CSB340H1- Midterm Exam Guide - Comprehensive Notes for the exam ( 63 pages long!)

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sangriamule773

11 Oct 2017

School

UTSG

Department

Cell and Systems Biology

Course

CSB340H1

Professor

alanmos

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Document Summary

Genomic techniques for large scale gene expression analysis. Basic genomic techniques is genome sequence sequence: determine order of atgc. 3 billion base pair we determine the order of atgc. Next generation sequencing (a. k. a massively parallel sequencing, high-throughput sequencing) (cid:149) refers to new sequencing technologies that appeared in the mid-00s. Dna sequencing technology did not change much between 1980 and 2000. (cid:149) for the last 10 years, Dna sequencing technology has been improving extremely rapidly (cid:149)new methods are still appearing, indicating that we have not nearly reached the theoretical limit. Graph- improvement in dna sequencing technology y axis is cost x axis year cost has dramatically decreasing. Annual review of genomics and human genetics (2008) 9: 387-402. Dna microarrays and dna chips (cid:149) original functional genomics technology. Actual scan of an old fashioned microarray made in the lab (by a robot) Comparing two (or more) transcriptomes (cid:149) making any measurement for 1000s of different genes is technically challenging.

Related Questions

1. Compare and contrast the three kinds of transposable element.Consider their structure and method of transposition. Describe theeffects that transposons can have on bacterial gene expression andon bacterial evolution.

2. Discuss the importance of homologous recombination andtransposition in natural horizontal gene transfer and evolution.Consider how these two process affect genome structure and theeffect they have depending on what kind of sequence the insertinto.

3. Describe one strategy you could use to find a gene involvedin BPA degradation in the genomic library of a bacterium that cangrow using BPA as a sole source of carbon and energy. Include anyassumptions you made about the gene you are looking for of thelibrary you are screening. Propose a second experiment to confirmyour result. You do not have the genome sequence of thisorganism.

periwinkleeagle966

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

Section A â Edit distance (8 points total)

1. Find the edit distance between ACGAAGTTC and ACCAAGGTGTTC. How do you know it is at most that much? How do you know it is at least that much?

Section B â More on sequence alignment (52 points total)

2. (6 points) Which of the following would be the two most appropriate to assess whether there are regions of homology between two distantly related proteins? Select two.

A) Needleman-Wunsch Algorithm

B) edit distance

C) local sequence alignment

D) global sequence alignment

E) dot plot

3. Consider the following alignment: ACACCAGGTCCCC ACA-----TGACC

a) (3 points) With a match score of +3, and mismatch score of -1, and using a gap penalty of -2, what is the score of the above alignment?

b) (3 points) If we keep the same match score and mismatch score, but use a gap open penalty of -6 and a gap length penalty of -1 per additional residue, what is the score of the previous alignment?

c) (8 points) Using the scoring system used in part b (match +3; mismatch -1; gap open -6 ; gap extension -1), Which of the following alignments is best? Which is worst? How do you know?

A) ACACCAGGTCCC A---CATGA-CC

B) ACACCAGGTCCC ACA---TGACC

C) ACACCAGGTCCC ACATGA----CC

D) ACACCAGGTC-CC ACA-----TGACC

E) ACACCAGGTCCC ACA--TG-ACC

F) ACA---CCAGGTCCC ACATGACC-------

4. Look up the INS gene in Entrez Gene.

a) (2 points) What tag do you put inside square brackets after INS to search for INS as a gene symbol?

b) (2 points) Find the record for the human INS gene. Which chromosome is INS on?

c) (2 points) What is the accession number for the RefSeq record for the shortest mRNA for human INS?

d) (2 points) What is the accession number for the RefSeq record for the human geneâs genomic DNA?

d) (6 points) Using the Needleman-Wunsch program, how could you find the mRNA sequence within the genomic DNA version? Give the best strategy, including the scoring parameters.

e) (4 points) Find the next gene upstream of the INS transcription start site. Which neurotransmitter does the product of this next gene make? How many base pairs separate the two genes?

f) (2 points) Now look up ins-1 as a gene symbol in Entrez Gene. Choose the entry for the worm C. elegans. Which chromosome is it on?

g) (2 points) What is the URL for the Wormbase link on this Entrez Gene record? What is the gene accession number for ins-1 in the C. elegans genome database?

g) (10 points) Perform Needleman-Wunsch alignments between C. elegans INS- 1 protein and Homo sapiens INS preproprotein, using the default parameters, except changing the substitution matrix to PAM10 for one alignment, and PAM50 for the second alignment. Use the following online tool: https://www.ebi.ac.uk/Tools/psa/emboss_needle/ What are the number of identical residues, score, and number of gapped residues for each of the two alignments? What conclusions can you draw about which of the two matrices, PAM10 or PAM50, is more appropriate in this situation?

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CSB340H1- Midterm Exam Guide - Comprehensive Notes for the exam ( 63 pages long!)

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