BIOL1003 Study Guide - Final Guide: Bacteriophage, Cell Nucleus, Nuclease

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16 May 2018

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Molecular Genetics

Proteins and Nucleic Acids in Cells

• Viruses (bacteriophage) - no nucleus, no membrane, no "cell", protein capsule, nucleic acids

inside.

• Bacteria - cell surrounded by membrane and capsule, no nucleus (nucleic acids freely in cell),

proteins in membranes and capsule.

• Eukaryotes - cell surrounded by membrane, nucleus surrounded by membrane, nucleic acid

and protein in nucleus, proteins in membranes and proteins in organelles.

Griffith's

Experiment

• Griffith was looking for a vaccine for pneumonia by studying mice infected

with two strains of the causative bacterium: S strain = virulent, R strain = non-

virulent.

• He proposed that cells contained a non-living 'transforming principles' that

could transform one cell into another.

Avery,

MacLeod

and

McCarty's

Experiment

• Aimed to identify transforming principle.

• Grew large batches of S-strain bacteria and tested which of different cellular

components was the transforming principle.

• Treated cell extracts with:

o Proteinase - transformation still occurs.

o Nuclease - no transformation.

• Implies that DNA (nucleic acid) is the transforming principle, not proteins.

Hershey-

Chase

Experiment

• Add 35S (sulphur): Radioactive proteins

o Radioactive phage heads.

o No radioactivity in phage progeny.

• Add 32P (phosphorous 32): Radioactive DNA

o No radioactivity in phage heads.

o Radioactive phage progeny.

• The Hershey-Chase experiment provided powerful evidence that nucleic acids,

rather than proteins, are the hereditary material, at least for certain viruses.

Linking DNA to Proteins

• 1950: Chargaff's rules - composition of DNA varied from one species to another but in every

species, A = T, G = C.

• Early 1950s: Franklin and Wilkins - X-ray crystallogram of DNA suggests a helical structure.

• 1953: Watson and Crick - structure of DNA. Used X-ray data from Rosalind Franklin.

Molecular Structure of DNA

• The sugar-phosphate backbone is the same along the length of the

molecule.

• Information is encoded in the sequence of bases of a DNA molecule.

• Nucleotide mono-phosphates are joined by phosphodiester bonds.

• Base pairs are linked by hydrogen bonds.

o Hydrogen bonds are weak bonds and occur between an

electronegative atom and a hydrogen atom.

• Complementary base pairing - purine + pyrimidine.

• The two DNA strands are two antiparallel sugar-phosphate chains.

o One end of each strand has a free 5' phosphate group.

o The other end has a free 3' hydroxyl group.

o The two strands are linked by a hydrogen bond between bases.

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Document Summary

Molecular structure of dna: the sugar-phosphate backbone is the same along the length of the molecule. Genetic information is stored in dna: dna is a macromolecule, dna is a polymer - made up of monomers joined together. Each mutant had only one requirement, suggesting a mutation has a simple and limited effect. By examining different mutants requiring the same supplement they could identify which enzyme in a pathway was mutated: one gene, one trait beadle a(cid:374)d tatu(cid:373): (cid:862)o(cid:374)e ge(cid:374)e, o(cid:374)e e(cid:374)zy(cid:373)e(cid:863) one gene, one protein. Proteins are the link between genotype and phenotype: proteins are the products of genes: Proteins can be the molecules visible in the phenotype, eg: beta globin. Proteins can be enzymes that make molecules visible in the phenotype, eg: tyrosinase. Proteins can be the signals to make visible changes in phenotype, eg: sry gene: different alleles produce different proteins, different proteins produce different phenotypes.

Related Questions

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

#1 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?

A) Using an RNA probe complementary to the region not removed by the truncation

B) Using an RNA probe complementary to the region removed by the truncation.

C) Using an DNA probe complementary to the region not removed by the truncation.

D) Using an DNA probe complementary to the region removed by the truncation

E) Using an antibody complementary to the region not removed by the truncation.

F) Using an antibody complementary to the region removed by the truncation.

#2To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

A) As both are composed of nucleic acids, using either would result in identical results.

B) An RNA probe because RNA has uracil bases.

C) A DNA probe because it is more stable than RNA.

D) A DNA probe because RNA cannot bind to DNA.

#3Which of the following will lower the Tm of a given DNA strand?

A)Increasing the percentage of GC base pairs

B) Raising the pH of the solution from extremely acidic to neutral

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

#4 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?

A) No, because the RNA genome would not enter the bacteria upon infection

B) No, because while DNA and RNA nucleotides are similar, they are not identical

C) Yes, because DNA and RNA nucleotides are similar

D) Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

#5 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?

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

B) Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.

C) Heat killed mouse cells are added to a non-virulent bacteria.

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

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

B) It is faster to duplicate the genome when these are present.

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

D) It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

harlequineel21

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

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BIOL1003 Study Guide - Final Guide: Bacteriophage, Cell Nucleus, Nuclease

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