Suppose you are working in the laboratory and have been utilizing a DNA polymera

Question

Suppose you are working in the laboratory and have been utilizing a DNA polymerase in an experiment. The polymerase catalyzes the synthesis of DNA from individual nucleotides. In order for the polymerase to function, it requires a single-stranded DNA template, a short DNA primer, and nucleotides (ATP, GTP, CTP, and TTP). If everything is OK, the polymerase will then synthesize a new strand of DNA that is complementary to the template strand.

After a bit of trial and error, you find conditions where the synthesis reaction goes well: 1 µg of template; 100 pmols of primer; 5 units of polymerase; and 25 µM of each of the nucleotides.

Just when the Novel Prize looms clearly on the horizon, tragedy strikes: the reaction ceases to work well. The only difference between when the reaction was working and now is that the DNA polymerase is from a different source (isolated from a different organism). [NOTE: enzymes like polymerases used in most laboratories are purchased in kits from companies.] The Km of the "old" polymerase was 20 µM nucleotide, and the Km of the "new" polymerase is 100 µM. What is the first thing you would do to change the reaction conditions (using the new polymerase) in your attempt to get the reaction to work?

A) Increase [nucleotides] to 125 M

B) Increase [nucleotides] to 80 M

C) Decrease [nucleotides] to 10 M

D) Decrease [nucleotides] to 20 M

E) Increase [nucleotides] to 50 M

Explanation / Answer

This question deals with the Michaelis-Menten kinetics of enzyme kinetics. Km is a value which refers to half of the substrate concentration at the maximum rate of the reaction. Km is also an inverse measure of the affinity of the substrate for the enzyme. Higher Km indicates lower affinity and lower Km indicates higher affinity. Since the old enzyme had a lower Km than the new polymerase enzyme, the old enzyme was more efficient in catalyzing the reaction and had more affinity for the substrate. The new enzyme has a Km that is 5 times higher than the old one which means that the affnity of substrate is 5 times lower. Thus, in order to be successful we have to increase the substrate concentration by 5 times i.e. 5 x 25uM of nucleotides = 125 uM. Hence the correct answer is A) Increase [nucleotides] to 125 uM.

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