length gene. | |
The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify. | |
The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene. |
Question 4
1 pts
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?
As both are composed of nucleic acids, using either would result in identical results. | |
An RNA probe because RNA has uracil bases. | |
An RNA probe because it could also be used in a translation experiment. | |
A DNA probe because it is more stable than RNA. | |
A DNA probe because RNA cannot bind to DNA. |
Question 5
1 pts
Which of the following will lower the Tm of a given DNA strand?
Increasing the percentage of GC base pairs. | |
Raising the pH of the solution from neutral to basic. | |
Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl. | |
None of the above. |
Question 6
1 pts
One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?
To collect the bacteria at the bottom of the tube. | |
To break open the bacteria to release the genome. | |
To separate the bacteria from the bacteriophages. | |
To be able to detect the radioactivity. |
Question 7
1 pts
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?
No, because only DNA can be labeled with radioactivity. | |
No, because the RNA genome would not enter the bacteria upon infection. | |
No, because while DNA and RNA nucleotides are similar, they are not identical. | |
Yes, because DNA and RNA nucleotides are similar. | |
Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly. |
Question 8
1 pts
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?
Heat killed non-virulent bacteria is added to a live virulent bacteria strain. | |
Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain. | |
A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain. | |
Heat killed mouse cells are added to a non-virulent bacteria. |
Question 9
1 pts
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. | |
It is faster to duplicate the genome when these are present. | |
The existence of introns can lead to multiple variations of proteins encoded by a single gene. | |
It is unlikely transposons would exist in the genome if there was too much protein coding DNA. |
Question 10
1 pts
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?
Media containing histidine | |
Media containing adenine | |
Media lacking adenine | |
Media lacking histidine |
QUESTION 10
If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:
The mutation changes the number of domains in the enzyme, which makes it work more efficiently | ||
The mutation changes the amino acid sequence of the lactase protein | ||
The mutation increases the number of copies of the lactase gene that will be found in your genome | ||
The mutation changes whether the lactase sequence is found in an intron or exon | ||
The mutation affects the expression of the lactase gene |
1.2 points
QUESTION 11
A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.
The substrate will increase the reaction rate by binding to the allosteric site | ||
The substrate will increase the reaction rate by competing with the inhibitor for the active site | ||
The reaction rate will not change unless the inhibitor can be removed | ||
The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind | ||
The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme |
1.2 points
QUESTION 12
What determines where in the genome a transcription regulator will bind?
Transcription regulators bind to the 5' UTR region of a gene | ||
Regulators bind via complementary base-pairing to certain DNA molecules | ||
Covalent bonds form between the transcription regulator and the atoms of the DNA backbone | ||
Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene | ||
Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions |
1.2 points
QUESTION 13
What is the basic premise of cell theory?
DNA -> RNA -> protein | ||
All cells arise from pre-existing cells | ||
DNA provides the complete instructions to create a cell | ||
The identity of a cell is determined through gene expression patterns | ||
All cells contain the same four basic macromolecules |
1.2 points
QUESTION 14
What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?
Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment | ||
Enzymes to digest lactose are only made when energy is low and lactose is available | ||
The activator can override the inhibition of the lac operon by the repressor | ||
The repressor can control the enhancer, while the activator can control the promoter | ||
When neither the lac activator or repressor is present, expression of the lac operon is too high |
1.2 points
QUESTION 15
What is the histone code used for?
Phosphorylation and acetylation of DNA affect its ability to be compacted | ||
Changes to the sequence of DNA change whether DNA will wrap around histone proteins | ||
Covalent modifications of histones affect the ability of the transcription initiation complex to form | ||
Histones provide the codon sequences needed for translation to occur | ||
The histone code affects which amino acids will get added to a polypeptide |