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Question 1: Restriction Enzymes, 5 points As you know, bacteria express restrict

ID: 215950 • Letter: Q

Question

Question 1: Restriction Enzymes, 5 points As you know, bacteria express restriction enzymes to protect themselves from invading viruses called bacteriophages. Below are the restriction sites of three restriction enzymes. The arrows show where the enzymes cut each of the strands of DNA. Not EcoRV Acil GCGGCCGC. .3 . GATATC...3 3...CGCCGGCG...5 3...CTATAG..5 3... GGCG...5 A. 1 point. The E. coli genome is about 6 million base pairs long. On average, how many EcoRV sites would you expect in a genome of this size? To get credit you must show how you solved this problem. B. 1 point. Explain how E. coli that express the EcoRV restriction enzyme are able to keep their own genome intact? C. 3 points. Which of the enzymes shown above, if any, do you think would provide the most protection to bacteria against random invading viruses? Or do you think these sites would be equally effective? Explain your answer here.

Explanation / Answer

a. The answer is
As the DNA is composed with four nitrogen bases, at a time, at one position one nitrogen bases will occupay the position out of four. so, the proportion or probability of one nitrogen base = 1/4.

As the EcoRV has six bases in its restriction site, the probability of occurance of these sequence = 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = 1/4096.

As, the DNA has 6 million base pairs, the freequency = 6000000/ 4096 = 1464.84 ~ 1465.

b. As RM modification mechanism (in this system methylases added to nitrogen bases, the restriction enzymes cannot initiate the clevage reaction) protects the own, these resctriction enzymes cannot cleave her own DNA.

c. The restriction enzyme freequency is depends on the number of nitrogen bases in the recognition site. If the no of bases is less, the freequnecy will be high, so,
the enzyme which has less bases in its recognition sequence, that will be most effective one.
So, here the most effective one is Aci1.

The probability of Aci1 site = 4 bases in recognition sequnce = 1/4 x 1/4 x 1/4 x 1/4 = 1/256.

The probability of Aci1 site = 4 bases in recognition sequnce = 1/4 x 1/4 x 1/4 x 1/4 = 1/256.

The probability of EcoRV site = 6 bases in recognition sequnce = 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = 1/4096.

The probability of Not1 site = 8 bases in recognition sequnce = 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = 1/65536.

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