Hi, I\'m looking for answers for these questions. If you can help me with them a

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Hi, I'm looking for answers for these questions. If you can help me with them all I would HIGHLY appreciate it. Explanations are valued :) thank you sooo much. (looking for a good answer, test prep!)

03. A specific mutation in the promoter of the gene that encodes the enzyme lsocitrate dehydrogenase causes the promoter to become non-functional, and unable to bind to RNA Polymerase. take equal numbers of wild type cells and cells that are heterozygous for the mutation, and measure the kinetics of lsocitrate dehydrogenase in the cells. My most likely prediction for the observed Km and Vmax is: T OF O? A) Km and Vmax will be the same in both wild type and mutant cells T OF B) Km and Vmax will higher in wild type compared to mutant cells D T C) Km will be the same; Vmax will be higher in the wild type cells T F D) max. will be the same, Km will be higher in the mutant cells 04. Imagine that I start off with 10 molecules of oxaloacetate in a mitochondria. After running the TCA to generate 20 molecules of CO2, how many molecules of oxaloacetate will be left in the mitochondria (assuming oxaloacetate is being used only for TCA in these mitochondria)? Explain your answer. 05. I have mutant cells that make no PFK2 enzyme. I measure rates of oxygen consumption in these mutant cells and compare them to wild type cells. What is the MOST LIKELY outcome of this experiment

Explanation / Answer

Ans. 03. Correct option. C. Km will be the same; Vmax will be higher in the wild type cells.

The wild type cell has both the normal promotors, so it expresses the functional enzyme.

The mutant cell is heterozygous, that is contains one allele with normal promotor and the other allele with mutated promotor. So, the mutated cell also has one copy of normal promotor, so it also expresses functional copy of the enzyme isocitrate dehydrogenase.

Now, note the following points-

I. The alleles of gene for isocitrate dehydrogenase exhibit co-dominance, that is both the alleles are expressed in the same cell simultaneously. This is the reason that gives the production of isoforms of the enzyme in same cell/ organism.

II. So, normal cell produces two functional copies of the enzyme.

III. Mutant cell produces only one functional copy of the enzyme.

IV. So, concentration of enzyme in normal cell is twice that of the mutated cell.

That is, [E] in normal cell = 2 x [E] in mutated cell.

V. Conclusion:

V. A. Note that Vmax is directly proportional to enzyme concentration. So, normal cell has twice the Vmax of that of the mutant cell.

            That is, maximum velocity of normal cell = 2 Vmax

                        maximum velocity of mutant cell = Vmax

            Thus, 2 Vmax (normal cell) > Vmax (mutant cell)

            V. B. Note that Km of an enzyme is independent of enzyme concertation. So

Km remains constant in both cell types. That is-

                        Km (normal cell) = Km (Mutant cell)

Ans. 04. OAA is regenerated at the end of citric acid cycle. So it’s number remains constant to 10 after generation of CO2 (decarboxylation steps).

Ans. 05. PFK2 is an enzyme of glycolysis and gluconeogenesis and helps in maintaining glucose homeostasis in the cell.

Note that none of these pathways directly use O2. O2 is used up during electron transport chain only.

A cell with non-functional copies of PFK2 still use citric acid cycle (CAC) and CTC pathways with same efficiency. Even if glycolysis is halted, citric acid cycle can use lipids and other non-glucose derived substrates. NADH and FADH2 produced during CAC is oxidized through ETC and consuming O2 ultimately.

Therefore, the mutation is most likely to leave O2 consumption rate unaffected in both cell types.

Ans. 06. Biotin is used as cofactor for carboxylases.

Pyruvate carboxylase, an enzyme of gluconeogenesis pathway, will become catalytically non-functional if biotin is not available at all.

Biotin deficiency (lower biotin concertation than optimal value) will, therefore, reduce cell’s ability to carry out gluconeogenesis because only a small fraction of total available enzyme molecules get biotin and show catalysis.

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