Group Discussion #2

A. Carbohydrates

1. Glucose

a) Draw the Fischer projections for the linear forms of D and L glucose.

b) Draw the chair conformations of a-D-glucopyranose, b-D-glucopyranose, a-L-glucopyranose, and b-L-glucopyranose.

c) Draw the Haworth projections for a-D-glucopyranose, b-D-glucopyranose, a-L-glucopyranose, and b-L-glucopyranose.

Make sure to be able to do this for any of the monosaccharides that you’ve learned, as well as for a given Fischer projection.

2.  In an aqueous solution, a-D-glucose shows an initial specific optical rotation of +112o while a b-D-glucose solution shows an initial specific optical rotation of +19o.  Over time, however, both solutions show a specific rotation of +52.5o.

a) What phenomenon is illustrated by the change in optical rotation?

Mutarotation

b) Knowing that a solution of glucose that shows an optical activity of +52.5o, is composed
of an equilibrium mixture of both a and b glucopyranose, calculate the percentage of each
of the two forms present.

112 (%a)  +  19 (%b)  = 52.5      %b = (1 - % a)  let's call %a: x

112 (x)  +  19 (1-x) = 52.5

Solve for x;  x= 0.36 or 36%   and %b = 1-.36 + .64 or 64%

Extra - on board

a) Name the oligosaccharide using an abbreviated format.

man b(1--> 4) glc a (1-->4) gal

b) Draw out the products of the reaction between the oligosaccharide and CH3I followed by acid        hydrolysis.

c) Name an exoglycosidase that would free the monosaccharide at the non-reducing end.

b- mannosidase

3.The nutrition labels from a package of pasta and a carton of milk are shown:

 Nutrition Facts -- Pasta Amounts/serving Calories 200 Calories from Fat 10 % Daily Value* Total Fat 1g 2% Saturated Fat 0g 0% Polyunsaturated Fat 0.5g Monounsaturated Fat 0g Cholesterol 0mg 0% Sodium 0mg 0% Total Carbohydrate 40g 14% Dietary Fiber 2g 7% Sugars 2g Protein 10g

 Nutrition Facts – whole milk Amounts/Serving Calories 150 Calories from Fat 70 Total Fat 8g 12% Saturated Fat 5g 25% Cholesterol 35mg 12% Sodium 125mg 5% Total Carbohydrate 12g 4% Dietary Fiber 0g 0% Sugars 11g Protein 8g

a) The sum of the masses of total fat, cholesterol, sodium, total carbohydrate, and protein can be considered to be the total mass of the solids provided in a serving of these foods. Calculate the total number of grams of soilds that make one serving of each of these foods.

Pasta: 1g + 40. g + 10. g = 51 g

Milk: 8 g + .035 g + .125 g + 12 g + 8 g = 28 g

b) Consider the carbohydrate component of these foods and note that the subset of carbohydrates called “sugars” refers to the monosaccharides and dissacharides found in the food. Calculate the mass percent “sugars” found in a serving of pasta, and a serving of milk.

Pasta: 2g/51g x 100 = 3.9%

Milk: 11 g/ 28 g x 100 = 39%

c) Which of these foods supplies more "sugar?"

Milk is more sugary per serving and per calorie.

(This makes me feel better about eating pasta!)

B. Phospholipids and Membrane Asymmetry

The lipid composition of healthy cell membranes can be described as “asymmetric” – in other words, some glycerophospholipids and sphingolipids are more prevalent in the outer leaflet while others are more prevalent in the inner leaflet. Interestingly, cells that undergo apoptosis (programmed cell death) lose their membrane asymmetry as part of the process of their dying. Therapeutically, it might be useful to have a drug that could induce lipid asymmetry in problematic cells such as cancer cells. Clearly an important step in the development of such a drug involves understanding the mechanism behind the asymmetric distribution. In fact, this has been an interesting area of recent research.

In one technique used to study the mechanism behind establishing lipid asymmetry, researchers synthesize liposomes with one kind of phospholipid. These liposome preparations are then mixed with red blood cells in an appropriate buffer and there is an incorporation of the liposome phospholipids into the red blood cell membranes that results in an overall change in shape that correlates with the leaflet into which the phosphlipid has been added. Red blood cells normally have a biconcave disk shape. If an excess amount of phospholipids is added to the outer membrane leaflet, the cells are said to become echinocytic or “spiky.” If the phospholipids are transported into the inner membrane leaflet, the red blood cell surface appears to be covered in “craters” and are said to be stomatocytic.  In order to quantitatively describe the extent of changes in red blood cell shape, the morphological index (MI) scale was developed. With this index, biconcave cells are assigned a value of 0, echinocytic cells are given positive values (up to +5), and stomatocytic cells are given negative values (down to -4).

1. Liposomes composed of one of the glycerophospholipids, PS, PC, or PE were incubated with red blood cells and the MI values were assigned at time points shown in the table below.

 0 min 10 min 20 min 30 min 60 min 180 min PS 0.0 + 4.0 + 4.8 + 5.1 - 3.5 - 3.8 PC 0.0 + 2.5 + 3.8 + 3.8 + 3.8 + 3.8 PE 0.0 + 1.5 + 2.5 + 3.0 - 0.5 - 1.5

a) Described the distribution of each of the types of phospholipids into the red blood cell membrane.

The data shows that both PS and PE can be translocated into the inner leaflet -- PS to a larger extent.

PC is not translocated into the inner leaflet -- Therefore it is predominant in the outer leaflet

b) Describe the distribution of charge on each side of the membrane.

PC and PE each carry a neutral charge while PS carries a net negative charge at pH ~7. Because PC is predominant in the outer leaflet (There is only a small amount of PE in the inner leaflet.) And because PS is found predominantly in the inner leaflet we could describe the inner leaflet as more negative. So one could say that all carboxy groups of and all negative charges offered by membrane phospholipids are found on the inner leaflet.

2. The hydrolysis of the ester linkage at C2 of phosphatidylserine produces lyso-phosphatidylserine. Lyso-PS was introduced into red blood cells and the following MI values were noted at the same time intervals as in the experiment described in B1.

 0 min 10 min 20 min 30 min 60 min 180 min Lyso-PS 0.0 + 4.0 + 3.8 + 3.2 +2.0 +1.6

a) Describe this observation – What does it reveal regarding the mechanism of translocation in the red blood cell membrane?

Lyso-PS is missing a fatty acyl group (likely an unsaturated one). Perhaps this group is necessary for the mechanism of translocation of PS to the inner leaflet. Also, the hydrolysis leaves and extra - OH group that increases the polarity of the lipid -- making it harder to pass through the non-polar core of the membrane.

3.  Additional experiments were carried out to ascertain the involvement of ATP in the mechanism of translocation. A batch of cells was treated with with iodoacetamide and inosine, a process that results in ATP levels dropping about 60%. The treated cells were incubated with liposomes of PS and the following data was generated.

 0 min 10 min 20 min 30 min 60 min 180 min PS + ATP depleted cells 0.0 + 2.5 + 3.8 + 4.2 +4.0 +1.4

a) Interpret these results.

ATP seems to be necessary for the mechanism of translocation of PS.

4. It is proposed that there is an intrinsic protein at least partially responsible for the translocation of PS. It is nicknamed “flippase.” In order to begin understanding how a flippase might act to move PS from the outer leaflet to the inner leaflet of red blood cell membranes, cells were pretreated with diamide, a reagent that is known to react with free sulfhydryl groups.

 0 min 10 min 20 min 30 min 60 min 180 min PS + diamide  treated  cells 0.0 + 0.5 + 0.6 + 1.4 +3.0 +3.5

a) Are these data consistent with the possible existence of a flippase? Explain.

Yes, the 3-D folding of many proteins is dependent on cystine residues. The loss of these cystines in the putative flippase may result in compromising the conformational changes that need to take place in the translocation mechanism. (The data in 2 and 3 suggest that flippase may recognize the fattty acyl group of PS, and that flippase activity is ATP dependent.

5.  An ionophore acts to create “holes” in a membrane through which ions can leak out. (The ions are then bound to a chelating agent to prevent their re-entering the cells.) The following data tracks changes in the cells with the incorporation of PS with the addition of an ionophore from time 0 min to 30 min. At 30 min, the chelating agent was removed and either Na+, K+, Mg 2+, or Ca 2+ were added back to the cells.

 0 min 10 min 20 min 30 min 60 min 180 min Na+ 0.0 + 1.2 + 2.4 + 3.3 +3.5 +3.6 K+ 0.0 + 1.5 + 2.7 + 3.5 +3.5 +3.4 Mg 2+ 0.0 + 1.4 + 2.5 + 3.5 -0.4 -1.6 Ca 2+ 0.0 + 1.4 + 2.4 + 3.6 +0.5 -0.3

a) What can be said regarding the metal ion requirement of a putative flippase?

Flippase seems to be most active in the presence of Mg 2+ and a little active in the presence of Ca 2+. Therefore,  flippase can be said to require a divalent cation for activity-- probably Mg 2+.

(Flippase is not active when no divalent cation is present.)

C.  Coenzyme –Thiamine Pyrophosphate

Thiamine pyrophosphate is a derivative of vitamin B1, thiamine. A deficiency of thiamine in the diet is linked to a disease known as “beriberi.” Beriberi is characterized by an accumulation of body fluids, pain, paralysis, and ultimately death. Thiamine pyrophosphate is important in reactions involving the cleavage of bonds adjacent to carbonyl groups. There are several such reactions in carbohydrate metabolism.

The people of Malaysia belong to three communities: the native peoples known as Malays, immigrants from South India called Tamils, and immigrants from China.  During the early 20th century, the Malays lived mainly in coastal villages, the Tamils worked on rubber estates, and the Chinese worked in tin mines.  In general, the people were very poor and they consumed a limited variety of foods.

Rice was the staple diet of all three communities at that time and therefore the key source of caloric and nutrient content.  Malays typically grew their own rice and dehulled it before consuming it; dehulling partially removes the hard outer layer of the rice seed.  Following their South Indian food habits, the Tamils ate parboiled rice, a process of partially boiling rice that makes it easier to dehull and drives thiamine from the soft bran layer to the core of the grain. The Chinese ate polished white rice in which the outer bran layer of the seed was completely removed.  It was found that among the three Malaysian communities, the Chinese community was much more prone to suffer from some degree of sub-optimal thiamin intake.  Out of every 1,000 Chinese inhabitants, about 120 had mild thiamin deficiency, about 80 had severe thiamin deficiency, and about 16 people had died of it.  Sub-optimal intake of thiamin was rare in Malays and was not reported at all in Tamils.

1. Where do you suppose the main source of thiamine was in the diets of the Malays and Tamils of the early 20th century?

Thiamine is found in the bran layer of rice.

The human body cannot store thiamin and therefore requires a continuous supply of the vitamin.  The thiamin in an individual on a thiamin-free diet typically depletes in fewer than two weeks.  Thiamin absorption is affected by compounds, called thiamin antagonists or anti-thiamin factors, that alter the structure of thiamin.  These compounds include alcohols, polyphenols, and flavonoids. (Polyphenols and flavanoids are found in coffee and tea.) Thiamine is somewhat unstable in alkaline solutions and when exposed to UV radiation. Thiamine is also heat labile

2. Moderate thiamine deficiencies have been observed in industrialized countries in the 21st century. Explain.

Many individuals living in the 21st century have lifestyle habits that interfere with thiamine absorption: excessive drinking of alcoholic beverages, coffee, and tea.

Commercial preparation of foods may be subject to heat and UV radiation (sunlight) that deplete their thiamine content.

The minimum amount of thiamin needed by the body can increase because of increased physiological or metabolic demands arising from pregnancy and lactation, heavy physical exertion, illnesses like cancer, liver diseases, hyperthyroidism, and surgery.

3. What characteristic of a cancer cells demands extra thiamine?

Cancer cells are characterized by high metabolic rates. In other words, they consume and combust carbohydrates at a rate that much higher than most non-dividing cells of an adult living organism. Thiamine is necessary for carbohydrate metabolism because it is a precursor to TPP, a coenzyme required by some enzymes of carbohydrate metabolism.

4. Why is there a negative correlation between the need for thiamine and a high-fat diet?

TPP is involved in some reactions of carbohydrate metabolism. But, lipid metabolism  is not dependent on TPP. If one is eating a diet high in fat and low in carbohydrate, there is little need for thiamine.

D.  Consider the kinetics data collected for an enzyme in the absence of an inhibitor and in the presence of inhibitors “X” and “Y.” The concentration of enzyme in each reaction = 0.40 mmol/mL and each reaction volume was 1.0 mL

TABLE A:

 [Substrate] mM Velocity ( mmole / min ) No inhibitor 2 mM X 100 mM Y 3.0 10.4 4.1 2.1 5.0 14.5 6.4 2.9 10.0 22.5 11.3 4.5 30.0 33.8 22.6 6.8 90.0 40.5 33.8 8.1

a) Generate a Lineweaver-Burk plot (graph paper on next page) with each set and determine the Vmax and KM values

 Vmax KM Uninhibited enzyme 47.6 umole/min 1.1 x 10 -5 M Enzyme + X 47.6 umole/min 3.1 x 10 -5 M Enzyme + Y 9.5 umole/min 1.1 x 10 -5 M

b) With the parameters calculated above, classify X and Y by inhibitor class.

X is a competitive inhibitor; Y is a mixed inhibitor -- the subclass of mixed inhibitor known as non-competitive.

c) Calculate the kcat of this enzyme when no inhibitor is present.

Vmax = kcat [E]

[E] is given as 0.4 umol/mL and 1.0 mL is used in reaction.  (0.4 umol of Enz total)

Vmax is found to be 47.6 umole (product amt.)/min

So kcat = 119 min -1

d) Calculate the specificity constant for this enzyme when no inhibitor is present.

Kcat / Km = specificity constant = 1.1 x 10 7 min -1 M -1

Therefore , the efficiency of this enzyme is about 100x  less than "perfect." Not bad!

e) On a second sheet of graph paper generate a plot of Vo vs. Vo/ [S] for each set of kinetics data in Table A.  How can one use this plot to find the Vmax and Km values for the enzyme?  Hint: The equation of the line of the plot is  Vo =  - KM (Vo/[S]) + Vmax ; this equation is known as the Eadie- Hofstee plot.

A general Eadie-Hofstee plot is a line with a negative slope:

Vmax = y-intercept

Km = slope of the line --but, change sign to positive value.

The values determined for Vmax and Km via Eadie-Hofstee should, of course, match those found with Lineweaver-Burk.