Princeton bio

Author's Introduction
MCAT Biology in Context

Admission to medical school is a nebulous challenge. Every facet of our being is scrutinized,
from temperament to intellect, personal history to professional goals. Each credential in the
portfolio is an objectifying representation of complex life events that have taken place over
years. The process is frustrating because this objectification seeks to encapsulate all
elements of our being, save one, which is excluded at the crucial moment: will. We must use
all of our creative powers to analyze and integrate what we have done and who we are
becoming with the ultimate goal of ripping off our skin and laying passive on the chopping
block. The task is at once noble and demeaning, intellectual and bestial. In the few months
of our lives when we compile the application, we are forced to dissect our past and present,
prognosticate our future, and humbly present the case to physicians who will make a
diagnosis and either accept us as students or cut us out like tumors.
The MCAT is your only opportunity to take action during these crucial months. It is the
only part of the application that you can control after having committed yourself to applying.
The Biology component of the test is so important because it is the closest approximation
-------ofmea1cal school. lffi!fprffnary purpose of the MCAT is not to assess your potential as a
physician, your memory, or your ethics, but rather your ability to perform in medical school.
Hence, both Physical Sciences and Biological Sciences questions are designed to assess
your ability to integrate old and new information to solve problems like those faced by medical
students. Even the Physical Sciences section requires a minimum of equations and
mathematics. Problem solving in the Biological Sciences is the best practice for the MCAT
and for medical school.
Another crucial reason to put your best effort into studying biology in preparation for the
MCAT is that each and every insignificant factoid will become the cornerstone of a massive
edifice of knowledge during the first two years of medical school-the "basic science" years.
This means you can't study too much for the Biology section of the MCAT, as long as you
have plenty of time before the test.

Nature of MCAT Biology

The majority of questions are based on a Passage. This is text or graphics or both, presenting
information from the various fields within the broad category of Biology. The Passage is
representative of a textbook or article a medical student might have to read and understand.
There are three types of questions. Type one tests memory: you walk in the door of the
test room with the answer in your head. Type two tests your ability to process new
information: you must get the answer straight from the Passage. Type three tests your ability
to combine what you know with the new information given in the Passage and solve a unique
problem. The ratio of question types is approximately 1:2:3, meaning there are twice as
many type twos as type ones, and three times as many type threes. That means only one out
of six questions tests only straight-up factual knowledge.

What's in this Book

All factual knowledge required for the MCAT is contained within these pages. You will
note that the organization of MCAT Biology follows the topic outline of the Biology section
given in the AAMC's MCAT Student Manual. It is not necessary to use any more-advanced
text in preparation for the test. A difficulty is that it's not possible to include exactly 100% of
the information you need going into the test. I had to undershoot or overshoot. I overshot by
about 15%. The MCAT is a conceptual test, requiring little straight memorization, so you
might say I should have undershot, rather than risk burdening you with unnecessary details.
If this were a traditional text, that would be a good argument. But it's not. Although this book
contains a massive amount of factual information, its purpose is not to stuff you with details.
It is to make you think.

iii


The book contains two things that will help you perform well on the MCAT. It contains
115% of the factual information you'll need. This is good because if you read carefully, you'll

get all the type one questions right. But more importantly, consider this: the content of the
Passages is similar to the extra 15%. For example, you don't need to memorize the
differences between skeletal, cardiac, and smooth muscle for type one MCAT questions. But
what if I approached you the day before the test and offered you a copy of a Passage on this
subject? You'll find that Passage in the MCAT Science Workbook. The extra 15% is that
Passage.
The second thing the book has to prepare you is "grillage." That means it puts you on the
grill, by asking you questions on the material you're studying. In other words, the approach is
Socratic. Some of the questions test factual knowledge that has already been presented.
Others ask you to speculate, based on new information. Others force you to integrate factual
knowledge and speculation. Sound familiar?
The grillage takes the form of questions between paragraphs, but also rears its head as
queries that interrupt the flow of text. The idea is to wake you up and remind you that you're
not supposed to be memorizing, but rather thinking about the information flowing past your
eyes, speculating about it, integrating it with what you know, what you'd like to know, and
what you'd like to do with all that knowledge (help sick people).
It is crucial that you take advantage of the grillage. How? When the book asks you a
question, you'll usually find the answer in a footnote on the same page. DON'T READ THE
FOOTNOTE UNTIL YOU'VE ANSWERED THE QUESTION! Some of the answers are as
simple as "C" or "No," and others are complex conceptual explanations. In any case, take the
time to formulate a thorough answer before you go to the footnote. If you think you are too
rushed to 11Waste" time in this way, I've got news for you: you are studying the wrong way.
The waste of time is memorizing trivial details. The profitable time is pondering concepts, as
you'll do on the day of the MCAT. Though you shouldn't read the footnotes too soon, be sure
to read them eventually, as sometimes they contain important information or vocabulary not
given in the main body of the text.
The companion volume to this text is the MCAT Science Workbook, containing lots of
MCAT-style passages on a wide variety of subjects. Real MCAT passages aren't always so
cohesive, meaning they tend to integrate material from various different areas. The passages
in the Workbook also tend to be a bit harder, in the sense of factual and conceptual questions.

The factual questions are harder to help you study the information; the conceptual questions
so that you can practice. DO ALL OF THE PASSAGES. Be sure to read the solutions. They
address the confusions of thousands of past students.
The most important advice I can give you about the MCAT and the application process is
to stop thinking of it as "the MCAT and the application process," and start thinking of it as
"Becoming a doctor." That way you'll realize that all this study isn't just a hoop to jump
through, but rather a crucial foundation for the biomedical side of the doctor you are
becoming. I specify "the biomedical side" in recognition of the fact that there's much more to
doctoring than biomedicine. But don't get me started.
Once I made the book all-inclusive, I tried to make it as light and engaging as possible. If
at times you find the style flippant, please know that I don't intend any disrespect. I know that
I have the privilege of addressing the highest-level readership of any author, and I am
honored to share in your learning.

I don't wish you luck in your medical career because you don't need luck. All you need is
to work hard, play hard, and read some good novels. And take some time to travel before
medical school. And respect your patients.
DJP

iv


MCAT BIOLOGY
Table of Contents
Below each chapter title are listed the MCAT topics covered in that chapter. These topics correspond to those
given in the AAMC's MCAT Student Manual, in the section entitled MCAT Biological Sciences Topics,
Biology.
Note: Important biological molecules are discussed in Chapter 6 of the MCA T Organic Chemistry text in
this MCAT Biological Sciences Review. You may want to review that chapter before beginning-or refer to
during-your study of MCAT Biology.

1

Molecular Biology
Topic I

2

Microbiology
Topic II

3

Generalized Eukaryotic Cells
Topic Ill

4

Genetics and Evolution
Topic XI

5

Nervous and Endocrine Systems
Topics V and IVA

6

Circulatory, Lymphatic, and Immune Systems
Topic VI

7

Digestive and Excretory Systems
Topic VII

8

Muscle and Skeletal Systems
Topics VIII, IVB, and IVD

9

Respiratory and Skin Systems
Topics IX and IVC

10

Reproductive Systems and Development
Topic X

v


PERIODIC TABLE OF THE ELEMENTS

-

2

1

He

H

1.0
3

Li

6.9
11

Na

24.3
20

Ca

21

Sc

40.1
38

85.5

87.6

55

56

132.9
87

137.3
88

(223)

226.0 227.0

Cs
Fr

Sr

Ba
Ra

14.0

14

15

16

27.0

28.1

31.0

32.1

35.5

39.9

31

32

33

34

35

36

72.6
50

74.9

51

79.0
52

83.8
54

127.6

79.9
53
I
126.9

84

85

AI

39.1
37

Rb

12.0

8
0

16.0

13

12

Mg

19

7

10.8

45.0
39
y
88.9
57

La*

22

23

24

Cr

25

Mn

26

47.9
40

50.9
41

52.0
42

54.9
43

91.2

92.9

95.9

(98)

72

73

74

Ti

Zr

Hf

138.9 178.5
104
89

Act

Rf

(261)

v

Nb

Ta

w

Tc

75

Re

28

55.8
44

58.9
45

58.7
46

101.1

102.9

106.4

76

77

78

Fe

Ru

Os

Co

Rh

Db

Sg

186.2
107

190.2 192.2
109
108

(262)

(263)

(262)

(265)

58

59

140.1
90

Th

232.0

Pr

Bh

60

Nd

Hs

61

Pm

62

Sm

93

94

(231)

(237)

(244)

238.0

Np

30

Zn

Ga

63.5
47

65.4
48

69.7
49

107.9

112.4

114.8

80

81

Cu

Ag
79

Cd

In

Pt

Au

Hg

Tl

195.1

197.0

200.6

64

65

157.3

158.9

96

97

98

(247)

(247)

(251)

Ge

Sn

p

As

Sb

118.7 121.8
82

Pb

204.4 207.2

83

Bi

209.0

s

Se
Te

Po

F
19.0
17

Cl

Br

At

(209) (210)

Ne

20.2

18

Ar

Kr

Xe

131.3
86

Rn
(222)

Mt

(145)

u

Pd

29

Si

N

(267)

140.9 144.2
91
92

Pa

Ni

lr

180.9 183.9
106
105

Ce

t

Mo

27

c

9

4.0
10

6

B

Be
9.0

23.0

K

5

4

63

Eu

150.4 152.0

Pu

95

Am

(243)

vi

Gd

Cm

Tb
Bk

66

Dy

67

Ho

162.5 164.9

Cf

68

Er

69

Tm

70

Yb

71

Lu

167.3

168.9

99

100

101

102

103

(252)

(257)

(258)

(259)

(260)

Es

Fm

Md

173.0 175.0

No

Lr


MCAT BIOLOGY
TABLE OF CONTENTS

1 MOLECULAR BIOLOGY
PARTS 1 AND 2: BIOCHEMISTRY ................................ 1
PART 1: ENZYMES ................................................................................ 2
1 .1 The Kinetic Role of the Enzyme .............................................................. 2
1.2 Enzyme Structure and Function ............................................................ 10
1.3 Basic Enzyme Kinetics .......................................................................... 15

PART 2: ENZYMES AND CELLULAR RESPIRATION ......................... 19
2.1
2.2
2.3
2.4
2.5
2.6
2.7

2.8

1

Energy Metabolism and the Definitions of Oxidation and Reduction .... 19
Introduction to Cellular Respiration ....................................................... 21
Glycolysis .............................................................................................. 22
Fermentation ......................................................................................... 24
The Pyruvate Dehydrogenase Complex ............................................... 26
-rhe Krebs Cycle .................................................................................... 27
Compartmentalization of Glucose Catabolism in Eukaryotes ............... 31
Electron Transport and Oxidative Phosphorylation ............................... 32

MOLECULAR BIOLOGY
PARTS 3 AND 4: GENE EXPRESSION ....................... 37
PART 3: DNA AND THE GENETIC CODE ........................................... 38
3.1 DNA Structure ....................................................................................... 38
3.2 The Role of DNA ................................................................................... 46
3.3 DNA Replication .................................................................................... 51

PART 4: GENE EXPRESSION ............................................................. 58
4.1 RNA and Transcription .......................................................................... 58
4.2 Transcription .......................................................................................... 59
4.3 Translation ............................................................................................. 68

vii


2

MICROBIOLOGY ......................................................... 77
PART 1: VIRUSES ................................................................................ 78
1.1 Viral Structure and Function .................................................................. 79
1.2 Bacteriophage Life Cycles .................................................................... 82
1.3 Viral Genomes ....................................................................................... 87
PART 2: PROKARYOTES .................................................................... 90
2.1 Bacterial Structure and Classification ................................................... 90
2.2 Bacterial Growth Requirements and Classification ............................... 95
2.3 Bacterial Life Cycle ................................................................................ 99
2.4 Genetic Exchange Between Bacteria ................................................. 101

PART 3: FUNGI .................................................................................. 104
3.1 Fungal Structure .................................................................................. 104
3.2 Fungal Life Cycles ............................................................................... 105

*PART 4: RECOMBINANT DNA ....................................................... 108

3

GENERALIZED EUKARYOTIC CELLS ...................... 113
Introduction .................................................................................................... 114

PART 1: THE ORGANELLES .............................................................. 115
1.1
1.2
1.3
1.4
1.5
1.6

The Nucleus ........................................................................................ 115
Mitochondria ........................................................................................ 119
Endoplasmic Reticulum ....................................................................... 121
The Golgi Complex .............................................................................. 123
Lysosomes .......................................................................................... 124
Peroxisomes ........................................................................................ 125

PART 2: STRUCTURAL ELEMENTS OF THE ANIMAL CELL ........... 126
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8

Membranes of the Eukaryotic Cell ...................................................... 126
Passive Transmembrane Transport .................................................... 130
Active Transport ......................... :........................................................ 135
The Na+fK+ ATPase and the Resting Membrane Potential ................. 136
Endocytosis and Exocytosis ................................................................ 138
Cell-Surface Receptors ....................................................................... 139
The Cytoskeleton ................................................................................ 141
Cell Adhesion and Cell Junctions ........................................................ 143

PART 3: THE CELL CYCLE AND MITOSIS ....................................... 144

viii

4

GENETICS AND EVOLUTION ................................... 149
PART 1: GENETICS ........................................................................... 150
1 .1
1.2
1.3
1.4
1.5
1.6

The Language of Genetics .................................................................. 150
Meiosis ................................................................................................ 155
Mendelian Genetics ............................................................................. 160
Linkage ................................................................................................ 165
Sex Linkage and Pedigrees ................................................................ 171
Population Genetics ............................................................................ 174

PART 2: EVOLUTION ......................................................................... 178
2.1
2.2
2.3
2.4

5

Evolution by Natural Selection ............................................................ 178
The Species Concept and Speciation ................................................. 182
Taxonomy ............................................................................................ 183

The Origin of Life ................................................................................. 185

NERVOUS AND ENDOCRINE SYSTEMS ................. 187
PART 1: NEURONAL STRUCTURE AND FUNCTION ...................... 188
1.1
1.2
1.3
1.4

Structure of the Neuron ....................................................................... 188
The Action Potential ............................................................................ 190
Synaptic Transmission ........................................................................ 195
Summation .......................................................................................... 198

PART 2: ORGANIZATION OF THE HUMAN NERVOUS SYSTEM ... 199
2.1 Functional Organization ...................................................................... 199
2.2 Anatomical Organization ..................................................................... 202

PART 3: SENSATION ......................................................................... 208
3.1
3.2
3.3
3.4

Types of Sensory Receptors ............................................................... 208
Gustation and Olfaction ....................................................................... 209
Hearing and the Vestibular System ..................................................... 210
Vision: Structure and Function ............................................................ 212

PART 4: THE ENDOCRINE SYSTEM ................................................ 216

4.1 Hormone Types: Transport and Mechanisms of Action ...................... 216
4.2 Organization and Regulation of the Human Endocrine System ......... 218
4.3 Major Glands and Their Hormones ..................................................... 221

ix


6

CIRCULATORY, LYMPHATIC, AND
IMMUNE SYSTEMS ................................................... 225
PART 1: THE CIRCULATORY SYSTEM ............................................ 226
1.1
1 .2
1 .3
1.4
1.5
1.6
1.7

Components of the Circulatory System .............................................. 226
The Heart ............................................................................................. 228
Hemodynamics .................................................................................... 237
Components of Blood .......................................................................... 240
Transport of Gases .............................................................................. 243
Exchange of Substances Across the Capillary Wall ........................... 246
The Lymphatic System ................................................ :....................... 248

PART 2: THE IMMUNE SYSTEM ....................................................... 248
2.1

2.2
2.3
2.4

7

Innate Immunity ................................................................................... 248
Humoral Immunity, Antibodies, and B Cells ........................................ 249
Cell-Mediated Immunity and the T Cell ............................................... 252
Others Tissues Involved in the Immune Response ............................ 253

DIGESTIVE AND EXCRETORY SYSTEMS ............... 255
PART 1: THE DIGESTIVE SYSTEM .................................................. 256
1.1
1.2
1.3
1 .4
1.5

Overview of the G I Tract ..................................................................... 256
The Gastrointestinal Tract ................................................................... 260
The Gl Accessory Organs ................................................................... 266
A Day in the Life of Food ..................................................................... 269
Vitamins ............................................................................................... 271

PART 2: THE EXCRETORY SYSTEM ............................................... 272
2.1 Overview .............................................................................................. 272
2.2 Anatomy of the Excretory System ....................................................... 275
2.3 Endocrine Role of the Kidney .............................................................. 284

X


8

MUSCLE AND SKELETAL SYSTEMS ....................... 285
PART 1: MUSCLE ............................................................................... 286
1.1 Skeletal Muscle ................................................................................... 286
1.2 Cardiac Muscle Compared to Skeletal Muscle ................................... 294
1 .3 Smooth Muscle Compared to Skeletal Muscle ................................... 295

PART 2: THE SKELETAL SYSTEM .................................................... 298
2.1
2.2
2.3
2.4
2.5
2.6

9

Connective Tissue ............................................................................... 298
Macroscopic Bone Structure ............................................................... 298
Microscopic Bone Structure ................................................................ 299
Cartilage .............................................................................................. 300
Bone Growth and Remodeling: the Cells of Bone .............................. 300
Ligaments, Tendons, and Joints .......................................................... 301

RESPIRATORY AND SKIN SYSTEMS .................... 303
u

PART 1: THE RESPIRATORY SYSTEM ............................................ 304
1 .1
1.2
1.3
1.4
1.5

Functions ............................................................................................. 304
Anatomy of the Respiratory System .................................................... 304
Pulmonary Ventilation .......................................................................... 307
Gas Exchange ..................................................................................... 311
Regulation of Respiration .................................................................... 314

PART 2: THE SKIN ............................................................................. 315
2.1 Structure .............................................................................................. 315
2.2 Temperature Regulation by the Skin ................................................... 316

xi


10

REPRODUCTIVE SYSTEMS
AND DEVELOPMENT ............................................. 319

PART 1: THE REPRODUCTIVE SYSTEMS ...................................... 320
1 .1
1.2
1.3

1.4
1.5
1.6
1.7
1.8

The Male Reproductive System .......................................................... 320
Spermatogenesis ................................................................................ 323
Development of the Male Reproductive System ................................. 326
The Androgens .................................................................................... 328
The Female Reproductive System: Anatomy and Development ........ 329
Oogenesis and Ovulation .................................................................... 331
The Menstrual Cycle ........................................................................... 333
Hormonal Changes during Pregnancy ................................................ 337

PART 2: DEVELOPMENT .................................................................. 338
2.1
2.2
2.3
2.4
2.5

Fertilization and Cleavage ................................................................... 338
Implantation and the Placenta ............................................................. 340
Post-Implantation Development .......................................................... 341
Differentiation ...................................................................................... 343
Birth and Lactation .............................................................................. 344

xii

..c:Ju...___ , ____... __
L:::JIIJJ-II::rl 71::rc:::ll ·1 Ill 1!/

MCAT Bt.ology

Molecular Biology
Parts 1 and 2:
Biochemistry

Part 1: Enzymes I 2
1.1 The Kinetic Role of the Enzyme I 2
1.2 Enzyme Structure and Function I 10
1.3 Basic Enzyme Kinetics I 15
Part 2: Enzymes and Cellular Respiration I 19
2.1 Energy Metabolism and the Definitions of Oxidation and Reduction I 19
2.2 Introduction to Cellular Respiration I 21
2.3 Glycolysis I 22
2.4 Fermentation I 24
2.5 The Pyruvate Dehydrogenase Complex I 26
2.6 The Krebs Cycle I 27
2. 7 Compartmentalization of Glucose Catabolism in Eukaryotes I 31
2.8 Electron Transport and Oxidative Phosphorylation I 32


Chapter 1

2

MCAT Biology

Part 1: Enzymes
1.1: The Kinetic Role of the Enzyme
The notion of life refers to both the activities and the physical structures of living organisms.
(Chapter 6 in MCAT Organic Chemistry discusses the fundamental molecules of life and how
molecules are used to store energy and to form physical structures, which you may want to
review.) Both the storage/utilization of energy and the synthesis of structures depend on a large
number of chemical reactions that occur within each cell. Fortunately, these reactions do not
proceed on their own spontaneously, without regulation. If they did, each cell's energy would
rapidly dissipate and total disorder would result. Most reactions are slowed by a large barrier
known as the activation energy (EJ, discussed below. The Ea is a bottleneck in a reaction, like a
nearly-closed gate. The role of the enzyme is to open this chemical gate. In this sense, the enzyme
is like a switch: When the enzyme is on, the gate is open (low E), and the reaction accelerates.
When the enzyme is off, the gate closes and the reaction slows. Before we can see how enzymes
work, we must digress a bit to review the basics of thermodynamics.
Thermodynamics is the study of the energetics of chemical reactions. There are two relevant
forms of energy in chemistry: heat energy (movement of molecules) and potential energy (energy
stored in chemical bonds). [What is the most in1portant potential energy storage molecule in all
cells?l] The first law of thermodynamics, also known as the law of conservation of energy,
states that the energy of the universe is constant. It implies that when the energy of a systen1
decreases, the energy of the rest of the universe (the surroundings) must increase, and vice versa.
The second law of thermodynamics states that the disorder, or entropy, of the universe tends to
increase. Another way to state the second law is as follows: a reaction will occur spontaneously
only if it increases the disorder of the universe. The symbol for entropy is S, and "a change in
entropy" is denoted

/j_S,

where

/j_S =Satter- Sbetore·

[If the

/j_S

of a system is negative, has the

disorder of that system increased or decreased ?2]
A practical way to discuss thermodynamics is the mathematical notion of free energy
(Gibbs' free energy), defined by Josiah Gibbs as follows:3
/j_G =

Ml- TM

(1)

T denotes temperature, and H denotes enthalpy, which is defined by another equation:
Ml =M- Pfj_V

(2)

1 ATP, which stores energy in the ester bonds between its phosphate groups.
2 If jj.S is negative, then the system lost entropy, which means that disorder decreased.
3 As in /j.S, the Greek letter /j. (delta) indicates "the change in." For example, jj.Grxn = Gproducts- Greactants·


Biochemistry

MCAT Biology

3

Here E represents the bond energy of products or reactants in a system, P is pressure, and V is
volume. [Given that cellular reactions take place in the liquid phase, how is H related to E in a
cell?4] -~9 incre~~~s \Vit~ ~nc!easing~ QJo~~eneigy) ~(t!l,~ decreases with increasing entr• Given the second law of thermodynamics and the mathematical definition of ~G, which reaction
will be favorable: one with a decrease in free energy
energy

(~G

(~G

< 0) or one with an increase in free

> 0)?5

The change in the Gibbs free energy of a reaction determines whether the reaction is favorable
(spontaneous,

~G

negative) or unfavorable (nonspontaneous,

~G

positive). In terms of the generic

reaction

the Gibbs free energy change determines whether the reactants (denoted A and B) will stay as they
are or be converted to products (C and D).
Spontaneous reactions, ones that occur without a net addition of energy, have ~G < 0. They
occur with energy to spare. Reactions with a negative ~G are exergonic (energy exits the system);
reactions with a positive ~G are endergonic. Endergonic reactions only occur if energy is added.
In the lab, energy is added in the form of heat; in the body, endergonic reactions are driven by
reaction coupling to exergonic reactions (more on this later). Reactions with a

n~gative

Mf ~r~

called exothermic and liberate heat. Most metabolic reactions are exothermic (which is how
homeothermic organisms such as mammals maintain a constant body ten1perature). Reactions with
a positive ~ require an input of heat and are referred to as endothermic. (Thermodynamics will
be discussed in more detail in General Chemistry.)
The signs of thermodynamic quantities are assigned from the point of view of the system,
not the surroundings or the universe. Thus, a negative

~G

means that the system goes to a lower

free energy state, and a system will always move in the direction of the lowest free energy. As an
analogy, visualize a spinning top as the system. What happens to the top? Does it spin faster and
faster? No. It moves towards the lowest energy state. Let's expand the analogy, using an equation:
motionless top

~

spinning top

4Hz E, since the change in volume is negligible (il V z 0).
5 Favorable reactions have ilG < 0. We can deduce this from the second law and equation (1) because the second law
states that increasing entropy is favorable, and the equation has ilG directly related to -TilS.


Chapter 1

4

MCAT Biology

Here the "reactant" is the motionless top, and the "product" is the spinning top. Which is lower: the
free energy of product or reactant? Reactant. Is the reaction "spontaneous" as written? No; in fact,
the reverse reaction is spontaneous. Hence,
G spinning > G motionless

and thus,
Ll G reaction as written (motionless to spinning; left to right) > 0

So the reaction is nonspontaneous. In other words, it requires energy input, namely, energy from
your muscles as you spin the top. [If the products in a reaction have more entropy than the
reactants, and the enthalpy (H) of the reactants and the products are the same, can the reaction
occur spontaneously?6]
The value of ilG depends on the concentrations of reactants and products, which can be
variable in the body. Therefore, to compare reactions, biochemists calculate a standard free energy
change, denoted ilG 0 , with all reactants and products present at 1 M concentration. Furthermore,
the biochemist's standardized ilG determined at pH 7 is denoted LlG0 '.

ilGo' is related to the equilibrium constant for a reaction by the following equation:

where R is the gas constant (which would be given on the MCAT, along with the entire equation),
and K;q is the ratio of products to reactants at equilibrium:

K;q is the ratio of products to reactants when enough time has passed for equilibrium to be
reached. [When K;q = 1, what is ilGo' ?7]
But what if we wanted to calculate ilG for a reaction in the body? In this case, we need one
more equation:

ilG

= LlG

0
'

+ RTlnK,

where K

= [C][D]
[A][B]

(4)

Here, K is calculated using the actual concentrations of A, B, C, and D (for example, the
concentrations in the cell). Equation (4) is simply a conversion from ilGo' (the laboratory standard
6 Yes. If !l.S > 0 and A.H = 0, then according to the second law of thermodynamics, the reaction is spontaneous; see
equation (1).

7 Equation (3) says that !l.G 0 ' = 0 when K;q 1 since In 1 = 0.


MCAT Biology

Biochemistry

5

D,.G with initial concentrations at 1 M) to the real-life here-and-now D,.G. Note that if we put 1 M
concentrations of A, B, C, and D into a beaker (at pH 7), we have recreated the laboratory standard
initial set-up: K = 1, so InK= 0, which means D,.G = D,.Go'.
• You are studying a particular reaction. You find the reaction in a book and read D,.Go' from a table.
Can you calculate D,.G for this reaction in a living human being without any more information ?8
Remember that K and Keq are not the same. K is the ratio of products to reactants in any given
set-up; Keq is the ratio at equilibrium. Equilibrium is defined as the point where the rate of
reaction in one direction equals the rate of reaction in the other. At equilibrium, there is constant
product and reactant turnover as reactants form products and vice versa, but overall concentrations
stay the same. Theoretically (given enough time), all reactant/product systems will eventually reach
this point.
• How can D,.G be negative if D,.Go' is positive (which indicates that the reaction is unfavorable at
standard conditions) ?9
• Does Keq indicate the rate at which a reaction will proceed? 1O
• When Keq is large, which has lower free energy: products or reactants?ll
• When K is large, which has lower free energy: products or reactants?l2
• Which direction, forward or backward, will be favored in a reaction if D,.G = 0? (Hint: What does
equation (4) look like when D,.G = 0?)13]
8 No. You need to know the concentrations of A, B, C, and D in the human cell. For example, /:1G 0 ' might be
+14.8 kcal/mol, indicating that the reaction is very unfavorable under standard conditions. But, if the concentration
of reactants is much higher than the concentration of products in the cell, the reaction may be favorable in the cell

since K < 1 and /:1G may be less than zero. (Although 1'1G could still be positive if /:1G 0 ' is very large.) The
significance of K as an independent variable in equation (4) is that it accounts for Le Chatelier' s principle: a high
concentration of reactants will drive a reaction forward and a high concentration of products will drive it backward,
regardless of the intrinsic thermodynamics ( 1'1G0 ' ) of the reaction.
9 The reaction may be favorable (/:1G < 0) if the ratio of the concentrations of reactants to products is sufficiently
large to drive the reaction forward (that is, if RT ln K is more negative than 1'1G0 ' is positive, which would make
their sum-which, by equation (4), is /:1G-negative).
10 Keq indicates only the relative concentrations of reagents once equilibrium is reached, not the reaction rate (how
fast equilibrium is reached).
11 A large Keq means that more products are present at equilibrium. Remember that equilibrium tends towards the
lowest energy state. Hence, when Keq is large, products have lower free energy than reactants.
12 The size of K says nothing about the properties of the reactants and products. K is calculated from whatever the
initial concentrations happen to be. It is Keq that says something about the nature of reactants and products, si nee it
describes their concentrations after equilibrium has been reached.
13 If /:1G is 0, then neither the forward nor the reverse reaction is favored. Look at equations (3) and (4). Note that
when /:1G = 0, equation (4) reduces to equation (3 ), and thus K = Keq (which means K at this moment is the same as
Keq• measured after the reaction system is allowed to reach equilibrium). When K Keq• we. are by definition at
equilibirum. Understand and memorize the following: When L1G 0, you are at equilibrium; foJWard reaction
equals back reaction, and the net concentrations of reactants and products do not change.


Chapter 1

6

MCAT Biology

• Radiolabeled chemicals are often used to trace constituents in biochemical reactions. The
following reaction with /1G = 0 is in aqueous solution:
A::;;;:!'!:: B+C,

K

= [B][C]
eq

[A]

A small amount of radio labeled B is added to the solution. After a period of time, where will the
radiolabel most likely be found: in A, in B, or in both?14
In summary, then, there are two factors that determine whether a reaction will- occur
spontaneously (/1G negative) in the cell:
1. The intrinsic properties of the reactants and products (/1G 0 ) , and

2. The concentrations of reactants and products (RT InK).

Thermodynamics vs. Reaction Rates
The term spontaneous is used to describe a reaction system with /1G < 0. This can be misleading,
since the common usage of the word spontaneous has a connotation of rapid rate-this is not
what spontaneous means in the context of chemical reactions. For example, many reactions have a
negative /1G, indicating that they are "spontaneous" from a thermodynamic point of view, but they
do not necessarily occur at a significant rat~pontaneous Jile_ap._~J~~t a r~~~Ji<:>.l1_ll1ay pro~d
"

.

~.:

--~-....~

without ~gditiOP.
~~hermodynall!ic~_vyill. telli~<2£~h~~-~~~~!~~Jinishes_but;QtltingJ!Q_Qll!!h~.J>9th

~aveled

to get there. The difference in free energy in a reaction is only a function of the nature of

the reactants and products. Thus, /1G does not depend on the pathway a reaction takes or the rate of
reaction; it is only a measurement of the difference in free energy between reactants and products.
• How does the /1G for a reaction burning (oxidizing) sugar in a furnace compare to the /1G when
sugar is broken down (oxidized) in a human?15

14 The reaction is in dynamic equilibrium where reactions are occurring in both directions, but at an equal rate.
Because ~G = 0, we know that the forward reaction and the reverse reaction proceed at equal rates, even though we
don't know the actual value. Therefore, after a period of time, the radio label will be present in both A and B.
15 The ~G is the same in both cases. ~G does not depend on the pathway, only on the different energies of the
reactants and products.


Biochemistry

MCAT Biology

7

Kinetics and Activation Energy (E a)
The reason some spontanoeus (that is, themodynarnically favorable) reactions proceed very slowly
or not at all is that a large amount of energy is required to get them going. For example, the
burning of wood is spontanoeus, but you can stare at a log all day and it won't burn. Some energy

(heat) must be provided to kickstart the process.
The study of reaction rates is called chemical kinetics. All reactions proceed through a
transient intermediate that is unstable and takes a great deal of energy to produce. The energy
required to produce the transient intermediate is called the activation energy (EJ. This is the
barrier that prevents many reactions from proceeding even though the ~G for the reaction may be
negative. The match you use to light your fireplace provides the activation energy for the reaction
known as burning. It is the activation energy barrier that determines the kinetics of a reaction.
[How would the rate of a spontaneous reaction be affected if the activation energy were
lowered?16]
The concept of Ea is key to understanding the role of enzymes, so let's spend some time on it.
To illustrate, take this reaction:
Bobwithout ajob + job --t Bobwith ajob
Is this a favorable reaction, i.e., will the universe be better off, with less total (nervous) energy, if
Bob gets the job? Will things settle down? Let's assume yes. However, between the two states
(without/with) there is an intermediate state, namely, Bobapplying for Jot· So the reaction will look this
way:
Bobwithout a job+ job --t [Bobapplying for job]+ --t Bobwith a job
The middle term is the transition state (TS), traditionally written in square brackets with a doublecross symbol: [TS]+. It exists for a very very short time, either moving forward to form product or
breaking back down into reactants. The energy required for Bob to be job hunting is much higher
than the energy of Bob with a job or Bob without a job. As a result, he may not go job hunting,
even though he'd be happier in the long run if he did. In this model, we can describe the Ea as the
energy necessary to get Bob to apply for a job.
(

A catalyst lowers the Ea of a reaction without changing the L1G. The catalyst lowers the Ea

\, by stabilizing the transition state, making its existence less thermodynamically unfavorable. The

~',,,)second

important characteristic of a catalyst is that it is not consumed in the reaction; it is

fegenerated with each reaction cycle.

16 The rate would be increased, since lowering Ea is tantamount to reducing the energy required to achieve the
transition state. The more transition state intermediates that are formed, the greater the amount of product produced,
the more rapid the rate of reaction.


Chapter 1

8

MCAT Biology

In our model, an example of a catalyst would be a career planning service (CPS). Adding a
CPS won't make Bobwithout a job any happier or sadder, nor will it make Bobwith a job happier or
sadder. But it will make it much easier for Bob to move between the two states: without a job vs.
with a job. The traditional way to represent a reaction system like this is using a reaction
coordinate graph, as shown in Figure 1. This is just a way to look at the energy of the reaction

system as compared to the three possible states of the system: 1) reactants, 2) [TS]t, and
3) products. The x axis plots the physical progress of the reaction system (the "reaction
coordinate"), and they axis plots energy.

Figure 1: The Reaction Coordinate Graph
[TS]*
~----=="'":::::'""''"""'"''""""''"""''""'"''"'""'"""'"'""'"''"""'"''"'"'"""''""'"'""'"'"'"'"''""''"'"""''"""''"""''"'"'"""'""'""""'"''

Ea without catalyst

......_ . ,_........- .................................I.........................

Ea with catalyst
tu•

IO>HI .. INI»Hfl>Hnl!UtiOnOnHUOI<f.hJt»mmJ~-HIIUHOiltUtllllti#tfUH-lUUfUIUtl*fl\flltltlUU»fttf>tlHUtllni .. OI

110 ...1

lU!ft

reactants

ttfOt ..IOHOitUI>UIWI\t-UntHifiOIUlltUlttttU_.UHO-#
• .6.0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . -. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . ! . .
products
Reaction coordinate

Enzymes are catalysts. They increase the rate of reactions by lowering the activation energy
but do not affect ~G between reactants and products. As catalysts, enzymes have a kinetic role, not a
thermodynamic one. [Will an enzyme alter the concentration of reagents at equilibrium?I7]
Enzymes n1ay alter the rate of a reaction enormously: a reaction that would take a hundred years to
reach equilibrium without an enzyme may occur in just seconds with an enzyme. (Contrast the
kinetic role of enzymes with collision kinetics as discussed in MCAT General Chemistry.)

ATP as an Energy Source-Reaction Coupling

Enzymes increase the rate of reactions that have a negative

~G.

These reactions would occur on

their own without an enzyme (they are spontaneous) but far more slowly than with one. However,
there are many reactions in the body that occur which have a positive
macromolecules such as DNA and protein is not spontaneous

(~G

~G.

The biosynthesis of

> 0), but clearly these reactions

do take place (or we wouldn't be here). How can this be? Thermodynamically unfavorable

17 No.

It will only affect the rate at which the reactants and products reach equilibrium.


MCAT Biology

9

Biochemistry

reactions in the cell can be driven forward by reaction coupling. In reaction coupling, one very
favorable reaction is used to drive an unfavorable one. This is possible because free energy

changes are additive. [What is the favorable reaction that the cell can use to drive unfavorable
reactions?I8] In the lab, the l.lG0 ' for the hydrolysis of one phosphate group from ATP is -7.3
kcal/rnol, so it is a very favorable reaction. In the cell, l.lG is about -12 kcal/rnol, so in the cell it is
even more favorable. [What's the difference between the situation in vitro (lab) and in vivo
(cell)?19 ]

~H~w does ATP hJLdrill~~sis=JiriY~=!!nfavorable r_e_actions? There are many ways. One example
is b'Y>fausing a cQ!)formational change in a __g_Qtein; in this way.ATP hydrolysis

{ff be used to

power energy-costly events like transmembrane transport. Another example is by _Jransfer of a
phosphate group from ATPlo a substrate. Take the unfavorable reaction A + B
that Reactant A must proceed through an intermediate, APO

2
4 -

-7 C.

Let's say

in order to participate. Let's say

llG =+7 kcal/rnol for the overall reaction. What if the two partial reactions have l.lGs as follows:
A+ P0 42-

-7

APO/- + B

AP0 42-

-7

llG

C +PO/-

= +2 kcaVrnol

llG = +5 kcaVrnol

Total llG

= +7 kcaVmol

These reactions will not proceed, because the overall l.lG will be +7 kcal/mol. What will be the

overallllG if we couple the reaction A + B -7 C to the hydrolysis of one ATP? All we have to do is
add up all the l.lG values, as follows:
ATP

ADP + P042-

-7

A + P0 4

2
-

-7

APO/- + B

AP04

-7

2

llG

=-12 kcal/mol

-

llG = +2 kcaVrnol

C +PO/-

llG = +5 kcaVrnol

Total llG = -5 kcal/mol
Now the overall reaction, shown below, is thermodynamically favorable. We have coupled the

unfavorable reaction A + B -7 C to the highly favorable hydrolysis of ATP:
A+ B + ATP

18 ATP hydrolysis!
19 K (cell) :t:. Keg_· This

-7

C + ADP +PO/-

llG = -5 kcal/rnol

means that the relative concentrations of ATP and ADP + Pi are not at equilibrium levels in
the cell. Actually, K (cell) << Keq because the cell keeps a high concentration of ATP around-we'll see how in
Part 2.


Chapter 1

10

MCAT Biology

Note that we first stated that the enzyme has only a kinetic role (influencing rate only), not a
thermodynamic one (determining favorability). Then we went on to discuss reaction coupling,
which allows enzymes to promote otherwise unfavorable reactions. There is no contradiction,
however. The only difference is viewing reactions in an isolated manner or in the complex series of
linked reactions more commonly found in the body. The same rule applies in either case: AG n1ust
be negative for either a single reaction or a series of linked reactions to occur spontaneously.
one reaction in a test tube:

enzyme = catalyst
kinetic role only-influences rate, not outcome

many "real life" reactions in the cell: enzyme controls outcomes by selectively promoting
unfavorable reactions via reaction coupling

1.2: Enzyme Structure and Function
Most enzymes are proteins that must fold into specific three-dimensional structures to act as
catalysts. (Some enzymes are RNA or contain RNA sequences with catalytic activity. Most
catalyze their own splicing, and the rRNA in ribosomes helps in peptide-bond formation.) An
enzyme may consist of a single polypeptide chain or several polypeptide subunits held together in a
_20

(primary? secondary? etc.) structure. The reason for the importance of folding in enzyme

function is the proper formation of the active site, the region in an enzyme's three-dimensional
structure that is directly involved in catalysis. [What shape are enzymes more likely to have:
fibrous/elongated or globular/spherical?21 ] The reactants in an enzyme-catalyzed reaction are called
substrates. (Products have no special name; they're just "products.") What is the role of the active
site, that is, how do enzymes work? The enzyme is like a career planning service (CPS), and the
active site is like the room where Bob can sit and do his job hunting. The CPS has counselors,
books, and job lists, all of which make job hunting easier. The active site has amino acid residues
that stabilize the transition state of a reaction. [For example, if a transition state intermediate
possesses a transient negative charge, what amino acid residues might be found at the active site to

20 quaternary
21 Globular. Structural proteins such as collagen tend to be fibrous, but proteins that act as catalysts tend to be
roughly spherical to form an active site in a cleft in the sphere.

11

Biochemistry

MCAT Biology

stabilize the transition state? 22 ] This lowers the activation energy barrier between reactants and
products.
• Is it possible that amino acids located far apart from each other in the primary protein sequence
may play a role in the formation of the same active site?23

• If during an enzyme-catalyzed reaction an intermediate forms in which the substrate is covalently
linked to the enzyme via a serine residue, can this occur at any serine residue or must it occur at a
specific serine residue ?24

• Compound A converts into Compound Bin solution: A
equilibrium constant:
for the reaction A

~

Keq

~

B. The reaction has the following

= [B]e/[A]eq = 1000. If pure A is dissolved in water at 298 K, will /lG

B be positive or negative? Is it possible to answer this question without

knowing /lGo' ?25

• Regarding the reaction described in the previous question, if pure B is put into solution in the
presence of an enzyme that catalyzes the reaction between A and B, which one of the following
will be true?26
A. All the B will be converted into A, until there is 1000 times more A than B.
B • All of the B will remain as B, since B is favored at equilibrium.

C . The enzyme will have no effect, since enzymes act on the transition state and
there is no transition state present.

D. The reaction that produces A will predominate until /lG = 0.

22 A positive charge would stabilize the negative charge in the intermediate. Such a charge might be contributed by
His, Arg, or Lys. Alternatively, the hydrogen of the -NH group in glutamine or asparagine could hydrogen bond
with the negative charge.
23 Yes, the amino acids at the active site may be distant from each other in a polypeptide's primary sequence but be
near each other in the final folded protein. This is why protein folding is crucial for enzyme function.
24 It must occur at a particular serine residue which sticks out into the active site.
25 You don't need to calculate b.G 0 ' ; all you need to know is that with a Keq of 1000, there will be 1000 times
more B than A in solution at equilibirum. If we create a solution with only A, the reaction must move
SQOntaneously toward B.
25 A: No. Based on the Keq given above, we expect the opposite ratio to be reached. B: No. A small amount of A
will be produced, and will continue to be produced. However the opposite reaction will keep pace, so that the ratio of
A to B remains at 1:1000 once equilibrium is reached. C: No. Enzymes do not act on the transition state; they
increase the formation of the transition state. D: Yes. If only B exists in solution, then the back-reaction producing
A will predominate until equilibrium is reached (b.G = 0), regardless of the presence or absence of enzyme.

Chapter 1

12

MCAT Biology

• The transition state intermediate for a reaction possesses a transient negative charge. The active
site for an enzyme catalyzing this reaction contains a His residue to stabilize the intermediate. If the
His residue at the active site is replaced by a glutamate which is negatively charged at pH 7.0, what
effect will this have on the reaction, assuming that the reactants are present in excess compared to
the enzyme ?27

A.

The repulsion caused by the negative charge in the glutamate at the altered
active site will increase the activation energy and make the reaction proceed
more slowly than it would in solution without enzyme.

B.

The rate of catalysis will be unaffected, but the equilibrium ratio of products
and reactants will change, favoring reactants.

C.

The transition state intermediate will not be stabilized as effectively by the altered enzyme, lowering the rate relative to the rate with catalysis by the normal
enzyme.

D•

The rate of catalysis will decrease, and the equilibrium constant will change.

The active site for enzymes is generally highly specific in its substrate recognition, including
stereospecificity (the ability to distinguish between stereoisomers). For example, enzymes which
catalyze reactions involving amino acids are specific for D or L amino acids, and enzymes
catalyzing reactions involving monosaccharides may distinguish between stereoisomers as well.
[Which configurations are found in animals ?28]
Many proteases (protein-cleaving enzymes) have an active site with a serine residue whose
OH group can act as a nucleophile, attacking the carbony1 carbon of an amino acid residue in a
polypeptide chain. Examples are trypsin, chymotrypsin, and elastase. These enzymes also usually
have a recognition pocket near the active site. This is a pocket in the enzyn1e' s structure which
attracts certain residues on substrate polypeptides. The enzyme always cuts polypeptides at the
same site, just to one side of the recognition residue. For example, chymotrypsin always cuts on
the carboxyl side of one of the large hydrophobic residues tyr, trp, phe, and met. Enzymes that act
on hydrophobic substrates have hydrophobic amino acids in their active sites, while
hydrophilic/polar amino acids will comprise the active site of enzymes with hydrophilic substrates.
27 Beware of long, complex-sounding questions. For instance, the phrase "assuming that the reactants are present in
excess compared to the enzyme" adds nothing to the substance of this question. A: No. It is true that the negative
charge of the glutamate will decrease the effectiveness of the active site, or even totally destroy it, but enzymes never
inhibit reactions. If the His and Glu substitution destroyed the enzyme's active site, the reaction would still be free
to proceed as fast as it would in pure solution without any enzyme. B & D: No. Enzymes do not alter reaction
equilibria. C: Yes. The transition state intermediate will not be stabilized as effectively, decreasing the rate of
catalysis.
28 L amino acids and D sugars. Remember the Lin aLanine.


MCAT Biology

Biochemistry

13

Regulation of Enzyme Activity
Metabolic pathways in the cell are not all continually on, but must be tightly regulated to maintain
health. For example, if glycogen synthesis and breakdown occur in the same cell at the same time,
a great deal of energy will be wasted without accomplishing anything. Therefore, the activity of
key enzymes
in metabolic pathways is usually regulated in one or more of the follo~ingj~l~
.
"~--~~---~

1) Covalent modification. The addition of a phosphate group by a protein kinase
to the hydroxyl of serine, threonine, or tyrosine residues is the most common
example. Phosphorylation of different sites on an enzyme can either activate or
inactivate the enzyme. Protein phosphorylation by kinases can be reversed by
protein phosphatases.
2) Proteolytic cleavage. Many enzymes (and other proteins) are synthesized in
inactive forms (zymogens) that are activated by cleavage by a protease.
3) Association with other polypeptides. Some enzymes have catalytic activity in
one polypeptide subunit that is regulated by association with a separate
regulatory subunit. Removal of the regulatory subunit results in continuous
rapid catalysis by the catalytic subunit, known as constitutive activity
(constitutive connotes continuous or unregulated).
4) Allosteric regulation. The modification of active-site activity through interactions of molecules with other specific sites on the enzyme (called allosteric
sites).

Allosteric Regulation
If the cell is to make use of the enzyme as a biochemical switch, there must be a way to tum the
enzyme ON or OFF. One mechanism of regulation is the binding of small molecules to particular

sites on an enzyme that are distinct from the active site; this is allosteric regulation. This name
comes from the fact that the particular spot on the enzyme which can bind the small molecule is not
located close to the active site; allo means "other," and steric refers to a location in space (as in
"steric hindrance"), so allosteric means "at another place." The binding of the allosteric regulator to
the allosteric site is generally noncovalent and reversible. When bound, the allosteric regulator can
alter the conformation of the enzyme to increase or decrease catalysis, even though it may be bound
to the enzyme at a site distant from the active site or even on a separate polypeptide.


14

Chapter 1

MCAT Biology

Feedback Inhibition
Enzymes usually act as part of pathways, not alone. Rather than regulate every enzyme in a
pathway, usually there are one or two key enzymes that are regulated, such as the enzyme that
catalyzes the first irreversible step in a pathway. The easiest way to explain this is with an example.
Three enzymes (E 1, E2, and E3) catalyze the three steps required to convert Substrate A to Product
D. When plenty of Dis around, it would be logical to shut off El so that excess B, C, and D are
not made. This would conserve A and would also conserve energy. Commonly, an end-product
such as D will shut off an enzyme early in the pathway, such as

. This is called feedback

inhibition.

2: Feedback Inhibition

, ....
f

------- ---

E2
A El.., B

)II

c

- -......
...

E3 ..

.'

n-"' '

There are examples of positive feedback ("feedback stimulation"), but feedback inhibition is
by far the most common example of feedback regulation. On the other hand, feedforward

stimulation is common. This involves the stimulation of an enzyme by its substrate, or by a
molecule used in the synthesis of the substrate. For example, in Figure 2, A might stimulate E3.
This makes sense because when lots of A is around, we want the pathway for utilization of A to be
active.
Allosteric regulation can be quite complex. It is possible for more than one small molecule to
be capable of binding to an allosteric site. For example, imagine a reaction pathway from A

through Z, where each step (A
allosteric enzyme called

---1-

B, B

---1-

C, etc.) is catalyzed by an enzyme. Let's say that an

5 catalyzes the reaction 0

---1-

P. It would be possible for A to

allosterically activate E15 (feedforward stimulation) and for Z to allosterically inhibit E15 (feedback
inhibition). This may sound complex, but it's quite logical. What it means is that when lots of A is
around, E15 will be stimulated to use the molecules made from A (B, C, D, etc.) to make P, which
could then be used to make Q, R, S, etc., all the way up to Z. On the other hand, if a lot of excess
Z built up, it would inhibit

5, thereby conserving the supply of A, B, C, etc. and preventing

more build-up of Z, Y, X, etc. Hence, in addition to acting as switches, enzymes act as valves,
because they regulate the flow of substrates into products.


Biochemistry

MCAT Biology

15

1.3: Basic Enzyme Kinetics
Enzyme kinetics is the study of the rate of formation of products from substrates in the presence of
an enzyme. The reaction rate (V, for velocity) is the amount of product formed per unit time, in
moles per second (molls). It depends on the concentration of substrate, [S], and enzyme.29 If there
is only a little substrate, then the rate Vis directly proportional to the amount of substrate added:
double the amount of substrate and the reaction rate doubles, triple the substrate and the rate triples,
and so forth. But eventually there is so much substrate that the active sites of the enzymes are
occupied much of the time, and adding more substrate doesn't increase the reaction rate as much,
that is, the slope of the V vs. [S] curve decreases. Finally, there is so much substrate that every
active site is continuously occupied, and adding more substrate doesn't increase the reaction rate at
aq. At this point the enzyme is said to be saturated. The reaction rate when the enzyme is saturated
is denoted V max; see Figure 3. This is a property of each enzyme at a particular concentration of
enzyme. You can look it up in a book for the common ones. [If a small amount of enzyme in a
solution is acting at V max' and the substrate concentration is doubled, what is the new reaction
rate?30]

Figure 3: Saturation Kinetics

t

After a certain [S] has been reached,
many active sites are occupied, and
adding more substrate increases the
reaction rate little.

Adding more substrate does
not increase reaction rate

'

linear portion of curve:
Vis proportional to [S]

Substrate concentration, [S]

~

29 Usually the concentration of enzyme is kept fixed, and [S] is taken as the only independent variable (the one the
rate depends on). This is applicable to biological systems, where substrate concentrations change much more than
enzyme concentrations.
30 If the enzyme is acting at Vmax' it is saturated with substrate; adding more substrate will not increase the reaction
rate-the rate is still Vmax.