BIOL 200 Lecture Notes - Lecture 8: Minisatellite, Dna Profiling, Genetic Disorder

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erinshark255

2 Oct 2016

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The c-value enigma- the lack of correlation between genome size and the biological complexity of an organism. Gene density is variable among eukaryotes due to the size of intragenic (introns) and intergenic (exons) regions. Difference in genome size is due to the amount of non-coding dna. The entire nucleic acid sequence that is necessary for the synthesis of a functional gene product (polypeptide or rna) Organized in operons transcribed into polycistronic (sing. cistronic) Mrna is polycistronic so it has multiple sites for protein synthesis because there are many proteins. Single mrna: 3" and 5" end are untranslated regions. Polycistronic are rare in eukaryotes: separate genes (sometimes on different chromosomes) Translated into separate polypeptides: instead are polyadenylated (sequence of a"s) At the end of every transcriptional sequence. Before it there is an polyadenation signal (a sequence) that tells the cellular machinery that"s outside the nucleus to add the a tail on the mrna.

Related Questions

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

Multiple answers to each question might be possible!

	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.

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.

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.

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.

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.

Explain the 5â to 3â directionality of a DNA stand.

	It is due to the fact that the free 5â carbon is on one end and the free 3â carbon is on the other
	It is due to the fact that new nucleotide are added to the 5â carbon of the previous nucleotide
	It is due to the fact that there are 3 phosphate groups attached to the 5â carbon
	It is due to the fact the complementary strand is 3â to 5â
	More than one of the above explain the 5â to 3â directionality

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Why does DNA polymerase utilize an RNA primer?

	DNA polymerase is unable to initiate strand synthesis but RNA polymerase can
	DNA polymerase can only add a dNTP to an rNTP
	DNA synthesis proceeds in the 3â to 5â when initiating strand synthesis
	Chromosomal DNA contains interspersed RNA fragments
	The RNA primer increases stability of the newly synthesized strand

rosehippopotamus950

22. What is probably the single most important distinction between bacteria and eukaryotes?

a. transcription

b. the separation of the cell's genetic material from the surrounding cytoplasm

c. the existence of ribosomes

d. RNA processing in bacteria

e. antibiotics

23. Which one does NOT characterize an open reading frame (ORF)

a. Uses six reading frames during translation

b. Contains a start codon

c. Contains a stop codon

d. Its length is in multiples of 3 nucleotides

e. It should have the number of codons for a protein or on average ~250 codons in length.

24. Nested intronic genes lie within the non-coding intronic region of a larger gene, and occur in the human genome.

a. True

b. False

25. How would you characterize this statement: EF-Tu will hydrolyze GTP only if the codon:anticodon & tRNA:amino acid matchings are correct

a. Transcription; initiation steps

b. Transcription; termination steps

c. Translation: initiation steps

d. Tanslation: elongation steps

e. Spliceosome

26. Regarding the transcription unit architecture of the Escherichia coli genome, which one is FALSE?

a. Each of the 4,661 transcription units comprised an average of 1.1 modular units.

b. 3,010 transcription units (~65%) are monocistronic.

c. 1,652 transcription units (35%) contain more than one ORF (polycistronic)

d. The largest transcription unit is equivalent to 16 ORFs with numerous introns.

e. 398 transcription units (~9%) are nested within each other in opposite transcriptional orientation

27. Regarding RNA interference (RNAi), which one is FALSE?

a. It is a common and conserved mechanism in eukaryotic cells.

b. Originally discovered in bacteria and plant viruses.

c. A form of primitive immunity to protect the genome.

d. Pre microRNA (miRNA) genes have been found in introns.

e. Recent research shows that exogenous plant miRNAs in food can regulate the expression of target genes in mammals.

28. Aminoacyl-tRNA synthetases

a. Recognize the 3âUTR

b. The genes for these enzymes fall into the family of inducible genes

c. The enzymes are imported into the nucleus

d. Covalently attaches an amino acid to the appropriate tRNA. This takes place in the cytoplasm.

e. Forms the peptide bond during protein synthesis

29. What is the process by which complementary RNAs are made from a DNA template?

a. translation

b. transcription

c. protein synthesis

d. transduction

e. transliteration

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BIOL 200 Lecture Notes - Lecture 8: Minisatellite, Dna Profiling, Genetic Disorder

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