• Search in book...
• Toggle Font Controls
157Zero to Genetic Engineering Hero - Chapter 6 - Processing Enzymes
Advanced Catalase Calculations Going Deeper 6-3
The catalase enzyme will complete the Four B’s inside the hydrogen peroxide container. Catalase will bump
around until two hydrogen peroxide molecules bump into the catalase and bind. The catalase will cause
a burst, meaning it will create a chemical reaction that breaks the hydrogen peroxide molecules apart
and reforms them into one molecule of oxygen and two molecules of water. Once catalase completes the
chemical reaction, the water and oxygen will leave the enzyme and catalase continues to catalyze further
chemical reactions. A single catalase enzyme can complete the chemical reaction thousands of times
every second!
If all of the hydrogen peroxide is converted into end products, how much oxygen will be created? Lets
calculate a simple approximation:
The bottle of hydrogen peroxide purchased from the pharmacy is 3% hydrogen peroxide, this means that
for every 100 mL of liquid, there are 3 grams of hydrogen peroxide. Say we added 100 mL of hydrogen
peroxide solution to 200 mL of water, this means we added 3 grams of hydrogen peroxide.
For this reaction, assuming the conservation of mass in a chemical reaction where the mass of the reac-
tants equals the mass of the products, the mass of the oxygen gas (O
2
) and water (H
2
O) will also be 3 g. If this
assumption is valid, then the total molecular weight of the reactants will also equal the molecular weight
of the products. What percentage of the end products mass belongs to oxygen gas?
2 H
2
O
2
—-catalase—> O
2
+ 2 H
2
O
In each reaction:
• 1 oxygen gas molecule has a molecular weight of 31.998 g/mol
• 2 water molecules have a combined molecular weight of 36.030 g/mol (2 molecules x 18.015 g/mol)
The total molecular weight of the products is 31.998 g/mol + 36.030 g/mol = 68.028 g/mol. This is consis-
tent with our assumption in that the total molecular weight of the hydrogen peroxide reactants is similar
(68.028 g/mol).
Of the total molecular weight of the products, what percentage is oxygen gas?
O
2
/ (O
2
+ 2 H
2
O) x 100 = percentage oxygen molecular weight
31.998 g/mol / (31.998 g/mol + 36.030 g/mol) = 47.03 %
47.03% of 3 g is 1.41 g of oxygen gas! Let’s take it one step further. How much volume will 1.41 g of O
2
occupy? The density of oxygen gas at standard temperature and pressure (STP) is 1.429 g per litre. With
100 mL of store-bought hydrogen peroxide, you can create:
1.41 g / 1.429 g per L = 0.99 L of oxygen gas.
Can you calculate how many mL of water is created by catalase during the chemical reaction? Use your
Table 6-1. Be the RNA polymerase and the Ribosome to check your answer
5’ 3’
aac gaa ccg aac acc ttt att gtg gaa aac att aac gaa atg ctg
Template DNA (-)
3’ 5’
RNA
Amino Acids
O
Book _genetic engineering hero-AUG2021.indb 157Book _genetic engineering hero-AUG2021.indb 157 8/18/21 12:03 PM8/18/21 12:03 PM
158 Zero to Genetic Engineering Hero - Chapter 6 - Processing Enzymes
on completing your fth set of experiments!
Book _genetic engineering hero-AUG2021.indb 158Book _genetic engineering hero-AUG2021.indb 158 8/18/21 12:03 PM8/18/21 12:03 PM
159Zero to Genetic Engineering Hero - Chapter 6 - Processing Enzymes
Fundamentals: Diving into enzymatic processing
The basics of enzymatic chemical
reactions
The basic model of a chemical reaction involving an
enzyme (E), a substrate (a), and a product (b) is as
follows. Recall that the substrate is the starting mole-
cule and the product is the end material resulting
from the chemical reaction (Figure 6-8):
E + a <—> Ea —> *Ea* —> Eb —> E + b
1.
The enzyme (E) and substrate (a) are independently
bumping around the cell (E + a). There may be
hundreds, thousands or millions of each in the cell
at any given time. The numbers depend on the cell
metabolism and the environment the cell is in.
2.
They bind because of complementary molecular
shapes and chemical bonding (Ea). The “lock and
key” mechanism is illustrated in Figure 6-8.
3.
The chemical reaction occurs (*Ea*). The asterisks
denote the energy change resulting in the chem-
ical reaction happening.
4.
Through the chemical reaction, the substrate
molecule is changed into the product that is a
different molecule and has a different shape and
therefore chemical bonding (Eb).
5.
The product has a different shape and chemical
structure than the substrate. It no longer binds
strongly to the enzyme, and they separate (E + b).
The enzyme (E) and product (b) continue to bump
around the cell to complete their functions.
A common way to describe how an enzyme and the
substrate bind, is called the “lock and key” mecha-
nism (Figure 6-8). A key (substrate) can “bump” into
many different locks (enzyme binding site), but only
one key has a very specic pattern that will match the
lock mechanism and be able to turn. The molecule
structures of the enzyme and substrate are simi-
lar to the shape of the lock mechanism and the key.
However, unlike how a person has the intention to put
the key into a lock, there is “no one” and no “inten-
tion” to put the substrate molecule into the enzyme.
This is why the Four B’s of bumping around the cell
and chemical bonding are so important in causing the
molecule to rst encounter the enzyme, bump its way
into the enzyme, and if the t is proper and it binds, a
chemical reaction can occur.
In the previous example there was a single enzyme (E)
and a single substrate (a). What is far more common in
chemical reactions in the cell is that there will be more
than one substrate. You experienced this in Exercise
1 (isoamyl alcohol + acetyl-CoA) and Exercise 2 (X-gal
+ H
2
O). The enzyme will take one or more atoms from
one substrate and transfer it to the other substrate.
This results in two different products (Figure 6-9). The
same basic principles hold:
E + a + b <—> Eab —> *Eab* —> Ecd —> E + c + d
1.
The enzyme (E) and the substrates (a) and (b) are
independently bumping around the cell (E+a+b).
There might be hundreds, thousands or millions
of each at any given time in the cell.
2.
They bind together because of complementary
molecular shapes and matching chemical bonding
(Eab). The “lock and key” mechanism illustrated in
Figure 6-9.
3.
The chemical reaction occurs (*Eab*). The aster-
isks denote the energy change resulting in the
chemical reaction happening which may or may
not include the enzyme transferring some atoms
from one substrate to the other. This results in two
new molecule products (c) and (d): Ecd
4.
The products have different composition and
shapes than the substrates.
5.
Because the products have different shapes and
chemical structure than the substrates, they no
longer bind strongly to the enzyme, and they sepa-
rate (E + c + d). The enzyme (E) and products (c + d)
continue to bump around the cell to complete their
functions.
The Four B’s and enzyme function
In Chapter 4 you learned about the basic operating
environment of a cell. You learned that molecules do
not have intentions, nor do they have intelligence.
Instead, three primary factors drive the operation and
decision making” in cells:
• The number of individual or groups of molecules.
The rate at which molecules bump into other mole-
cules and move around the cell.
• The strength of interaction between molecules.
Book _genetic engineering hero-AUG2021.indb 159Book _genetic engineering hero-AUG2021.indb 159 8/18/21 12:03 PM8/18/21 12:03 PM
160 Zero to Genetic Engineering Hero - Chapter 6 - Processing Enzymes
Figure 6-8. “Lock and key” illustration: the Four B’s involving an enzyme, a substrate, and the product of a chemical reaction.
a
E aE E b
a
E
b
*
Bump Bind After-Burst Bump
1
2 43 5
Figure 6-9. A protein enzyme can also bind to multiple substrates and process them into multiple different products. The shape of
the molecule is like the shape of a key, and the shape of the binding region of the enzyme is the like the lock.
*
Bump Bind After-Burst Bump
1
2 43 5
a
E
b
a
c
b
d
Eab Ecd E
In this section, we are going to look more closely at this
concept in the context of protein enzymes.
The number of molecules: The number of protein
enzymes and the number of substrate molecules
(reactants) are essential factors in determining
how frequently a chemical reaction will happen.
If you have twice as many enzymes or substrate
molecules in the cell, this increases the probability
that they will encounter one another while bump-
ing around.
In Exercise 2, you added extracted beta-galacto-
sidase enzymes to your tube of X-gal substrate
reaction, there would have been more substrate
molecules bumping around and more substrate
bumping into the enzymes resulting in an quicker
color change. Similarly, an increased number
of enzymes would increase the chance of the
substrate bumping into the enzymes, resulting in
a chemical reaction and a color change happening
faster.
The rate at which molecules bump into other
molecules: The bigger a molecule is, the slower
it moves around the cell. If a protein enzyme is
very large, it will bump around the cell slower, and
therefore encounter the substrate molecules less
frequently than a small protein enzyme. Similarly,
a large substrate molecule will move more slowly
around a cell than a very small molecule.
So, how often does a small molecule like X-gal
bump into a protein such as beta-galactosidase?
While scientists haven’t figured out the rate for
every molecule and many factors are at play in
determining the rate at which molecules move,
Book _genetic engineering hero-AUG2021.indb 160Book _genetic engineering hero-AUG2021.indb 160 8/18/21 12:03 PM8/18/21 12:03 PM
161Zero to Genetic Engineering Hero - Chapter 6 - Processing Enzymes
X-gal is considered a “normal” small molecule
with a “normal” rate of movement for a small
molecule. A “normal” amount can be estimated
at up to 10
20
collisions per second. In the world
of biochemistry, each molecule can have up to
100,000,000,000,000,000,000 or 100 quintillion
collisions with other molecules each second! If you
were an atom or small molecule, this is the equiv-
alent of you brushing elbows with every person on
earth 17 billion times each second. In other words,
every protein enzyme in the cell, such as beta-ga-
lactosidase, has MANY opportunities to interact
with its substrate, X-gal, hundreds or thousands
of times each second. This is why an enzyme can
cause a chemical reaction to happen. You might
recall RNA polymerase can typically add 50 ribo-
nucleotides to an RNA transcript each second.
A single catalase enzyme can turn thousands of
hydrogen peroxide molecules into oxygen and
water every second.
This can be simplified into knowing that mole-
cules typically interact (“bump”) a billion times a
second with other molecules within the cell. This
means that often two molecules will bump into
one another countless times per second and it
might take millions of bumps before a substrate
“key” bumps into the enzyme “lock, resulting in
a chemical reaction.
The strength of interaction between molecules:
The most important factor in determining the
operation or decision making in a cell is bonding.
The specic interaction between molecules is the
“logic” that drives specic events in cells. In the
X-gal example, the protein enzyme beta-galacto-
sidase has the right structure and composition
to be able to bind quite specically to X-gal, and
X-gal alone. It does not bind to other molecules in
the cell with any notable strength or specicity. It
is ‘beta-galactosidase’s job’ to bind specically to
X-gal and catalyze the chemical reaction that cuts
off a sugar ring allowing X-gal to dimerize and
become colorful (Figure 6-6).
If the interaction is strong, the substrate is more
likely to get “locked” in place in the protein enzyme
active site, the site at which the chemical reac-
tion happens. If the bonding interaction between
the enzyme and the substrate is weak, then the
substrate will not “lock” into place in the enzyme
active site. The strength of an interaction between
molecules is dened by its dissociation constant,
discussed in the breakout below.
The same principles apply to the thousands of other
protein enzymes in a cell. But what is an interaction?
What does it mean when a substrate gets “locked into
place”? In the next sections, we’ll dig a little deeper
into atoms, molecules, and the bonding that makes
it all happen so that we may answer these questions.
Atoms
Before learning what causes the interactions between
substrates, protein enzymes and having a more
in-depth discussion about chemical reactions, we
need to see what an atom is.
An atom is very small and generally considered the
smallest unit of matter. A single atom is usually about
300 picometers in width, which is 0.0000000003
meters, or about a million times thinner than a
human hair. An atom has two essential attributes:
Nucleus: a very dense core called a nucleus,
contains sub-atomic particles called protons
(which are positively charged), and neutrons
(which have no charge). The nucleus makes up
more than 99% of the mass of each atom and has
a positive charge.
Electron clouds: electrons are very small nega-
tively charged atomic particles that orbit the
nucleus very fast in “cloud-like” patterns. The
electron clouds are called orbitals, and they are
negatively charged.
Dissociation constant Web Search Breakout
Theequilibrium dissociation constant (K
d
) is used to evaluate and rank strengths of molecular interac-
tions. You can search for this, as well as “Enzyme Kinetics”, “Michaelis-Menten Kinetics” and “Specicity
Constant” online to learn more in-depth. Each of these topics discusses how the strength of bonding
between a substrate and an enzyme relate to the specicity of the chemical reaction (how well a substrate
“locks in place”), as well as the speed of the chemical reaction.
Book _genetic engineering hero-AUG2021.indb 161Book _genetic engineering hero-AUG2021.indb 161 8/18/21 12:03 PM8/18/21 12:03 PM
• No Comment
..................Content has been hidden....................