The human ABO blood group system was first discovered in 1901 , and in 1924 this trait was shown to be inherited at a single gene locus with three alleles. In $1960,$ W. T. J. Morgan published a paper summarizing what was known at that time about the structure of the ABO antigen molecules. When the paper was published, the complete structures of the $A, B,$ and $O$antigens were not yet known; this paper is an example of what scientific knowledge looks like "in the making."
In any attempt to determine the structure of an unknown biological compound, researchers must deal with two fundamental problems: (1) If you don't know what it is, how do you know if it is pure? (2) If you don't know what it is, how do you know that your extraction and purification conditions have not changed its structure? Morgan addressed problem 1 through several methods. One method is described in his paper as observing "constant analytical values after fractional solubility tests" (p.312). In this case, "analytical values" are measurements of chemical composition, melting point, and so forth.
(a) Based on your understanding of chemical techniques, what could Morgan mean by "fractional solubility tests"?
(b) Why would the analytical values obtained from fractional solubility tests of a pure substance be constant, and those of an impure substance not be constant?
Morgan addressed problem 2 by using an assay to measure the immunological activity of the substance present in different samples.
(c) Why was it important for Morgan's studies, and especially for addressing problem
$2,$ that this activity assay be quantitative (measuring a level of activity) rather than simply qualitative (measuring only the presence or absence of a substance)?
The structure of the blood group antigens is shown in Figure $10-14$. In his paper, Morgan listed several properties of the three antigens, $A, B,$ and $O,$ that were known at that time $(p .314):$
1. Type $B$ antigen has a higher content of galactose than $A$ or $O$
2. Type A antigen contains more total amino sugars than B or O.
3. The glucosamine:galactosamine ratio for the A antigen is roughly $1.2 ;$ for $B$, it is roughly 2.5
(d) Which of these findings is (are) consistent with the known structures of the blood group antigens?
(e) How do you explain the discrepancies between Morgan's data and the known structures?
In later work, Morgan and his colleagues used a clever technique to obtain structural information about the blood group antigens. Enzymes had been found that would specifically degrade the antigens. However, these were available only as crude enzyme preparations, perhaps containing more than one enzyme of unknown specificity. Degradation of the blood type antigens by these crude enzymes could be inhibited by the addition of particular sugar molecules to the reaction. Only sugars found in the blood type antigens would cause this inhibition. One enzyme preparation, isolated from the protozoan Trichomonas foetus, would degrade all three antigens and was inhibited by the addition of particular sugars. The results of these studies are summarized in the table below, showing the percentage of substrate remaining unchanged when the $T$. foetus enzyme acted on the blood group antigens in the presence of sugars. $$\begin{array}{lccc} & \ {\text { Unchanged substrate (\%) }} \\ \text { Sugar added } & \text { A antigen } & \text { B antigen } & \text { O antigen } \\\hline \text { Control - no sugar } & 3 & 1 & 1 \\\text { L-Fucose } & 3 & 1 & 100 \\\text { D-Fucose } & 3 & 1 & 1 \\\text { L-Galactose } & 3 & 1 & 3 \\\text { D-Galactose } & 6 & 100 & 1 \\N \text { -Acetylglucosamine } & 3 & 1 & 1 \\N \text { -Acetylgalactosamine } & 100 & 6 & 1 \\\hline\end{array}$$
For the $\mathrm{O}$ antigen, a comparison of the control and $\mathrm{L}$ -fucose results shows that $\mathrm{L}$ -fucose inhibits the degradation of the antigen. This is an example of product inhibition, in which an excess of reaction product shifts the equilibrium of the reaction, preventing further breakdown of substrate.
(f) Although the $\mathrm{O}$ antigen contains galactose, $N$ -acetylglucosamine, and $N$ acetylgalactosamine, none of these sugars inhibited the degradation of this antigen. Based on these data, is the enzyme preparation from $T$. foetus an endoglycosidase or exoglycosidase? (Endoglycosidases cut bonds between interior residues; exoglycosidases remove one residue at a time from the end of a polymer.) Explain your reasoning.
(g) Fucose is also present in the $A$ and $B$ antigens. Based on the structure of these antigens, why does fucose fail to prevent their degradation by the $T$. foetus enzyme? What structure would be produced?
(f) and (g) are consistent with the structures shown in Figure
(h) Which of the results in
$10-14 ?$ Explain your reasoning.