6. A specific mutation in the promoter of the gene that encodes the enzyme Isoci

ID: 208299 • Letter: 6

Question

6. A specific mutation in the promoter of the gene that encodes the enzyme Isocitrate dehydrogenase causes the promoter to become non-functional, and unable to bind to RNA Polymerase. I take equal numbers of wild type cells and cells that are heterozygous for the mutation, and measure the kinetics of Isocitrate dehydrogenase in the cells. My most likely prediction for the observed Km and Vmax is: T F ? A) Km and Vmax will be the same in both wild type and mutant cells T T F F F ? ? ? B) Kn and Vmax will higher in wild type compared to mutant cells C) Km will be the same; Vmax will be higher in the wild type cells D) Vmax will be the same; Km will be higher in the mutant cells T

Explanation / Answer

Ans. Given, cells are heterozygous in mutation.

Note the following points-

I. A heterozygous individual consists of 2 copies of the allele – here, one dominant (wild type) allele and the other allele with mutated promotor. So, though the allele with mutated promotor does not express/encode Isocitrate dehydrogenase enzyme; the wild type allele would produce the enzyme naturally.

II. Homozygous wild-type cells have 2 copies of normal promotors whereas the mutated cell has only 1 copy of the normal promotor. So, the quantity of enzyme produced by wild type cells is twice that of the mutated cells. That is (hypothetically), if wild type cell produces 100 enzyme molecules, the mutated cell would produce only 50.

III. Km is the measure of affinity of the enzyme for substrate. It is an intrinsic property of the enzyme for the specified substrate under given set on conditions. So, Km remains the same.

IV. Vmax is an extrinsic property of the enzyme for the substrate under given conditions. That is, if you double the amount/number of enzyme molecules, Vmax or Vo would also be doubled because there are 2time the number of enzyme molecules catalyzing the substrate.

So, Vmax in wild-type cell extract would be greater than (2 times) that of Vmax observed in the mutated cell extract.

# So, correct option is – C. Km will be the same; Vmax will be higher in the wild type cells.

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6. A specific mutation in the promoter of the gene that encodes the enzyme Isoci

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