Solutions for Selected End of Chapter 6 Problems Class 1 1

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Solutions for Selected End of Chapter 6 Problems

Class 1

1. Store bought vs fresh picked corn. Fresh corn, just picked, is sweet (hence the name sweet corn) but it looses its sweetness as it travels from farm to supermarket. After a day after picking, up to 50% of the free sugar in the corn is converted to starch which is not sweet. To preserve sweetness, the husked ears of corn can be immersed in boiling water for 1-2 minutes, then rapidly cooled. These “blanched” ears keep their sweetness in the refrigerator or freezer. Why?

Heating inactivates enzymes, the enzymes that convert sugar to starch.

2. Calculating the intracellular concentration of the average bacterial enzyme from a bunch of assumptions (data):

a. There are 1,000 different enzymes in the cytoplasm.

b. Each enzyme has a MW of 100,000 daltons.

c. Bacterial cell is a cylinder with a height of 2.0 μm and a diameter of 1.0 μm.

d. Cytosol has a specific gravity of 1.20 is 20% soluble protein by weight.

There are three different strategies to do the calculation. The simplest:

The total concentration of protein in the cytosol:

= 2.4 x 10-6 mole/mL = 2.4 x 10-3 mole/L

For 1 enzyme of the thousand types of enzymes, then

2.4 x 10-3 M / 1000 = 2.4 x 10-6 M

Note that this way did not use all the data provided. It is good to think out a solution rather than try to use all the data, just use the data that you need to solve the problem.

Another more complex way would be: calculate the moles of each enzyme in cell / cell volume

First calculate the volume of the bacterial cell = πr2h = 1.6 μm3

Then calculate moles of each enzyme in the cell:

(0.20)(1.2 g/cm3)(1.6 μm3)(10-12 cm3/μm3) / (100,000 g/mole)(1000) = 3.8 x 10-21 mole

Then the average molar concentration: 3.8 x 10-21 mole / 1.6 x 10-15 L = 2.4 x 10-6 M

There is another way that even take more steps. They all are valid and come out with the same answer. A good take from this problem is that enzyme concentrations in cells are low. Each is not present at the same concentration. For some proteins there are only <10 molecules/cell, for others it can be as high as 200,000 molecules/cell. The actual numbers are highly regulated: that is all about molecular biology.

3. Rate enhancement. Most enzymes specifically enhance the rate of chemical reactions. Urease rate enhancement is 1014. Data: a given quantity of urease can convert a given quantity of urea in 5 min at 20oC and pH 8.0. How long would that take the non-enzyme catalyzed reaction to go to completion?

It would take 1014 times longer, lets express this in years:

(5 min x 1014 ) / (60 min/hr)(24 hr/day)(365 days/yr) = 9.5 x 108 years = 950 million years !

WOW…enzymes are quite spectacular! And the Power of logs!

4. This problem could be better stated. It says that hexokinase “when incubated at 45oC in the presence of a very large concentration of one of its substrates” could lead to two alternative explanations: 1) The substrate bound to the enzyme and held it in one conformation, which makes sense because as shown in the text and “lecture”, hexokinase was the first enzyme to show a conformational change when binding to its substrate glucose. Or, 2) the concentration was so high is removed waters from the enzyme making it more stable. Heat makes water molecules move around much faster. And, it is known that dehydrated proteins are much more stable than proteins in solution.

They should have simply stated that “when incubated at 45oC in the presence of 2 times the KM concentration of one of its substrates”. This would be concentration that would not cause dehydration.

5. Amino acids at the active site do not have to be right next to each other in primary structure.

a. Here with one of the many carboxypeptidases, arg145 would be how far away from glu270 if the protein were all α-helix? This is simple because we know the properties of α-helices: 3.6 amino acids/turn and one turn is 5.4Å long. So from position 145 to 270, it is125 amino acids away. This would be 125 aa / (3.6 aa/turn) = 34.7 turns. And, 34.7 turns x 5.4Å/turn = 188 Å or 19 nm. It doesn’t seem like much, but compare this distance to a whole polio virus (containing many proteins wrapped around it’s RNA genome) which is only 28 nm in diameter.

But, we know proteins are not a straight line of amino acids, but rather have all sorts of neat folding such that amino acid 145 can be right next to amino acid 270. Our model enzyme, chymotrypsin’s active site is dependent on ser195 being very close to his57.

b. That folded protein simply gets the reactive amino acids (mainly R groups) in the right orientation, being very close, to catalyze the reaction.

6. To measure Lactate Dehydrogenase, we can use a simple spectrophotometer and a watch. In the spectrophotometer tube, add the buffer, the right amount of pyruvate and NADH. After putting it into the spectrophotometer, the Optical Density at 340 nm (due to the absorbance of NADH) should he at the high end of the OD scale. Then at time zero, add the enzyme and put the tube back into the spectrophotometer. The OD will decrease over time as the enzyme catalyzes the reaction converting NADH to NAD+ and pyruvate to lactate. What you need to do is record the OD at specific times, like every 10 sec or so until the OD goes close to zero. Then, the change in OD / min or sec can be easily determined.

9. Good ole Happyase an enzyme that can convert: HAPPY SAD.

a. When [Et] = 4 nM, the Vmax is 1.6 μM sec-1, what is the kcat? Have to convert either [Et] to μM or Vmax to nM/sec. Either one works.

kcat = = = 400 sec-1

b. In another experiment, using [Et] at 1 nM and [Happy] = 30 μM the vo was 300 nM/sec which is the same as 0.3 μM/sec. What is the measured KM? This can be solved by getting the new Vmax (because the concentration of of enzyme was changed.

So, kcat = can be changed to Vmax = [Et] kcat = [1 nM] (400 sec-1) = 400 nM/sec which is 0.4 μM/sec.

Now we can use the Michaelis-Menten equation if all the units are similar (all molar conc in nM or in this case μM:

Vo = = 0.3 μM/sec =

Then solving for KM, first cross multiply to get:

(0.3 μM/sec) (Km + 30 μM) = 0.4 μM/sec (30 μM)

0.3 μM/sec(Km) + 9 μM2/sec = 12 μM2/sec

0.3 μM/sec(Km) = 3 μM2/sec

Km = 10 μM

Another way to do this would be to first rearrange the Michaelis-Menten equation to:

Vo/ Vmax = [S] / (Km + [S])

(300 nM/sec) / (400 nM/sec) = ¾ = [S] / (Km + [S])

4 [S] = 3 Km + 3[S]

Km = [S]/3 = 30 μM / 3 = 10 μM

c. OMG an inhibitor in the enzyme preparation, and ANGER was it’s name! REMOVE it and to further purify the enzyme. After removal of ANGER, the Vmax increased to 4.8 μM/sec and the Km became 15 μM. We really do not have to calculate α and α’ to know what sort of inhibitor ANGER is: has to be mixed because it is affecting both Vmax and KM.

But you can calculate the two alphas: α = 2 and α’ = 3.

11. Estimating Vmax and KM just by looking at the data:

[S] in M

vo in μM/min

2.5 x 10-6


4.0 x 10-6


1 x 10-5


2 x 10-5


4 x 10-5


1 x 10-4


2 x 10-3


1 x 10-2


Look at the data…the substrate concentrations are going from ~10-6 to 10-2, that is a ten thousand times change. This is enormous, so the Vmax is likely around 140 μM/min. Half of the Vmax would be 70 μM/min…which is at [S] = 1 x 10-5 M which is then the KM (defined as [S] at half Vmax. Neat, how easy is this!!!

13. This problem is just like part of the Practicase problem in the “lecture”. We will be doing a similar graphing problem in Class 1 of Chapter 6. So this a GREAT THING to do BEFORE CLASS. Practice Practicase!

Lineweaver-Burke Plots: Be sure to make the ordinate (y-axis) near the middle of the page NOT on the left edge. This is important because you need to extrapolate the curve through the y-axis and and on to the x-axis (abscissa). This is a simple exercise in first converting the data number (x) into reciprocals (1/x)…and then plotting the reciprocals.


12. This is a classic, and it has medical application: how ibuprofen works. Plus it is a great idea to practice graphing Lineweaver Burke plot. The enzyme is a cyclooxygenase that converts a normal 20 carbon fatty acid, arachidonic acid, to PGG2 (an intermediate in the formation of prostaglandins (local hormones, some of which initiate the pain response on nerves) and thromboxanes. In this example, ibuprofen is used as the enzyme inhibitor, but it could have been aspirin or naproxen.

Here’s the data:

[arachidonic acid]


Rate of PGG2 Formation


Rate of PGG2 Formation in presence

of ibuprofen (10 mg/mL),

















a. For the first two columns, plot it as a Lineweaver-Burke plot and determine the KM and Vmax.

To do that first convert all the data to reciprocals. It should come out like:

Here’s the data converted to 1/[S] and 1/(mM/min) which is 1/vo:

[arachidonic acid]

mM (1/[S])

Rate of PGG2 Formation

mM/min (1/vo)

Rate of PGG2 Formation in presence

of ibuprofen (10 mg/mL),

mM/min (1/vo)

0.5 (2.0)

23.5 (0.043)

16.67 (0.060)

1.0 (1.0)

32.2 (0.031)

25.25 (0.0396)

1.5 (0.67)

36.9 (0.027)

30.49 (0.0328)

2.5 (0.40)

41.8 (0.024)

37.04 (0.0270)

3.5 (0.28)

44.0 (0.023)

38.91 (0.0257)

Now plot this out….it should NOT look like: In this copied graph, the data is plotted correct, but the cartoonist forgot to put in the units for each axis. The x-axis should have following 1/[S], 1/mM, and the y-axis should have following 1/vo, 1/(mM/min). Graphs are incomplete without all the information needed to interpret the graph.

a. Now calculate the KM and Vmax: this is calculated ONLY from the rate without inhibitor.

-1/KM = -1.7 KM = = - = 0.59 mM

Vmax = = = 50 mM/min

b. This is obviously a competitive inhibitor, KM is increased (-α/KM is closer to the origin).

-1 / α KM = -1.0 α =1.69 α = 1 + Ibuphrofen 10 mg/mL is10 g/L MW = 206 g/mole

[I] = 48.5 mM So: 1.69 = 1 + So, Ki = 70.2 mM

This inhibitor binds the enzyme with less affinity than the natural substrate!

19. Inhibition of carbonic anhydrase by acetozaolamide which is a diuretic (increases urine formation) to lower blood pressure. The graph below shows the effect of acetazolamide then explain how acetazolamide works:

1. Acetazolamide inhibits achieving Vmax, thus it is interacting with ES. Here the Acetazolamide inhibiton Vmax is about half of the un-inhibited Vmax.

2. The overall shape of both curves is almost identical. It is likely not affecting the KM = substrate binding (both curves have identical [S] for their ½ Vmax values).

3. So, this data if plotted on a Lineweaver Burke plot, both curves would intersect at the x-axis.

4. This is a special (rare) form of mixed inhibition called non-competitive inhibition. PLEASE do not CONFUSE this with UN-COMPETITIVE inhibition they are completely different.
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