CHEM 200 INTRODUCTION BIO EXAM QUESTIONS WITH
COMPLETE SOLUTIONS|UPDATED VERSION|2025/2026
Intro Bio Questions and answer archive. Use the find function to search for
topics of interest
1. A: No, allosterics are in a separate class. They do not obey the simple
Michaelis-Menten kinetics that we have presented here. That is, even the
curves for Vo vs. S are different. The mechanism is more complex. Non-
competitve inhibitors on the other hand do follow Michaelis-Menten
kinetics, in the way shown in lecture. Indeed they are defined this way, by
their kinetic behavior. A mechanism by which a non-competitive inhibitor
blocks a catalytic site without blocking substrate binding and without
distorting the enzyme would explain this kinetic behavior (lower Vmax,
same Km).',1),(###-09-22&&
2. Q: One of my Chemistry professors made a specific point during our
discussion that catalysts don\'t alter the activation energy, they alter the
mechanism by which the reaction proceeds, and this alternate pathway
corresponds to a lower activation energy. He was very emphatic about
how
,incorrect it was to state that \"catalysts act by reducing the activation
energy\" - yet this is what is in your notes and the text. Can you reconcile
these seemingly contradictory statements for me
3.A: I\'m glad you brought this up and got an expert opinion from a chemist. I
have always felt a little uncomfortable talking about the activation energy
being \"lowered,\" even though that\'s what it says in most biochemistry
texts. I think your chemistry professor\'s explanation is much more
reasonable and provides a better insight into what\'s going on. The energy
needed to reach a transition state by the simple collision of the two molecules
represents the activation energy for that situation and nothing can lower it
for that situation. The catalyst brings the two molecules together by a
different mechanism, (binding them side by side often) and a transition state
can be reached in this situation with much less input of energy. We don\'t
need the great kinetic energy necessary in the uncatalyzed reaction. So I like
his phrasing. Nevertheless the fact remains that in the presence of the
catalyst, the activation energy needed is less, it is lower, than in the
uncatalyzed situation. So I can understand why one might say \"In the
presence of a catalyst the activation energy to reach a transtition state is
lower.\", and then \"Catalysts work by allowing a lower activation energy\"
and then it\'s short step to \'Catalysts act by lowering the activation energy\"
which could even be considered to be literally if loosely true, but I agree is
misleading. Indeed, I plan to incorporate this way of explaining catalysts into
future years\' lectures. Thanks. *1),(###-09-22&&
3. Q: I want to know more about the mechanism behind noncompetitive
inhibition. It apparently doesnt seem to distort the enzyme in the slide, yet
, it\'s binded to another site. Does it repel substrate at the binding site? It\'s
just unclear to me how it\'s lowering reaction velocity.&&
A: No. The substrate binds just fine in non-competitive inhibition. The inhibitor does not interfere
with the substrate binding, it does not compete with it for the substrate binding site, thus it iscalled
non-competitive, when this situation is seen. It could interfere with a side chain that is importantfor
catalyzing the reaction once the substrate is already bound. Thus the catalysis would not take place,
even though the substarte is bound. ',1),(###-09-19
4. Q: In problem 2-11b, why are the 26 spots if in part A, leu -> ala and both
are nonpolar
8 A: Leu and ala are both non-polar. They both appear in the non-polar
category of pages in books showing the categories of amiono acids. Why are
they non-polar? Because they have side chains made up of hydrophobic pure
hydrocarbons (-CH2- or -CH3). But leucine has a lot more of these non- polar
groups than ala,which has but one. And the side chains must \"compete\"
with the polar end of the molecules to establish the overall character of the
molecule. So leu with its many hydrocarbon carbon atoms is a lot more non-
polar than ala. All amino acids with predominantly non-polar side chains are
non-polar but some are more non-polar than others. ',1),(###-09-19
9 Q: In problem 1-23 (16th ed.rev.), how do you generate the chair
form of galactose from the Haworth ring forms given in the texts
9 A: It is not necessary to do this to answer the question (part b). You
know from the Haworth or even from the straight chain (Fisher) projections
that glucose and galactose differ only in the orientation of the OH at C4. The
chair form of glucose shows that the orientation of the OH at C4 in glucose is
, equatorial and that the H is axial up. . Therefore in galactose it must be the
other way around: now it is the OH that is axial up. There are rules and
conventions for following the H\'s from a Fisher projection all the way to a
chair, but they require knowing (memorizing) the conventions and
COMPLETE SOLUTIONS|UPDATED VERSION|2025/2026
Intro Bio Questions and answer archive. Use the find function to search for
topics of interest
1. A: No, allosterics are in a separate class. They do not obey the simple
Michaelis-Menten kinetics that we have presented here. That is, even the
curves for Vo vs. S are different. The mechanism is more complex. Non-
competitve inhibitors on the other hand do follow Michaelis-Menten
kinetics, in the way shown in lecture. Indeed they are defined this way, by
their kinetic behavior. A mechanism by which a non-competitive inhibitor
blocks a catalytic site without blocking substrate binding and without
distorting the enzyme would explain this kinetic behavior (lower Vmax,
same Km).',1),(###-09-22&&
2. Q: One of my Chemistry professors made a specific point during our
discussion that catalysts don\'t alter the activation energy, they alter the
mechanism by which the reaction proceeds, and this alternate pathway
corresponds to a lower activation energy. He was very emphatic about
how
,incorrect it was to state that \"catalysts act by reducing the activation
energy\" - yet this is what is in your notes and the text. Can you reconcile
these seemingly contradictory statements for me
3.A: I\'m glad you brought this up and got an expert opinion from a chemist. I
have always felt a little uncomfortable talking about the activation energy
being \"lowered,\" even though that\'s what it says in most biochemistry
texts. I think your chemistry professor\'s explanation is much more
reasonable and provides a better insight into what\'s going on. The energy
needed to reach a transition state by the simple collision of the two molecules
represents the activation energy for that situation and nothing can lower it
for that situation. The catalyst brings the two molecules together by a
different mechanism, (binding them side by side often) and a transition state
can be reached in this situation with much less input of energy. We don\'t
need the great kinetic energy necessary in the uncatalyzed reaction. So I like
his phrasing. Nevertheless the fact remains that in the presence of the
catalyst, the activation energy needed is less, it is lower, than in the
uncatalyzed situation. So I can understand why one might say \"In the
presence of a catalyst the activation energy to reach a transtition state is
lower.\", and then \"Catalysts work by allowing a lower activation energy\"
and then it\'s short step to \'Catalysts act by lowering the activation energy\"
which could even be considered to be literally if loosely true, but I agree is
misleading. Indeed, I plan to incorporate this way of explaining catalysts into
future years\' lectures. Thanks. *1),(###-09-22&&
3. Q: I want to know more about the mechanism behind noncompetitive
inhibition. It apparently doesnt seem to distort the enzyme in the slide, yet
, it\'s binded to another site. Does it repel substrate at the binding site? It\'s
just unclear to me how it\'s lowering reaction velocity.&&
A: No. The substrate binds just fine in non-competitive inhibition. The inhibitor does not interfere
with the substrate binding, it does not compete with it for the substrate binding site, thus it iscalled
non-competitive, when this situation is seen. It could interfere with a side chain that is importantfor
catalyzing the reaction once the substrate is already bound. Thus the catalysis would not take place,
even though the substarte is bound. ',1),(###-09-19
4. Q: In problem 2-11b, why are the 26 spots if in part A, leu -> ala and both
are nonpolar
8 A: Leu and ala are both non-polar. They both appear in the non-polar
category of pages in books showing the categories of amiono acids. Why are
they non-polar? Because they have side chains made up of hydrophobic pure
hydrocarbons (-CH2- or -CH3). But leucine has a lot more of these non- polar
groups than ala,which has but one. And the side chains must \"compete\"
with the polar end of the molecules to establish the overall character of the
molecule. So leu with its many hydrocarbon carbon atoms is a lot more non-
polar than ala. All amino acids with predominantly non-polar side chains are
non-polar but some are more non-polar than others. ',1),(###-09-19
9 Q: In problem 1-23 (16th ed.rev.), how do you generate the chair
form of galactose from the Haworth ring forms given in the texts
9 A: It is not necessary to do this to answer the question (part b). You
know from the Haworth or even from the straight chain (Fisher) projections
that glucose and galactose differ only in the orientation of the OH at C4. The
chair form of glucose shows that the orientation of the OH at C4 in glucose is
, equatorial and that the H is axial up. . Therefore in galactose it must be the
other way around: now it is the OH that is axial up. There are rules and
conventions for following the H\'s from a Fisher projection all the way to a
chair, but they require knowing (memorizing) the conventions and
