BIOL 102 Chapter Notes - Chapter 14: Dna Ligase, Replisome, Reverse Transcriptase

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22 Jun 2016

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The chromosome theory of inheritance proposed that chromosomes are made up of genes. Since it has been discovered that chromosomes are made up of a complex of dna and protein, the question of what genes are made of came down to dna or proteins. Capsid stays attached to the host cell as a. Their strategy was based on 2 facts: proteins contain sulphur but not phosphorus, dna contains phosphorus but not sulphur. Result: all of the radioactive protein was in the ghosts, while all the radioactive dna was inside the host cells. After these results were published, proponents of the protein hypothesis accepted that dna must be the hereditary material. Recall from watson & crick"s model for the secondary structure of protein: dna is a long, linear polymer made up of monomers called deoxyribonucleotides, deoxyribose sugar + phosphate group + nitrogenous base= deoxyribonucleotides. Thus, the molecule has a 3" end and a 5" end.

Related Questions

The bacterium E. coli, which is a major model system for the study of genes and DNA, has a singular circular chromosome of about 4x106 bp. In contrast, human diploid cells have __1__ chromosomes, of __2__ different types (donât forget that the X Â¹ Y!). The total amount of DNA found in a haploid cell such as a sperm or egg is called the organismâs âC-valueâ; the human C-value is __3__.

Also in contrast to the E. coli circular chromosome is the fact that eukaryotic chromosomes are not circular, but instead are __4__. Eukaryotic chromosomes are complexed heavily with protein to form a DNA-protein complex generically called __5__. The proteins found can be considered to fall into two general classes: __6__ and __7__. The first class is made up of 5 major proteins called __8__, __9__, __10__, __11__, and __12__. On a weight to weight basis, these proteins make up about __13__% of the total mass of a chromosome. Since these proteins must associate relatively non-specifically with DNA, they are rich in the basic amino acids __13__ and __14__, which gives the proteins a net __15__ charge.

One goal of packaging the DNA into chromatin is to compact the genome so it will fit into a nucleus. The basic unit of chromatin structure is the __16__, which has __17__ base pairs of DNA wrapped around a âcore particleâ composed of __18__ copies each of the core histones. Each nucleosome is separated by a stretch of âlinkerâ DNA so that the repeating unit is approximately 200 bp. Lets do a little arithmetic: If we consider that the nucleosome is approximately __19__ nm in diameter, that there is about 200 bp in this structure, and that 200 bp of DNA would normally occupy __20__ nm if stretched out straight (recall that each base is spaced about 0.34 nm from the next), we can see that the nucleosome alone allows for a __21__-fold compaction ratio.

The nucleosome structure is further compacted by histone __22__, which is not a part of the core particle. Instead, this histone __23__ (does what?) to cause the nucleosomal 10 nm fiber to condense into a higher-order structure referred to as the __24__ nm fiber. In the nucleus, chromosomes do not just float around freely: they are tethered to a proteinaceous structure called the __25__ at specific DNA regions called âSARsâ, which stands for __26__. SARs are rich in topoisomerases, to allow the winding and unwinding of the DNA that accompanies replication and transcription to occur, despite this constraint.

The 30 nm fiber discussed above is the form in which transcriptionally active DNA is believed to exist in the nucleus; however, further condensation leads to a highly condensed inactive form called __27__; in this terminology, the active form is referred to as __28__. The inactive condensed form can be divided into two types: one type is essentially invariant between cell types, and since it is âon all t timeâ, it is referred to as __29__. The other type comes and goes, depending on the expression of the genes present in that region of the chromosome; it is referred to as __30__.

Eukaryotic chromosomes carry the genes, of course, but also have other functional regions. One such region is the place where the spindle apparatus attaches during mitosis and meiosis; this region of the chromosome is called the __31__. Another important part of the linear eukaryotic chromosome is the ends; these are called __32__. In humans, the end of each chromosome is composed of many repeats of the short sequence __33__. One more type of functional region on each chromosome is the place where DNA replication starts; in humans, each chromosome has several of these __34__. Thus, eukaryotic chromosomes carry several different types of functional sequences, which together allow for their __35__ during S phase, __36__ during mitosis and meiosis, and the __37__ of their genes

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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BIOL 102 Chapter Notes - Chapter 14: Dna Ligase, Replisome, Reverse Transcriptase

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