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#Question id: 13094


You are studying a specific gene in yeast, and you want to express that yeast gene in E. coli. Your task is to design a strategy to insert the yeast gene into the bacterial plasmid. Below is a map of the area of the yeast genome surrounding the gene in which you are interested.
 
The distance between each tick mark placed on the line above is 100 bases in length
Below are the enzymes you can use, with their specific cut sites shown 5’-XXXXXX-3’ 3’-XXXXXX-5’
 
The plasmid is 5,000 bases long and the two farthest restriction enzyme sites are 200 bases apart. The plasmid has an ampicillin resistance gene somewhere on the plasmid distal from the restriction cut sites.
                                     
You transform your ligation planned in which two restriction enzymes would you use to design a way to get the insert into the vector if you had to use two enzymes simultaneously, into bacteria and plate the bacteria on Petri plates containing ampicillin. (You actually transform six different ligation mixtures, which are described below, into six different populations of cells, and plate each transformation onto a different plate, because you want to do all of the correct controls.) The next day you come in to lab to look at how many colonies of bacteria are on each plate. You are really excited, because the number of colonies you see on each plate tells you that the entire procedure worked! Which of the three following patterns of number of colonies did you see in order to conclude that you had a successful transformation?
In this table, DV = digested vector. DYG = digested yeast genome.
 
a) Pattern-1, DV only + Ligase→No colonies b/c you have digested with 2 different restriction enzymes that can’t ligate together 
b) Pattern-2, DYG only + Ligase→ No colonies because all you transformed is the digested, linear yeast DNA.
c) Pattern-3, Water + Ligase→ No plasmid with the ampicillin resistance gene (or any DNA) was transformed into the bacteria and so it won’t grow in the presence of ampicillin.
d)Pattern-3, DV + DYG + Ligase→Colonies. The plasmid and yeast gene can ligate together to form a functional plasmid that will express the ampicillin resistance gene.
e) Pattern-1 and 2 only, DV + DYG (No Ligase) →No colonies because, although you have both digested plasmid and a digested yeast gene with complementary sticky ends
Which of the following statements about these ligations and their pattern is correct?

#Unit 13. Methods in Biology
  1. A and C
  2. C and D
  3. D and E
  4. B, D and E
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TLS Online TPP Program

#Question id: 4256

#Unit 2. Cellular Organization

With respect to CpG islands in our genomic DNA, which, if any, of the following descriptions do not apply?

1) frequently occurring (there are about 30,000 in the human genome).

2) long DNA sequences (typically from 10 kb to 100 kb in length).

3) low CG dinucleotide frequency.

4) frequently associated with transcriptional start sites.

TLS Online TPP Program

#Question id: 4257

#Unit 3. Fundamental Processes

An original section of DNA has the base sequence AGCGTTACCGT. A mutation in this DNA strand results in the base sequence AGGCGTTACCGT. What type of mutation does this change represent?

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#Question id: 4258

#Unit 3. Fundamental Processes

A single base substitution mutation is likely to have a less deleterious effect when the base change exhibits which of the following results?

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#Question id: 4259

#Unit 3. Fundamental Processes

Rank the following one-base point mutations with respect to their likelihood of affecting the structure of the corresponding polypeptide (from most likely to least likely).

1. insertion mutation deep within an intron

2. substitution mutation at the third position of a codon in an exon

3. substitution mutation at the second position of a codon in an exon

4. deletion mutation within the first exon of the gene

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#Question id: 4260

#Unit 3. Fundamental Processes

Which of the following occur(s) when the eukaryotic translational machinery encounters the TAG codon?

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#Question id: 4261

#Unit 3. Fundamental Processes

Polyadenylation can promote translation

a. of the fragile-X gene.                     

b. of stored oocyte mRNAs.

c. by allowing for greater interaction at the 5´ end.

d. by allowing for greater interaction with translation initiation factors.