TBR Biology, Section 7, Passage 5, Question 26

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sillyjoe

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The question reads:

"In the absence of uroporphyrinogen synthase, uroporphyrinogen I (uroI) is synthesized in place of the biologically active uroporphyrinogen III (uro III)."

Conversion of uro I into uro III requires:

A. An isomerase, converting a symmetric molecule into an asymmetric molecule.
B. An isomerase, converting an asymmetric molecule into an asymmetric molecule.
C. A kinase, converting a symmetric molecule into an asymmetric molecule.
D. A kinase, converting an asymmetric molecule into a symmetric molecule.

VQ2Rvtj.png


A is correct. We have to look at the two structures given in the question. First, there is no difference in the chemical constituents of the two molecules. However, the atoms are arranged differently. For this reason, an isomerase is the enzyme that would be involved. Next, in looking at the picture one should arrive at the conclusion that uro I is a symmetric molecule, symmetric around the center point of the molecule (where the iron would be bound). By switching one set of constituents, uro III is produced. Switching these constituents produces an asymmetric molecule. The correct choice is A.

Can someone please help me understand why one structure (uro I) is symmetric and one structure (uro III) is asymmetric? I am having a hard time picturing it.

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Uro I is symmetric about the point where iron would be complexed (not shown). So for Uro I vs. Uro III, take a look at how the acetic acid and propionic acid acid groups are arranged on the four "arms" if you will. Starting clockwise, let's take the order of acetic acid (left) propionic acid (right) to be "X" and proprionic acid (left) acetic acid (right) to be "Y." In Uro I going around the ring you'd have an attachment pattern of X-X-X-X, making the molecule symmetric about a point. For Uro III, the left arm has the opposite rearrangement "Y" so starting from the top you'd get the attachment pattern X-X-X-Y, so the molecule is asymmetric. Did that help at all?
 
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Look at those COO chains on the outside.

The first molecule is symmetric about the center - you could rotate the molecule 90 degrees and have the same molecule.

At first glance, the second molecule might seem to have left-right symmetry. But look at the COO chains at the top and bottom of the molecule. They aren't left-right symmetric. No matter how you try to twist, fold, or mirror image the bottom molecule, there is some kind of asymmetry going on so that it is not symmetric about any point or line.
 
Ah, I remember this question. I was thinking of line symmetry, but it's point symmetry.
 
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