Answers to TBL1 and TBL2 ( ATTACHED AND COPY PASTED)

[Xiaolu Xu]

TBL 1 team 4:

“Define the terms hemiacetal, acetal, hemiketal and ketal. Explain why the cyclic form of the monosaccharide glucose is a hemiacetal, and that of the disaccharide maltose is an acetal. Using maltose as an example, describe the acetal (or o-glycosidic) linkage that it contains. Draw the structures of the disaccharides lactose and cellobiose. Why are some people lactose intolerant? Why can humans not digest cellulose?”

 

 

A hemiacetal is formed when an alcohol and the carbonyl group of an aldehyde react. The resulting hemiacetal molecule has the following four groups connected to the central carbon: -H, -R, -OH, -OR.

When the hemiacetal reacts wit another alcohol molecule, an acetal is formed. The resulting acetal molecule has the following four groups connected to the central carbon: -H, -R, -OR, -OR.

 

A hemiketal is derived from a ketone, which has two R groups attached to the carbonyl group. The hemiketal molecule has the following four groups connected to the central carbon: -R, -R, -OH, -OR. In short it is a hemiketal that has no Hydrogen in its R group.

 

A ketal is formed when the alcohol group is replaced by a second alkoxy group. The ketal molecule has the following four groups connected to the central carbon: -R, -R, -OR, -OR.

 

The cyclic form of the monosaccharide glucose forms an intramolecular hemiacetal due to the reaction between the aldehyde and hydroxyl group.

 

https://lh5.googleusercontent.com/OFs82DZIe-n27edWv2x98jk7csgL1BX6H5nX3INaNsm2nEcY9b2pKK6Q89vTPxexXgd53AHWz--PgKKBUqmaN-CwK4RgXP8Qy47dsh2GSYIJVy8B--Q

Monosaccharides can be linked via o-glycosidic bonds to another monosaccharide, forming o-glycosides and this is how dissacharides and polysaccharides are formed. The diasaccharide maltose is an acetal because the alcohol group gets replaced when the two glucose units join via an alpha(1 to  4) bond.Maltose consists of two glucose molecules, when C1 of the first glucose molecule forms a bond with the C4 of another glucose molecule.

 

 

 

 

Glycosidases are enzymes which break up O-glycosidic linkages. Cellulases and lactases break down cellulose and lactose, respectively. Humans cannot digest cellulose because we lack the glycoside hydrolase or cellulase. People who are deficient in the lactase experience a lactose intolerance because the enzyme lactase is not present to hydrolyze the lactose. Thus the lactose cannot be absorbed into the small intestine.

 

TBL 2 team 4:

Team 4.

“A patient with a history of juvenile-onset diabetes (type 1) is found in a coma. Analysis of serum             indicates:  pH = 6.8, PCO2 of 12 mm Hg and high levels of ketone bodies in the plasma (PCO2 x 0.03 = [H2CO3] and pKa’s: 6.1, 10.2). (a) Determine the acid-base status of the individual and the [HCO3-]. (b) Determine the expected ratio of [HPO4-2]/[H2PO4-1] in blood  (pKa's are:  2.0, 7.2, 11.8). (c) Calculate the direction and change in the anion gap. The anion gap in plasma is defined as [Na+] - ([HCO3-] + [Cl-]). The plasma [Cl-] and [Na+] remain unchanged at 102 and 138 mM, respectively.”

 

a) the individual has ketoacidosis and the bicarb concentration is 1.8 mM

ph= pka + log [hco3]/ (pCO2 x .03)

6.8 = 6.1 + log [HC03]/ .36

.7 = log [HCO3]/ .36

antilog of .7 = 5.01

5 = [ HCO3] / .36

 [HCO3] = 1.8 mM

 

b) when pH= 6.8 and pka= 7.2             6.8- 7.2= antilog -0.4= .398 ~ 4/10

pH= pka + log [HPO4]/[H2Po4]

6.8= 7.2 + log [HPO4]/[H2PO4]

6.8-7.2 = -.398

antilog -.4è.398

c) the anion gap has increased to 34.2 mM due to the huge drop in bicarb

138-(1.8 + 102)= 34.2