BIOLOGY 1A03 Lecture 5: Lecture 5

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1 Apr 2016

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BIOLOGY 1A03 Full Course Notes

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Lecture 5: theme 1 module 3 & 4 review. Transformation: a change in cell behaviour resulting from the incorporation of genetic material from outside the cell: he found that killing bacteria is a method of preventing outbreaks and effects of ammonia. He did this by heat-killing the bacteria so they are dead and no longer harmful (heated the s- strain) Combine heat killed s-strain and live r- strain: caused. Why did the heat-killed s-strain and live r-strain cause the death: no protein coating, was alive, dead s-strain fragmented everywhere, was able to acquire information from the r-strain. Conclusion: living benign cells acquired information from the dead virulent cells. Which macromolecule carries hereditary information: oswald avery, colin macleod, maclyn mccarty all tried to gure out the macromolecule that passed on this genetic information, they new it was one of 4 macromolecules. Virulent bacteria (killed): dna extracted from heat-killed virulent cells along with trace amounts of rna and protein.

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magine that the year is 1928 and the scientific world is in an uproar trying to determine what component of the cell isresponsible for genetic inheritance. Is it DNA, protein, perhaps even RNA? You are busy in your lab, trying to help solve this crucial mystery that still plagues (then) modern day biology. Your work focuses on a microorganism, a bacterium named Streptococcus pneumoniae, and its effects on mice. There are two strains of S. pneumoniae (which is responsible for pneumonia), the âRoughâ strain (R) and âSmoothâ strain (S). The strains are very similar except the S strain has a smooth polysaccharide coat that enables the microorganism to avoid the organismâs immune system and cause illness (and death in mice). The R strain has a rough coat which is detected and dealt with by the immune system (and as a result causes no death in mice). You decide to perform a series of experiments using the R and S strain of S. pneumoniae involving injections of these strains in mice. Your results are documented in the figure below: living S strain of S. pneumoniae mouse dies of infection living R strain of S. pneumoniae mouse lives mouse lives S strain heat-killed living R strain mouse dies of infection living, pathogenic S strain recovered S strain heat-killed From these results, you figure out that something from the heat killed S strain is transforming the R strain into the virulent strain. What is this âTransforming principleâ? Is it related to the cellular component that was passed from one generation to the next? As you know, this was the initial experiment performed by Dr. Frederick Griffith that eventually led to the discovery of the true cellular component that is the genetic inheritance molecule as well as the âtransforming principleâ (Hint: Itâs DNA!).

1. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells?

a. Griffith b. Watson and Crick c. Avery, MacLeod, and McCarty d. Hershey and Chase

2. From your results, which one of the cellular components can be determined NOT to be the component responsible for genetic inheritance? Why?

a. Proteins because these macromolecules become denatured with heat. b. DNA because these macromolecules form a single helix in solution. c. RNA because it's double stranded 100% of the time d. Sugars because glycolysis happens.

3. Which one of the following experimental trials below are considered to be controls for this experiment? a. S Strain grown in 10% ethanol medium. b. Living S strain; Living R strain; Heat Killed S Strain c. Living R Strain with heat killed S strain; Living S Strain d. Heat Killed S strain; Living R strain; R strain grown in 10% ethanol medium

4. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule? a. Griffith b. Watson c. Crick d. Hershey and Chase

5. Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information. What radioactive material was used to track DNA during this experiment?

a. 3H b. 14C c. 35S d. 32P

6. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.

a. True b. False

7. The polarity of a DNA strand results from the polarity of the nucleotide subunits.

a. True b. False

8. There are five different nucleotides that become incorporates into a DNA strand.

a. True b. False

9. Hydrogen bonds between each nucleotide hold individual DNA strands together.

a. True b. False

brownpanda894

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

tealcockroach755

LAB MODULE: BASIC MICROSCOPY â STUDY HELP

The information in this document is helpful in preparing for the Basic Microscopy Pre-Lab Quiz on Canvas and the Basic Microscopy quiz on Late Nite Labs.

Instructions:

â¢ Go to your Late Nite Labs course site and click on the Basic Microscopy link. The lab manual content will appear on the right side of the screen.

â¢ PRINT and READ all of the information found in the Basic Microscopy lab manual.

â¢ Use the information to answer the following questions.

Questions

â¢ Fill in the remainder of the chart.

Objective Lens	Objective Magnification	Ocular Magnification	Total Magnification
Scanning	4X	10X	40X
Low Power	10X	10X
High Power	40X	10X
Oil Immersion	100X	10X

â¢ Do all organisms consist of more than one cell?

â¢ What is the structural and functional unit of living things? In other words, what is considered the basic building block of LIFE?

â¢ What three structures are found in the cells of ALL organisms, both prokaryotic and eukaryotic? You may have to look at the pictures of the cells in the Basic Microscopy lab manual content to find all three structures. Note: plasma membrane = cell membrane

â¢ List the major differences in prokaryotic cells and eukaryotic cells? Include the location of genetic material in a eukaryotic versus prokaryotic cell.

â¢ List the organelles found in a eukaryotic cell that are not found in a prokaryotic cell. You may have to look at the pictures of the cells in the Basic Microscopy lab manual content to find all of these.

â¢ In eukaryotic cells, genetic information is found in the form of DNA. Is this true for prokaryotic cells?

â¢ Are flagella and cell walls unique to eukaryotes? Why or why not?

â¢ The __________is the part of a cell that allows the movement of substances into and out of the cell and protects the cell from its environment.

â¢ The__________is a substance that contains all of the dissolved and suspended ions and other molecules needed to keep the cell alive.

â¢ A__________is an organelle that serves as the site of photosynthesis in a eukaryotic cell.

â¢ Some organisms have cells that contain a__________that gives the cell a more structured shape and strengthens the cell by osmotic pressure.

â¢ Describe dark field microscopy.

â¢ Describe phase contrast microscopy.

â¢ Why do scientists use dark field microscopy and phase contrast microscopy when studying prokaryotic cells?

â¢ How does phase contrast microscopy help scientists visualize difficult specimens?

â¢ Fill in the remainder of the chart. The organisms studied during this lab are: Amoeba, Spirogyra, cardiac muscle, and bacteria.

Structure	Function	Organisms Studied
Plasma Membrane		Amoeba, Spirogyra, cardiac muscle, and bacteria
Cytoplasm
Nucleus	Contains the genetic material of a eukaryotic cell
Chloroplast		Spirogyra
Cell wall

BIOLOGY 1A03 Lecture 5: Lecture 5

BIOLOGY 1A03 Full Course Notes

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