Enzyme Biochemistry Practice Problem Solutions

You are a research scientist studying a novel enzyme X, and you want to characterize this new enzyme. You measure the velocity of the reaction with different substrate concentrations and get the following data:
[substrate] (mM) Initial Velocity (mmol/min) --------------------------------------------------------------------- 3.0 10.4 5.0 14.5 10.0 22.5 30.0 33.8 90.0 40.5
a) Graph the above data. From the graph, estimate KM.
Km = (approximately) 10mM
b) Calculate Vmax. Show any equations and calculations.
Using the equation:

Vmax= (approx) 40 mmol/min
c) Is X an allosteric enzyme? Explain.
No, X is not allosteric.
Allosteric enzymes do not follow Michaelis-Menton kinetics.
Allosteric enzymes would give a sigmoidal curve when plotting Vi vs. [S]
d) You decide to do this experiment again, but this time with only one third of the enzyme X concentration used in the first experiment. Draw a new graph on the same graph that you did the first graph on. Estimate Km and Vmax from the new graph.
Km stays the same as in part a. Vmax = 1/3 original Vmax = (approx) 13
e) You wish to find the amino acid sequence of the enzyme X. What methods might you use to determine this? Name at least three.
Acid hydrolysis, base hydrolysis, enzyme digestion (exopeptidase/endopeptidase), and Edman degradation.
a) Given the following kinetic data, estimate Km and Vmax for the enzyme:

The graphs below show two ways to estimate KM and Vmax from the above data:

OR

Both of these graphs, and a full explanation of their properties, is available on the Measuring Km and Vmax page.
b) What would be the effect on the initial reaction velocities if [enzyme] was reduced to 10% of the amount used above?
With 10% of the amount of enzyme, the reaction velocity at each substrate concentration will be reduced 10-fold. One way to look at this is: with 10 times less catalyst, the reaction will run 10 times slower. The resulting velocities would be:

c) How would this change in [E] effect the observed KM and Vmax?
There are two ways to look at this:
1) Empirical: Plot the new data. You will find that KM is unchanged, while Vmax is reduced 10-fold to 10 mmol/min.
2) Theoretical: KM will not change, because the expression for KM consists of rate constants only. Since rate constants are constant and do not depend on enzyme concentration, KM will not change. Vmax does depend on enzyme concentration:

So, if [E]total is reduced 10 fold, Vmax will be reduced 10-fold.
Lactose is a disaccharide found in milk. Although in the United States we are told that "milk, it does a body good," many adults throughout the world get sick from drinking milk because they cannot digest lactose. Lactose intolerance varies markedly among various human populations. (For example, only about 3% of people of Danish descent are lactose intolerant, compared with 97% of people of Thai descent.) When someone who is lactose intolerant ingests milk, the lactose accumulates in the lumen of the small intestine because there is no mechanism for uptake of the disaccharide. This causes abdominal distension, cramping, and watery diarrhea.
a) Why can't lactose diffuse across the membranes of the intestinal epithelial cells in the absence of a carrier-mediated uptake system?
Lactose is a disaccharide and has eight hydrophilic -OH groups which prevent it from passing through the highly hydrophobic bilipid layer.
b) Why does the accumulation of sugar (or any solute) in the intestinal lumen cause an influx of water that leads to watery diarrhea?
Osmosis. Water goes from areas of high concentrations of solute to low concentrations of solute to keep a certain water:solute ratio. With a large accumulation of sugar, water from the cells (intestinal epithelial cells) goes out into the lumen area to decrease the concentration of sugar.
c) Adults who can drink milk can do so because of the enzyme lactase which is located on the outer surface of epithelial cells lining the small intestine. Lactase hydrolyzes lactose into its two component monosaccharides, glucose and galactose. Both glucose and galactose can cross the epithelial cells, and therefore do not cause illness.
Based on your knowledge of transport across cell membranes, propose a mechanism by which galactose is transported into the intestinal epithelial cells. Include a diagram of your mechanism. (There are several possible solutions- you only need to propose one.)
One possible model is the Na-glucose symport, modified to transport galactose, and powered by K-Na antiports.
d) You decide to study lactase further, and see whether it can also cleave other common disaccharides, such as maltose. (Maltose = glucose + glucose.) You find that maltose is NOT cleaved by lactase, and furthermore, maltose appears to have some kind of inhibitory effect on lactase's ability to cleave lactose.
Is maltose a more likely candidate for competitive or noncompetitive inhibition of lactase? Explain.
Maltose is a more likely candidate for competitive inhibition, because of its similar structure. It can quite possibly fit the lactose active site. However, it is highly doubtful that lactase could be able to cleave a beta-linkage, whereas it can cleave an alpha linkage.
In order to confirm your hypothesis in part (d), you quantitatively study the kinetics of lactase with lactose alone, and in the presence of both lactose and maltose. You measure the initial velocity of the reaction (rate at which lactose is cleaved) at varying concentrations of substrate. The data is given below.
```
  [Lactose] moles/liter                   Velocity  (moles/min)
------------------------------------------------------------------            
                                       lactose only   with maltose
------------------------------------------------------------------
  0.3 x 10-5                              10.4          4.1
  0.5 x 10-5                              14.5          6.4
  1.0 x 10-5                              22.5          11.3
  3.0 x 10-5                              33.8          22.6
  9.0 x 10-5                              40.5          33
```
e) Make a Lineweaver-Burke plot for lactase both with and without maltose. Does your graph confirm or contradict your prediction in part (d)? Why?
A Lineweaver-Burke plot shows that they have the same Vmax, indicating competitive inhibition.
You notice one day a slimy patch of goo on your carpet. It seems as if the goo is eating away at your carpet. Being an awesome biologist, you figure out it's bacteria feeding off your carpet. But wait, your carpet is made of nylon! How can this be? Perhaps the landfill and nuclear power plant next to your house has something to do with it...
You set out to determine how the bacteria can live off of nylon since nothing known can. You scrape some off the bug-infested carpet and take it to your lab. You culture large quantities of the bacteria and painstakingly purify a protein with the ability to cleave the nylon. You name the newly discovered enzyme Leggsase. Nylon is a polymer made up of many repeating subunits (like the polysaccharides). It looks like this:

The squiggly lines at the ends indicate that this same unit is repeated many times in both directions. The arrow points to the bond that is cleaved to break up the nylon polymer.
a) Just how good is the enzyme? If we just put nylon in water, the rate at which this bond will cleave is about 1 per year. In the presence of the enzyme, it's about 100 per second. What is the increase in the rate of the reaction?
The increase in the rate of the reaction is equal to the rate of the reaction with enzyme divided by the rate of the reaction without enzyme. So we get {(100 molecules/sec.)(3600 sec./hr.)(24 hr./day)(365.25 days/year)}/(1molecule/year) = 3160000000 fold increase in reaction rate with the enzyme Leggsase.
b) You make a solution that is 0.1M in nylon. You add some enzyme and allow the reaction to reach equilibrium at 25oC. You determine the concentration of nylon at equilibrium is .0001M.
i) What is the equilibrium constant for the reaction?

ii)What is the change in free energy in kcal/mol? Is this an exergonic or endergonic reaction?

c) You set up you assay system and collect data on the rate of the reaction as a function of the substrate concentration. (Data on next page)
i) Graph rate vs substrate concentration for this enzyme reaction.
(The graph approaches an asymptote of ~ 20.0 sec.-1 after a single, rapid increase. The graph is not sigmoidal).
ii) Is this likely to be a single- or multi-unit enzyme (see page 128-9 in Purves)? Explain.
Leggsase is likely to be a single-unit enzyme because the graph is typical of a single-unit and does not have an inflection point characteristic of an allosteric enzyme.
d) Since the world probably wouldn't like to have this bacteria eating up all the nylon around, you decide to make a bunch of money by discovering an inhibitor to it. You discover that a dipeptide, Gly-Gly, is a fair inhibitor of Leggsase.
i)Draw the structure of the dipeptide and explain why it would inhibit Leggsase.

The site of the nylon molecule at which Leggsase cleaves looks like a peptide bond. The glycine dipeptide has a peptide bond, and the CH2's of the glycines mimic the CH2's in nylon that are next to the bond that is cleaved. Since the dipeptide is structurally similar to nylon, it could bind to the active site of Leggsase.
ii) Is this likely to be a competitive or non-competitive inhibitor? Explain.
The dipeptide is a competitive inhibitor, since it bears structural resemblance to the cleaved site of the nylon molecule and thus can bind non-productively to the active site of Leggsase.
iii) On your graph for part c), draw in a curve for a Rate vs. [Substrate] in the presence of Gly- Gly. (Use a dashed line or another color, but indicate which is which.)
(The line should be below the original line at lower [S], but eventually reach the same asymptotic level as the original line). This rate vs. [S] graph is typical of a competitive inhibitor.
e) You are so good, you even get crystals of Leggsase. Through the wonders of X-ray crystallography, you are able to determine the exact molecular structure of Leggsase. The active site seems to be in the middle of a long shallow grove on one face of the protein. What kinds of amino acids (polar, charged, hydrophobic, small, large, etc.) would you expect to line this groove? Why?
Small hydrophobic residues would line the groove since they would have to be able to interact with the hydrophobic CH2 groups to either side of the cleaved site of nylon.
A mutation that changes an alanine residue in the interior of a protein to a valine residue is found to lead to a loss of activity. However, activity is restored when a second mutation at another position changes an isoleucine residue to a glycine residue. How might this second mutation lead to a restoration of activity?
The first mutation destroys activity because valine occupies more space than alanine, and so the protein must take an different shape. The second mutation restores activity because of a compensatory reduction of volume: glycine is smaller than isoleucine.

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