BIOLOGY 100/101
LECTURE OUTLINE
AND
STUDY GUIDE
Second Edition
Written by
Daniel Sourbeer©
Department of Life Sciences, Palomar College
1140 W. Mission Rd.
San Marcos, Ca 92069-1487
760.744.1150. Ext. 2775
dsourbeer@palomar.edu
Edited by
Joel Alan Harrison, Ph.D, M.P.H.
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3
TABLE OF CONTENTS
Lesson #
0
1
2
3
4
5
6
7
8
9
10
11
12
16
17
18
19
20
21
Topic
Biology and the Characteristics of Organisms
The Nature of Science
Basic Chemistry: Atomic Structure
Basic Chemistry: Chemical Bonding and Reactions
Biochemistry: General Concepts and Important Inorganics
Biochemistry: Carbohydrates and Proteins
Biochemistry: Proteins and Lipids
Plasma Membranes and Solutions
Cells and Their Structures
Energetics: An Overview
Energetics: Pieces of the Energy Puzzle
Photosynthesis
The Cellular Oxidation of Glucose
Mitosis
Meiosis and Sexual Reproduction
Mendel and Single Trait Inheritance
Solving Problems and Multiple Trait Inheritance
Multiple Trait Inheritance and Epistasis
X-Linked Traits, Imprinting, and More
General Biology Study Guide
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Page #
5
7
15
23
33
39
45
51
57
63
69
73
81
105
109
113
119
121
127
211
5
Lesson 0: Biology and the Characteristics of Organisms
I.
II.
III.
Biology, a specific discipline of science, is the study of living things, called organisms, and includes the
investigation of:
A.
Their physical characteristics.
B.
Their evolutionary relationships.
C.
The processes that distinguish the living from non-living, i.e. metabolism, inheritance, etc.
D.
The interactions within species, between species, and between species and the physical world.
E.
Etc.
Most biologists observe the following characteristics in true organisms.
A.
Organisms show specific levels of organization depending on their complexity.
1.
Atoms form molecules.
2.
Molecules form cellular structures.
3.
Cellular structures form cells.
4.
Cells form tissues (groups of cells with a specific function, e.g. muscle tissue, nervous
tissue, etc.)
5.
Tissues form organs--e.g. heart, lung, liver, kidney, etc.
6.
Organs form organ systems--e.g. cardiovascular system, urinary system, etc.
B.
Organisms are composed of carbon-based molecules--e.g. proteins, carbohydrates, lipids, etc.
C.
Organisms metabolize--i.e. they take in, use, and alter energy.
1.
Plants absorb solar energy, converting it first to chemical energy, and eventually releasing
it as heat energy.
2.
Animals take in chemical energy, altering it, and eventually releasing the energy as heat.
D.
Organisms respond to the environment.
1.
Some respond in obvious ways, like movement.
2.
Some respond in more subtle ways, like a chemical response.
3.
All evolve, and are adapted to their environment.
E.
Organisms grow, i.e. they increase in size.
F.
Organisms reproduce themselves, i.e. they make new individuals or offspring in two possible
ways.
1.
Asexual reproduction does not involve a mate, and offspring are genetically identical to
the parent and other offspring.
2.
Sexual reproduction requires a mate, and offspring are genetically distinct from parents
and other offspring.
G.
The specific characteristics of an organism are encoded by genetic material in the form of nucleic
acid (DNA or RNA), which is heritable (going from generation to generation).
It can be important to determine whether something is an organism or not.
A.
It is extremely important in the study of disease to know whether or not it is caused by an
organism.
1.
Epidemiologists are biologists who study the occurrence of disease and other health
related conditions or events in defined populations.
a)
When people begin to die in an area of the world, the Center for Disease Control
and the World Health Organization send in epidemiologists to investigate.
b)
They must decide whether the deaths are caused by chemical or biological agents.
c)
If a biological agent is involved, they must determine what type of biological
agent it is, how it is transmitted, etc.
2.
Consider viruses--something the general public would consider “alive” because they
cause disease and can pass from person to person.
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a)
A summary of viral characteristics is listed below.
Characteristic
Organized
Organic composition
Metabolism
Respond to environment
Grow
Reproduce self
DNA/RNA
b)
c)
Present/absent
Present
Present
Absent
Present
Absent
Absent
Present
Comments
Adapted to host
Host makes viruses
As you can see, viruses have some, but not all, of the characteristics of life, and
many biologists would therefore not consider them true organisms.
(1)
Viruses were certainly derived from organisms.
(2)
Viruses function as complex gene carriers, in nature, transporting genes
from one cell to another.
(3)
Once infected with a virus, a cell’s reproductive machinery makes new
viruses, killing itself in the process, and liberating more viruses.
Because viruses are not classic organisms, they do not respond to treatments such
as antibiotics that are effective against organisms, like bacteria.
7
Lesson 1: The Nature of Science
I.
II.
Science is a way of knowing that has a dual nature, process and product (body of knowledge).
A.
Science has, at its core, a process, a method of generating data and interpreting that data that will
be discussed below.
B.
The data, hypotheses, theories, and principles generated by the scientific process, are the product
of science.
The process of science is generally known as the scientific method.
A.
The process begins with a period of gathering information about a subject that may include the
following methods.
1.
Reading previously published literature about the subject.
2.
Making personal observations about the subject.
3.
Performing trials--best summarized by the expression “I wonder what would happen
if...?”
B.
This leads to the posing of a question that cannot be answered with existing data, and the
proposal of a possible explanation, or answer, to the question called an hypothesis.
1.
An hypothesis is a possible explanation to a question generated by existing data.
2.
An hypothesis is sometimes described as an “educated guess.”
3.
An hypothesis is an explanation that is disprovable.
C.
The hypothesis must be tested via experimentation or further observation for validity.
1.
Experimentation is a specific way of testing an hypothesis.
a)
An experiment typically has the following components:
(1)
A single independent variable.
(a)
The independent variable is the component that is being
manipulated by the scientist, it is what you are testing, the “cause”
in “cause and effect.”
(b)
There should be only one independent variable in a well
constructed experiment.
(2)
One or more dependent variable(s) which can be defined as the “effect” of
“cause and effect,” or as any change in initial conditions induced by the
independent variable.
(3)
One or more experimental (or test) group(s) in which the independent
variable is included in the procedure.
(4)
One or more control group(s), defined as a group that is treated the same
as the experimental group, except the independent variable is withheld.
(a)
The control group is compared to the experimental group to
determine if any changes are indeed caused by the independent
variable.
(b)
Without a control group you do not know if changes would have
occurred even without the independent variable.
(c)
Properly using control groups is key to experimentation and testing
hypotheses, and is fundamental to good science.
(d)
When the independent variable cannot be withheld, as in the case
of temperature, then one of the experimental groups is designated
as the control group, and is used for comparison to other
experimental groups.
(5)
An experiment will have numerous controlled variables.
(a)
Controlled variables are all factors that are common to both the
experimental and control groups.
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(b)
b)
Controlled variables must be held constant for both experimental
and control groups so there will be only one variable that is
different in the two groups, that being the independent variable.
The above terms are best appreciated by using an example, so consider the
following:
After reviewing some data on birth disorders near a site identified as a
dumping area for a pesticide, you observe that human birth disorders and cancer
rates are higher than normal, while longevity is lower than normal. You suspect
the pesticide is responsible for high incidence of developmental disorders and
mortality. You question whether the pesticide causes developmental disorders
and increased mortality in animals, but can find no information on the subject.
You wish to concretely test the effect of the pesticide on development and
mortality in non-human test animals. You choose chickens as your test subjects.
You hypothesize that the pesticide does cause developmental disorders
and increased mortality in chickens. You must now test your hypothesis. You
gather 100 chicken eggs to serve as your experimental group. You chip a small
hole in the shells, and inject a specific volume of pesticide mixed with dilute ethyl
alcohol solution (the pesticide dissolves better in the alcohol solution than pure
water) into each egg. You then you seal the hole with glue, and incubate the eggs
at the appropriate temperature and humidity. You take another group of 100 eggs
and treat them exactly the same (chip, inject alcohol, seal with glue, incubate)
except you do not include the pesticide in the alcohol solution you inject into the
eggs. Consider the following questions and answers.
You candle the eggs daily, removing and autopsying the eggs that die.
Those that hatch are evaluated for health and tracked to their death. Your study
gathers information concerning developmental disorders and longevity.
(1)
What is the independent variable?
Answer: A quantity of pesticide.
(2)
What is the dependent variable?
Answer: Chick development and longevity (mortality).
(3)
Why are the control eggs injected with alcohol? Why are they
manipulated at all? Wouldn’t eggs that had not been chipped and injected
serve as a reliable control group?
Answer: Remember, in an experiment you want only one variable that is
not held constant in both the experimental and control groups, and that
variable is the independent variable. Does alcohol cause developmental
disorders? Does chipping hole in the shell and covering it with glue cause
developmental disorders? Maybe, but you do not want to test whether
alcohol, or chipping, is a problem, but whether the pesticide is causing
problems. If there is a difference in the development between the control
and experimental groups we want that difference to be the result of only a
single variable, in this case, exposure to pesticide. If you do not control all
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other variables, we do not know whether differences are due to the
alcohol, or chipping a hole in the shell, etc.
(4)
What are some examples of controlled variables in this experiment?
Answer: Using eggs from same breed of chicken, incubating the eggs at
same temperature and humidity, injecting the same volume of alcohol
solution, etc.
2.
Classic experimentation may not be appropriate for testing some hypotheses, in which
case, further observation may determine their validity.
a)
What is an example of a situation in which further observation is more appropriate
than experimentation?
You and others have observed and catalogued the diet of a troop of
chimpanzees, and conclude that their vegetarian diet is too low in protein to
account for their robust condition, and reproductive success. You hypothesize
that they occasionally prey on animals to supplement their vegetarian diet with
concentrated protein. After months of further observation you observe a small
group of chimpanzees cooperatively hunting, killing, and eating a monkey.
b)
c)
D.
Observation may be directly observed or recorded as it is happening.
Observation may be indirect as was recently done in a grizzly bear census in
which hair samples were collected, and DNA analysis was used to determine the
number of individuals, without ever capturing or seeing any of the bears.
d)
Observation may involve the use of control groups: It is typical in
epidemiological studies of disease and nutrition, for example, to compare
populations that differ in the incidence of a disease, and look for an “independent
variable” that causes the disease.
A specific test (experimentation or observation) generates data, and that data is used to evaluate
the hypothesis.
1.
The test data must be statistically (objectively) analyzed.
a)
There are methods to evaluate differences in experimental and control groups to
determine whether chance can duplicate the differences in the groups, and how
often chance can duplicate the results (as opposed to some cause for the difference
in an experimental and control group).
b)
To accept a “cause” as the difference between experimental and control groups,
biologists require a standard of “95% confidence,” which means that if you
repeated the experiment 100 times, you would get similar results more than 95
times.
c)
If biologists have a “95% confidence” in their experimental results, then the
difference between the experimental and control group is described as a
“significant difference.”
d)
So if a physician said “There is a significant difference between the chance of
recovery from this disease using medicine A rather than medicine B,” it means
researchers have met the standard of “95% confidence” in experimental results.
e)
Let’s return to our pesticide and chicken egg development example.
(1)
Lets say that the experimental group had more abnormal egg development
than the control group, but statistical analysis determined that similar
10
E.
results would occur “only” 85 times out of 100 repetitions of the
experiment.
(2)
Your results do not meet the standard of “95% confidence” so there is no
significant difference between the two groups.
(3)
You cannot say “with confidence” that the pesticide causes developmental
disorders in chickens.
f)
As you can see from this example, the standard of “95% confidence” is a strict
one.
2.
For any test results to have validity they must be repeatable, i.e. others must be able to get
the same results under the same conditions.
3.
As part of evaluating the validity of the hypothesis you may come to one of the following
conclusions.
a)
You may choose to accept your hypothesis.
b)
You may choose to reject your hypothesis.
c)
You may decide to modify your hypothesis, and test again.
d)
You may decide to modify your test, if you see design flaws, and test your
hypothesis again.
e)
In our egg development example, the statistical analysis requires you to reject the
hypothesis, but you would be wise to consider possible flaws in your experimental
design and try again; for example:
(1)
You might consider using a different test animal.
(2)
You might try injecting different concentrations of pesticide into the eggs.
(3)
You might modify your procedure so that the embryo suffers multiple or
continuous exposure to the pesticide.
Once a scientist tests and evaluates an hypothesis, it is important to communicate the results with
other scientists.
1.
There are several reasons for communicating results.
a)
Others may see flaws in your experimental design, or analysis, and make
suggestions.
b)
Others may be able to take what you have done and build on it.
c)
Others may see applications for your work that you had not imagined.
d)
Others may want to collaborate or work with you on your projects.
2.
There are several ways to communicate results, and they include the following:
a)
Informal discussions among peers.
b)
Publishing results in scientific journals
(1)
Articles published in scientific journals have been “peer reviewed” by a
small panel of scientists who have expertise in the subject area discussed
by a particular article.
(2)
An article may be published if the panel is in general agreement with, or at
least has no major problems with, the design and analysis of the work.
(3)
“Scientific” articles in non-peer reviewed magazines should be viewed
with a great deal of skepticism, until the work is peer reviewed.
c)
Making formal presentations at scientific meetings of various types.
3.
Communication, and even the process of science, has suffered, in my opinion, over the
past 30 years, as funding for science has moved from the public to private sector.
a)
Science in the private sector is profit based, so sharing information can help the
competition.
b)
Universities and other public entities have likewise become more profit sensitive,
even forming partnerships with industry.
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c)
III.
IV.
Ownership of patents on scientific discoveries, such as transgenic animals(can life
forms be patented?), and DNA sequence information(who owns genes?), has
become financially important, and litigious.
(1)
On the one hand, this leads to secrecy about the specifics of research and
all things related to specific lawsuits.
(2)
On the other hand, it can actually cause a rush to publish incomplete work,
as publishing dates are traditionally used to determine “who was first.”
d)
Some scientific research has been impacted by its profit potential.
(1)
The National Park Service has recently entered profit sharing agreements
with companies wishing to do research in our national parks.
(2)
Investigation of genetic disease and evolutionary origins has been slowed
by suspicions concerning the profit motive of such research.
(3)
Etc.
The process of science is ongoing, as one conclusion leads to a new question, and it employs different
types of logic in building hypotheses and evaluating data.
A.
Inductive logic builds from specific information to a broad conclusion.
1.
For example, let’s say you observe a variety of plants and animals microscopically, and
determine they are all composed of cells and products secreted by cells.
2.
From these numerous, specific observations, you use inductive reasoning to reach the
broad conclusion that all organisms are composed of cells and their products.
B.
Deductive logic takes a broad conclusion and predicts specific outcomes, and is often phrased as
an “if....then...” statement.
1.
Continuing the example from above: From the broad conclusion that all organisms are
composed of cells and cell products, you deductively make the specific prediction that
brown kelp is composed of cells and cell products, even though you have never
microscopically observed brown kelp.
2.
Phrased as an “if...then” statement : “If all organisms are composed of cells and their
products, then the brown kelp is also composed of cells and their products.”
3.
You could test your hypothesis via microscopic observation of brown kelp.
The product of science includes the data generated in scientific tests, as well as the scientific ideas and
conclusions derived from data.
A.
An hypothesis is a possible answer to a proposed question, and has the following characteristics.
1.
It is specific in scope.
2.
It is testable.
3.
It is disprovable.
4.
It gains credibility only when verified by repeated testing.
B.
Related accepted/verified hypotheses are combined to generate a theory.
1.
A theory addresses the mechanism by which something happens.
2.
A theory is predictive.
3.
Theory is accepted when it accurately and repeatedly predicts outcomes.
4.
An example of accepted theory is the cell theory in biology.
a)
Hypotheses testing the cellular composition of organisms, and the behavior of
cells led to the “cell theory” which states that organisms are composed of cells
and their products, and that cells come from other cells.
b)
The cell theory gained acceptance as it repeatedly and accurately predicted the
cellular composition and origin of organisms.
5.
In the biological sciences accepted theory is considered “factual” because of the many
hypotheses from which it is composed, and the many hypothetical outcomes it accurately
predicts.
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a)
V.
In this course we will be considering only accepted theory, as credible scientific
information.
b)
We will, in fact, use the term theory to mean accepted theory, unless otherwise
noted.
6.
Theory is confusing for the general public because they use the word synonymously with
hypothesis or uncertainty.
7.
Biologists contribute to the public’s confusion by not contrasting hypothesis with theory,
or distinguishing accepted theory from untested theory when talking about their ideas in
public forums.
C.
The terms “Principle” and “Law” are commonly used in physics and chemistry.
1.
Principles and laws are generated from accepted hypotheses and theories.
2.
Natural laws and principles do not tend to address a mechanism by which something
happens, but rather state a relationship at specific experimental conditions.
3.
The terms principle or law are used when there is abundant supporting data, and they are
considered “factual.”
4.
These terms are rarely used in biology which creates a public perception that biological
theory is “uncertain” when this is not the case, as accepted theory is likewise “factual.”
5.
Consider this example: The Law of Conservation of Matter.
a)
Specific tests of hypotheses led to atomic theory, and other theories of chemistry.
b)
These theories led to the Law of Conservation of Matter which states that matter
is neither created nor destroyed in chemical reactions.
c)
The law states the relationship or circumstance, that matter is always
conserved(saved) in chemical reactions, but it is the underlying theories that
explain the means (reasons) by which the matter is conserved.
D.
The data generated by tests is also a product of science.
Science has a number of characteristics that distinguish it from non-science or pseudoscience.
A.
Science is mechanistic.
1.
It asks the question “By what mechanism does that happen?”
2.
“It’s magic” or “God did it” are not mechanistic answers.
B.
Science works in the natural world.
1.
To answer questions mechanistically one must work in the natural world.
2.
Supernatural explanations are not testable.
3.
Phrases like “God did it,” “It is not for us to know,” “It just...happened,” and “It’s
magic,” are not mechanistic explanations, and are not testable.
C.
Scientific ideas are testable.
1.
In science, one may not assume untested ideas to be true, i.e. because something is
possible does not mean it is so.
2.
Testable scientific ideas should be disprovable.
3.
Disproving one idea does not verify another.
D.
Science is objective.
1.
Scientific tests must generate data which anyone can see, examine, and (statistically)
evaluate.
a)
Conclusions about an hypothesis must be drawn specifically from the data
generated in testing that hypothesis.
b)
Science has no place for psychics, shamans, etc. who have “revelations”,
“visions,” or “knowledge” that is unavailable for objective analysis.
2.
Faith lacks objectivity, and is not appropriate in scientific discussions.
a)
Faith permits (or requires) belief in ideas that have either not been tested, or are
unsupported by test data.
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b)
c)
E.
F.
Faith is the polar opposite of objectivity, and is the antithesis of good
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ÿÿÿÿthe “placebo effect” in medical trials.
a)
The placebo effect is a real consequence of treatment, in which individuals who
are treated for a disorder are more likely to improve than those who are not treated
for the disorder, even if the treatment has no objective therapeutic value.
(1)
This is a real effect, thinking you can get better improves your chances of
getting better in a statistically significant way.
(2)
This is why physicians always treat individuals even in “hopeless”
situations: hope can heal.
b)
For example, if 1000 individuals have a disorder, and 500 are given a sugar pill
(placebo) and are told it is a treatment for the disorder, and 500 receive no
treatment at all, the 500 given the worthless sugar pills will typically show a
statistically significant rate of improvement over those not treated, i.e. thinking
you can get better can make you better.
c)
“Double blind” studies prevent prejudice by either scientists or patients involved
in a study.
(1)
Using the example from above and modifying it to a double blind study,
500 individuals would receive a possible medicine, and the other 500
would receive a placebo.
(2)
A third party (neither the scientists interpreting the work, nor the patients
involved in the study) randomly determines which 500 patients will
receive medicine, and which 500 patients will receive the placebo.
(3)
It is “double blind” because neither the scientists nor the patients involved,
know who is taking the medicine until statistical analysis of the data is
complete.
(4)
Since everyone may be getting treatment, treatment is reduced to a
controlled variable, so any significant difference in improvement between
the two groups can be traced to the medicine, rather than to a placebo
effect.
Science is repeatable and verifiable, i.e. the results from scientific tests must be duplicated by
others to have any credibility or veracity.
Science has no absolute truths.
1.
Scientists must be open to scientifically credible challenges to established ideas.
2.
Scientific conclusions are based on data, and if data changes, so too may the conclusions.
3.
There is no supreme authority in science to guarantee truth.
14
a)
VI.
Scientists must reason for themselves the veracity of scientific conclusions, as
there is no authority to say, “That’s correct!”
b)
I remember asking a colleague the “best way” to capture bats under certain
conditions. He replied, “ I don’t know,” and after a pause, “but I can tell you what
I do.” It was a subtle, but pointed, reminder about scientific authority.
Many factors have led to the general public’s confusion about science, and outright hostility towards
science.
A.
A general lack of understanding of the scientific process.
B.
The general public confuses reasoned conclusion, or logic, with science.
1.
Scientific conclusions require the scientific process described above.
2.
Being “careful” or “reasoned” is not enough to define something as science.
3.
A conclusion may be logical, but neither scientific nor accurate.
C.
The general public confuses “being open” with “acceptance.”
1.
As mentioned above scientists should be open to credible, scientific challenges to
established ideas.
2.
Many in the general public feel ideas should be accepted because they are “possible,”
until proven otherwise.
a)
Scientists do not accept hypotheses in the absence of testing, but rather accept
hypotheses only when verified by credible testing.
b)
For example: it may be possible that trolls live under bridges, but there is no
credible evidence for their existence, so I do not scientifically conclude that they
do: I may want to believe they do, but such a belief is clearly non-scientific.
3.
The scientist must always be open to exceptions and modifications to accepted theory.
D.
Ego, emotion, and truth-- “I believe it, therefore it is true (I don’t care what you say!).”
1.
It is human nature to feel, “My conclusion is better than your conclusion.”
2.
Emotional attachment to an idea prevents objectivity.
3.
Emotional attachment leads to “absolute truths,” when the reality is that neither intense
belief, nor good science, guarantees the truth of an idea.
E.
Personal and political agendas are often at odds with scientific conclusions, and organizations
deliberately work to generate hostility towards science, and to confuse the public about the nature
of science and its value to society, by appealing to fear, emotion, and prejudice.
1.
Some techniques to discredit valid scientific conclusions include the following.
a)
Equating testimonials with well designed, controlled studies.
b)
Making no attempts to distinguish between preliminary results and confirmed or
accepted results.
c)
Equating results of studies that have not undergone peer review, with those that
have.
d)
Attempting to “prove” one idea by discrediting another.
e)
Manipulating the public through emotional appeals to religious prejudice, and
possible economic and social consequences of scientific studies or theory.
2.
Theory and technology from a variety of biological sub-disciplines have been the subject
of such assaults, including: evolution, ecology, genetics, medicine, nutrition, and others.
3.
These efforts are dangerous to an enlightened electorate in several ways.
a)
Economic loss--tremendous amounts of money spent on worthless placebos.
b)
A scientifically ignorant electorate --it is hard to make sound judgments
concerning science and technology if one does not understand or appreciate
science.
c)
Loss of natural habitat, and quality of life--development proceeds unchecked,
often flying in the face of sound ecological conclusion.
15
d)
F.
Loss of life--some improperly tested supplements or procedures may be harmful
or take the place of truly beneficial treatments.
Most Americans lack the ability to examine events from different perspectives.
1.
My children, for example, are biological entities--I can explain their existence,
development, etc according to biological theory.
2.
I, however, refuse to think of my children as biological entities, to me they are miracles of
life.
3.
The difference between me and most Americans is that I know my perspective is not
scientific.
16
17
Lesson 2: Basic Chemistry-Atomic Structure
I.
II.
III.
IV.
You may be wondering why we are learning about chemistry in a biology course.
A.
The study of the processes of biology has led to questions that are answered increasingly at the
molecular level.
B.
The traditional lines that separate the disciplines of chemistry, biology, and physics have become
increasingly blurred, and a more comprehensive understanding of life processes is more
important than ever.
Chemistry is a discipline of science that studies the properties and interactions of the elementary
particles of matter, and the atoms and molecules they form.
A.
Matter has mass and volume.
1.
Mass is a measure of the substance of something, when gravity pulls on mass it generates
weight, so for our purposes we will consider mass and weight as interchangable terms.
2.
Volume is a measure of space.
3.
Matter can exist in different forms or “states,” and we will be concerned with three states
of matter.
a)
Gaseous--the least dense form of matter, it takes the shape of its container.
b)
Liquid--matter in an intermediate state of density, it also takes shape of its
container.
c)
Solid--the most dense state of matter, solids have a shape of their own.
B.
Elements are “pure substances” that cannot be purified into simpler forms using chemical
methods--e.g. Hydrogen, Oxygen, Carbon, etc.
C.
The basic unit of an element is the atom, and elements are composed of like atoms.
Atomic and chemical theory tell us about the composition and behavior of atoms.
A.
Atoms are composed of three important subatomic particles, which are listed below along with
some of their characteristics.
1.
Protons(p) are positively charged particles with a mass of approximately 1 dalton or amu
(atomic metric unit).
2.
Neutrons(n) are particles of neutral charge, with a mass of approximately 1 dalton or amu.
3.
Electrons(e-) are negatively charged particles, with a mass of approximately 1/1800
dalton or amu.
B.
The protons, neutrons and electrons of an atom are arranged as follows.
1.
Protons and neutrons are found in the “center” of an atom forming its nucleus.
a)
The nucleus contains virtually the entire mass of the atom (because the mass of
electrons is minimal), but makes up only about 1/100,000 of its diameter
(comparable to a fly in the middle of a football field).
b)
Protons and neutrons are held together in the nucleus by a gravity-like force called
strong force.
2.
The electrons are in constant motion and form an “electron cloud” by moving around the
nucleus.
a)
The electrons are found within regions called orbitals and shells--these are
important in chemical bonding, and will be discussed later.
b)
Electrons are held in their orbitals by two forces.
(1)
The electromagnetic attraction generated by the opposing magnetic forces
of the positively charged protons and negatively charged electrons.
(2)
A gravity-like force called weak force.
The Periodic Table is an assemblage of the known elements that includes information about the atoms of
which these elements are composed--refer to your periodic table as you consider the information below.
18
A.
V.
The Periodic Table is normally composed of squares, with information about an element within
each box.
1.
In the center of the box is the atomic symbol.
a)
The atomic symbol is an abbreviation of the element name,usually taken from the
first one or two letters of the element’s name in English, although some
abbreviations are taken from the element’s Latin name.
(1)
H = Hydrogen.
(2)
C = Carbon
(3)
Na = Sodium (Latin name = Natrium)
(4)
Au = Gold (Latin name = Aurum)
b)
The atomic number is the number listed below the atomic symbol (in our
example).
(1)
The atomic number represents the number of protons in the nucleus of the
atoms of this element, and it is the number of protons that determines the
element name.
(2)
Notice that the elements are arranged in ascending atomic number from
left to right, forming rows called periods.
c)
The atomic mass (or mass number, or atomic weight) is the number above the
atomic symbol, and is an approximation of the mass of an atom of this element.
2.
If an atom has a neutral charge, it will have the same number of protons as electrons.
a)
The charges of the protons and electrons neutralize one another.
b)
The number of electrons is initally assumed to be same as the atomic number.
The periodic table can be used to determine the physical and chemical characteristics of the atoms of
which an element is composed.
A.
The number of neutrons in the nucleus of an atom can be calculated using the atomic mass and
atomic number.
1.
Atoms are typically depicted as follows (when not represented in a periodic table):
Y(atomic mass)
12
X
2.
3.
C
for example,
Z(atomic number)
6
Where X is the atomic symbol, Y is the atomic mass, and Z is the atomic number.
The atomic number is the number of protons in the nucleus.
The atomic mass is the mass of the atom, but it is also the number of protons and
neutrons in the nucleus of the atom.
a)
You may remember that the mass of protons and neutrons is approximately 1
dalton, while the mass of electrons is only 1/1800 of a dalton, an insignificant
mass for our purposes.
b)
As a result, the mass number (atomic mass) is not only the mass of the atom, but
an approximation of the number of protons and neutrons combined in the nucleus.
c)
The number of neutrons in the nucleus can be calculated by subtracting the atomic
number (the number of protons) from the the atomic mass(the number of protons
and neutrons, rounded off to the nearest whole number).
#Neutrons = atomic mass - atomic number
d)
Examples:
19
Symbol
H
K
Cu
B.
at.#
1
19
29
mass#
1
39
64
#protons
1
19
29
#neutons
0
20
35
All atoms of a particular element have the same number of protons, but they may not have the
same number of neutrons(and therefore different atomic masses).
1.
Atoms with differing numbers of neutrons(and atomic masses) are called isotopes.
2.
Isotopes are atoms of the same element (because they have the same number of protons),
and they have similar chemical behaviors.
3.
Consider the element Hydrogen.
a)
Hydrogen is element number 1, so all isotopes of Hydrogen have one proton.
b)
There are three isotopes of Hydrogen: H1 with a mass of 1 dalton, H2 with a mass
of 2 daltons and H3 with a mass of 3 daltons.
20
c)
Isotope
Hydrogen 1
Hydrogen 2
Hydrogen 3
Carbon 12
Carbon 13
Carbon 14
Nickname
Deuterium
Tritium
Consider the information in the table below.
at#
1
1
6
6
6
mass#
1
2
1
12
13
14
#protons
1
1
3
1
6
6
6
4.
VI.
#neutrons
0
1
6
7
8
Radioactive?
no
yes
2
yes
no
yes
yes
The atomic mass for an element in the periodic table is an average mass from a sampling
of atoms in nature, not an average of the masses of the types of isotopes found in nature,
so the periodic table therefore reflects the abundance of a particular isotope in nature.
a)
If you average the masses of H1, H2, and H3 you get a mass of 2 daltons.
b)
However, if you look at the atomic mass of Hydrogen in the periodic table you see
a mass of 1.008 daltons listed, which means that of all the Hydrogen atoms in
nature, H1 is by far the most abundant because the average of the Hydrogen atoms
sampled from nature is very close to a mass of 1 dalton.
C.
Some isotopes have a combination of neutrons and protons which makes the atom unstable or
radioactive.
1.
Radiation or radioactivity is the spontaneous loss of matter and/or energy by an atom.
2.
Some examples of radiation include the following:
a)
Alpha radiation is a Helium nucleus (= 2 protons and 2 neutrons) lost from a
radioactive isotope.
b)
Beta radiation is composed of electrons lost from an atom.
c)
Neutron radiation is composed of neutrons lost from an atom.
d)
Electromagnetic radiation is composed of extremely small particles called photons
that travel in a wave-like trajectory (wave energy).
(1)
Shorter wavelength = higher frequency = higher energy.
(2)
Gamma radiation is photons traveling in a shorter wavelength emitted
from the nucleus.
(3)
X ray radiation is photons traveling in a longer wavelength emitted from
electron orbitals.
3.
Radioactive isotopes have similar chemical properties to non-radioactive isotopes, but are
easily tracked by different imaging techniques (X-rays, CAT scans, etc) and have many
uses in medicine and biology.
The chemical properties of an atom, i.e. how an atom behaves in chemical reactions, is largely
determined by the distribution of electrons around the nucleus of the atom.
A.
As mentioned previously, atomic theory predicts that electrons form an electron cloud around the
nucleus, however, it is not a completely random distribution of electrons.
B.
Electrons are found in discreet regions of energy distribution called shells around the nucleus,
with each shell capable of holding a certain number of electrons.
C.
The shells are composed of subshells, which are, in turn, composed of orbitals.
D.
Orbitals have specific shapes and orientations around the nucleus, and are the smallest unit of
electron distribution, but we will not be concerned with electron distribution at this level of
detail.
21
E.
Shells are numbered beginning closest to the nucleus(n = shell number in the diagram below).
etc.
n=3
n=2
n=1
Maximum of
18 electrons
nucleus
Maximum of
8 electrons
Maximum of
2 electrons
Electrons are distributed in the shells according to the formula 2n2 where n is the number of the
electron shell, with shell number one, being closest to the nucleus.
1.
The first shell can hold a maximum of 2 electrons.
2.
The second shell can hold a maximum of 8 electrons.
3.
The third shell can hold a maximum of 18 electrons.
4.
The fourth shell can hold a maximum of 32 electrons.
5.
Etc.
When determining the number of electrons around the nucleus of an atom we assume that the atom is of
neutral electromagnetic charge, so we will initially assume there are as many electrons as protons (same
as the atomic number).
A.
Examples: Hydrogen(H) has one electron, Helium(He) has two electrons, Sodium(Na) has eleven
electrons, Uranium(U)has ninety-two electrons, etc.
B.
Specific factors beyond the scope of this course, affect how electrons will be distributed around
the nucleus, so shells do not necessarily fill from an innermost to outermost order.
C.
Because of the factors cited above, electrons fill electron shells in the following order:
F.
VII.
Electrons
Electron shell
Number of electrons
1-2
1
2
3-10
2
8
11-18
3
8
19-20
4
2
21-30
3
10
31-36
4
6
We will not be concerned with electron distribution beyond this point.
1.
2.
Electrons 11-18 and 21-30 constitute “subshells” of shell number 3.
Electrons 19-20 and 31-36 constitue a “subshell” of shell number 4.
22
3.
These subshells must be full for electron(chemical) stability (this will be discussed later).
23
D.
Some examples of electron distributions are diagrammed below.
1.
Hydrogen(H)
Sodium(Na)
1e8e1e1P
2e0N
11 P
12 N
2.
3.
Calcium (Ca)
2e-
Magnesium(Mg)
8e8e2e-
2e8e2e-
20 P
20 N
12 P
12 N
Fluorine(F)
Chlorine(Cl)
7e2e-
7e8e2e-
9P
10 N
17 P
18 N
24
4.
Neon(Ne)
8e2e-
Argon(Ar)
8e8e2e-
10 P
10 N
18 P
18 N
E.
F.
VIII.
IX.
The diagrams above are known as Bohr diagrams and are useful, but not particularly realistic.
The atoms diagrammed above all have a neutral electromagnetic charge as the positive charge of
the protons is offset by the negative charge of the electrons.
In nature, atomic neutrality is not as important to electron stability as having a full outer shell or
subshell, i.e. a full outer shell (or subshell) means chemical stability.
A.
As a result, atoms will gain, lose, or share electrons to achieve electron (chemical) stability.
1.
An atom may gain electrons to fill an existing, partially filled shell.
2.
An atom may lose electrons, until that shell is empty, thereby exposing an underlying
shell that is full.
3.
Atoms may share electrons with another atom to fill electron shells.
B.
It is the gaining, losing, and sharing of electrons, in a way that leads to full outer electron shells,
that determines whether atoms can chemically react with one another.
C.
This outer shell, capable of gaining or losing electrons is called the valence shell.
1.
An atom is chemically stable if it has a full valence shell.
2.
An atom is chemically unstable, or chemically reactive if the valence shell is not full.
D.
If an atom gains or loses electrons there is an imbalance in the electromagnetic charges of the
atom, and the atom will have a net charge, as the number of protons and electrons will no longer
be equal.
E.
Atoms (and molecules) with an electromagnetic charge are called ions.
1.
If an atom loses electrons it will take on a positive electromagnetic charge(more protons
than electrons), and is called a cation.
2.
If an atom gains electrons it will take on a negative electromagnetic charge(more
electrons than protons), and is called an anion.
A reexamination of the atoms previously diagrammed should allow us to determine whether or not they
are likely to gain or lose electrons.
A.
Hydrogen is chemically reactive.
1.
Hydrogen has one electron in its only (valence)shell.
2.
It could gain an electron to fill its shell, or lose an electron emptying the (valence)shell.
3.
Typically Hydrogen will lose its only electron over gaining an electron (for reasons I will
not go into) and form a H+ cation.
B.
Sodium is chemically reactive.
1.
Sodium has one electron in the inner (valence) subshell of third shell.
2.
It could lose one electron to expose a full second shell, or gain seven electrons to fill the
inner subshell of shell number 3.
25
C.
D.
E.
F.
G.
H.
3.
It is intuitively “obvious” that it is easier to lose the one electron rather than add seven.
4.
Sodium typically loses one electron to form a Na+ cation.
Calcium is chemically reactive.
1.
Calcium has two electrons in a (valence)subshell of its fourth shell.
2.
It could lose two electrons to expose a full subshell of shell number 3, or gain many
electrons to fill the remaining subshell of shell number 3 and the remainder of the fourth
subshell(remember electrons 19-20 and 31-36 must be present to fill the subshell of shell
number 4).
3.
It is intuitively “obvious” that it is easier to lose the two electrons than gain the many
(sixteen).
4.
Calcium typically loses two electrons to form a Ca++ cation.
Magnesium is chemically reactive.
1.
Magnesium has two electrons in the inner (valence)subshell of third shell.
2.
It could lose the two electrons to expose a full shell number 2, or gain six electrons to fill
the third subshell.
3.
It is intuitively “obvious” that it is easier to lose the two electrons than add six.
4.
Magnesium typically loses two electron to form a Mg++ cation.
Fluorine is chemically reactive.
1.
Fluorine has seven electrons in the (valence)second shell.
2.
It could gain one electron to fill the shell, or it could lose seven to expose shell number 1.
3.
It is intuitively “obvious” that it is easier to gain one than lose seven.
4.
Fluorine typically gains one electron to form a F- anion.
Chlorine is chemically reactive.
1.
Chlorine has seven electrons in the inner (valence)subshell of shell number 3.
2.
It could gain one electron to fill the subshell, or lose seven to expose the full second shell.
3.
It is intuitively “obvious” that it is easier to gain one than lose seven.
4.
Chlorine typically loses one electron to form a Cl- anion.
Neon is chemically stable.
1.
Neon has eight electrons in the (valence)second shell.
2.
The second shell is full.
3.
Neon will typically neither gain nor lose electrons--it is already chemically stable(inert)
because its outer shell is full of electrons.
Argon is chemically stable.
1.
Argon has eight electrons in the inner (valence)subshell of the third shell.
2.
The subshell is full.
Argon will typically neither gain nor lose electrons-- it is already chemically stable(inert),
because its outer subshell is full of electrons.
26
27
Lesson 3: Basic Chemistry--Chemical Bonding and Reactions
I.
II.
Before discussing chemical bonds there are related topics I would like to consider first.
A.
A compound is formed when atoms of two or more elements are chemically bonded in definite
proportions.
B.
Compounds are expressed in chemical formulas composed of the atomic symbols and subscripts
that indicate how many of each type of atom is required to make the compound.
1.
NaCl is a compound Sodium Chloride, which is common table salt.
a)
Sodium Chloride is composed of 1 Sodium atom for each Chlorine atom.
b)
No subscripts are used if there is only one atom of a particular type required.
2.
C6H12O6 is the compound glucose.
a)
The subscripts 6,12, and 6 tell how many of each type of atom are required to
form the compound.
b)
Glucose is composed of six Carbon and Oxygen atoms for each twelve Hydrogen
atoms.
3.
CO2 is the compound Carbon Dioxide and is composed of two Oxygen atoms for each
Carbon atom.
4.
O2 is not considered a compound because it is composed of atoms from a single element,
Oxygen.
There are two types of chemical bonds, ionic bonds and covalent bonds.
A.
Ionic bonds result from electromagnetic attraction between cations and anions(holding them
together) that may result from a transfer(irreversible) of electrons from atom to atom.
28
1.
Sodium Chloride,NaCl, is an ionically bonded compound.
Before:
Transfer 1 of one
7e1eelectron
8e8e2e2e17 P
18 N
11 P
12 N
Na
Cl
Sodium has one electron in its valence shell, and as we have already
learned it will lose that electron easily to form a cation of Na+.
Chlorine, on the other hand, readily accepts an electron to fill its valence
subshell, to form the anion, Cl-. If Sodium and Chlorine are not already
in ionic form, Sodium will transfer one electron to Chlorine, forming
ions as shown below.
8eAfter:
8e8e2e2e17 P
18 N
11 P
12 N
+
-
The transfer of an electron converts the previously neutral atoms into ions.
The ions are held together by the electromagnetic attraction of opposing
charges--hence the term “ionic” bond. It is an association between ions of
opposite charge. If the Sodium and Chlorine were already in ionic form,
they likewise could associate to form the ionic bond. Note that the overall
charge is neutral, as the positive charge of the Sodium cation is neutralized
by the negative charge of the Chloride anion.
29
2.
Magnesium Fluoride,MgF2 , is another ionically bonded compound(composed of one
Magnesium atom for every two Fluorine atoms).
Before:
Transfer 1 electron
7e2e-
2e8e2e-
9P
10 N
12 P
12 N
F
Transfer one
electron
7e2e9P
10 N
F
Mg
Magnesium has two electrons in its valence shell, and as we have already learned
will lose those electrons easily to form a cation of Mg++. Fluorine, on the other
hand, readily accepts an electron to fill its valence subshell, to form the anion, F-. If
Magnesium and the Fluorines are not already in ionic form, Magnesium will transfer
one electron to each Fluorine, forming ions as shown below.
After:
8e2e-
8e2e-
12 P
12 N
9P
10 N
-
8e2e-
F
Mg
++
9P
10 N
F
-
The transfer of an electron from Magnesium to each Fluorine converts the
previously neutral atoms into ions. The ions are held together by the
electromagnetic attraction of opposing charges. Note that the overall charge is
neutral, as the (+2) positive charge of the Magnesium cation is neutralized by
the (-1) negative charge of the two Fluoride anions. -2 + 2 = 0.
3.
Characteristics of ionic bonds.
30
a)
B.
They tend to form between metals(and metalloids) and nonmetals.
(1)
On the periodic table you should see a dark line that separates the metals
and metalloids from nonmetals.
(2)
The metals and metalloids are to the left of the line, the nonmetals to the
right.
b)
They tend to form hard, brittle compounds with high melting points (all salts are
ionically bonded, for example).
c)
Many (but certainly not all) dissolve in water, dissociating to form free ions that
are no longer bonded.
d)
Compounds that dissolve to form ions in water are known as electrolytes.
Covalent bonds are associations between atoms in which electrons are shared.
1.
Sharing means that an electron from each atom orbits both atoms involved thereby
holding them together.
a)
This sharing of an electron pair constitutes a single covalent bond.
b)
Two atoms may share more than one pair of electrons forming double, triple, and
even quadruple bonds.
2.
Covalently bonded atoms are called molecules.
3.
Water(H2O)is an example of a covalently bonded molecule.
Before:
Electron of Oxygen
Electron of Hydrogen
2e8P
8N
1P
0N
1P
0N
O
H
H
Oxygen has six electrons in its outer shell, and needs two electrons to complete the
shell. Each Hydrogen has one electron in the only shell and needs one electron to
fill the shell. By Oxygen sharing one of its electrons with each Hydrogen, and each
Hydrogen sharing their only electron with Oxygen shells will be “full.” Oxygen
will have eight electrons orbiting shell two, although not necessarily all at the same
time, and each Hydrogen will have two electrons orbiting shell number 1, although
not necessarily at the same time. Notice that the molecule is neutral as positive and
negativeAfter:
charges cancel.
Electron of Oxygen
Electron of Hydrogen
2e8P
8N
1P
0N
O
1P
0N
H
H
We draw the after picture with the outer orbitals overlapping, with one
electron from each atom at the intersections of overlap. This represents one
31each Hydrogen, and each Hydrogen
electron from Oxygen will be shared with
will share one electron with each Oxygen. When drawn this way you can
count two electrons in the first orbital of each Hydrogen, and eight electrons in
4.
Please note: the the animation in the video lesson does not show the oxygen electrons
orbiting the Hydrogen nuclei, but it should--it is not as accurate as it could be.
32
5.
Methane (CH4) is another covalently bonded molecule.
Before:
Electron of Carbon
Electron of Hydrogen
1P
0N
H
2e6P
6N
1P
0N
1P
0N
H
H
C
1P
0N
H
After:
H
1P
0N
2e-
H
6P
6N
1P
0N
C
1P
0N
H
1P
0N
H
The after picture again shows the outer orbitals overlapping, with one
electron from each atom at the intersections of overlap. This represents
one electron from Carbon will be shared with each Hydrogen, and each
Hydrogen will share one electron with each Carbon. When drawn this
way you can count two electrons in the first orbital of each Hydrogen, and
eight electrons in the second orbital of Carbon. In this example Carbon
forms four covalent bonds, one with each Hydrogen, and each Hydrogen
forms one covalent bond with the Carbon atom.
33
6.
Please note: the the animation in the video lesson does not show the carbon electrons
orbiting the Hydrogen nuclei, but it should--it is not as accurate as it could be.
34
7.
Atmospheric Oxygen (O2 ) is yet another covalently bonded molecule (but it is not a
compound).
Before:
Electron of Oxygen
2e-
2e-
8P
8N
8P
8N
O
O
After:
2e-
2e-
8P
8N
8P
8N
To fill their second shells each Oxygen atom must share two
electrons with the other atom. This is a double (covalent) bond.
Two electrons from each Oxygen will not only orbit their own
nucleus, but the other Oxygen nucleus as well.
8.
Characteristics of covalent bonds.
a)
Covalent bonds tend to occur between nonmetal elements that are not strongly
prone to ionization.
b)
In future diagrams a line will represent a covalent bond (a sharing of a pair of
electrons).
O
H
O
35
H
(H2O)
O
(O2)
c)
d)
e)
f)
Covalent molecules can be quite large, composed of thousands of atoms (proteins
for example).
Not all covalently bonded molecules are compounds--e.g. O2, N2, H2 etc.
Covalent molecules can be small, and when soluble in water the atoms stay
bonded together.
(1)
When the ionically bonded table salt, NaCl, dissolves in water, the Na+
cations separate from the Cl- anions.
(2)
When the covalently bonded glucose sugar, C6H12O6 , dissolves in water it
dissolves as intact sugars, the Carbons, Hydrogens, and Oxygens stay
bonded together as C6H12O6 .
The sharing of electrons in a covalent bond may not be an equal sharing,
depending on the elements involved.
(1)
The unequal sharing of electrons produces a polar molecule.
(a)
One atom may hold the shared electrons tightly, and more
frequently than a covalently bonded partner, thus at any given time,
the electrons are not distributed evenly.
(b)
This produces a molecule where electromagnetic charge is
unequally distributed, producing one(or more) region(s) that is
slightly negative in charge, another(or more) region(s) that are
electropositive, even though the overall molecule is neutral.
(c)
This distribution of charge creating positive and negative “poles” is
like the distribution of magnetic charge in a bar magnet that has
positive and negative “poles,” hence the term polar molecule.
(d)
Water is an example of a polar molecule--the Oxygen tends to hold
electrons more strongly than the Hydrogens, so the Oxygen end of
the molecule is electronegative and the Hydrogens are slightly
electropositive.
O
H
+
(e)
(2)
H
+
Water molecule
+
Magnet
Common polar covalent bonds typically involve Oxygen, and
include C-O, H-O, and P-O.
(f)
Polar molecules tend to be soluble in water, and insoluble in oils,
the following terms are synonyms for water soluble: polar, fat
insoluble, hydrophilic (water loving), lipophobic (afraid of fat).
The equal sharing of electrons in which there are no positive or negative
regions produces a nonpolar molecule.
(a)
Methane is an example of nonpolar molecule.
(b)
The C-H covalent bond is a common nonpolar combination.
36
(c)
III.
IV.
Nonpolar molecules tend to be insoluble in water, and soluble in
oils, the following terms are synonyms for water insoluble:
nonpolar, fat soluble, hydrophobic (afraid of water), lipophilic (fat
loving).
(3)
Most of the compounds of which we and other living things are composed,
are covalently bonded.
Another factor that influences bonding is the electronegativity of the atoms involved.
A.
Electronegativity is a measure of the attraction an atom has for electrons--the greater the number
the more strongly it attracts, or holds, electrons.
B.
If the difference in electronegativity between two atoms is greater than 1.7, then an ionic bond is
possible between compatible atoms.
1.
NaCl is an ionic compound.
2.
Sodium has an electronegativity of 0.9, Chlorine has an electronegativity of 3.5 yielding a
difference of 2.6.
C.
If the difference in electronegativity between two atoms is between 0.5 and 1.7, then a polar
covalent bond is possible between compatible atoms.
1.
O-H bonds are polar covalent.
2.
Oxygen has an electronegativity of 3.5, Hydrogen has an electronegativity of 2.1 yielding
a difference of 1.4.
3.
The polar covalent bond can be visualized as a “semi-ionic” bond.
a)
One atom “steals” a shared electron for a time, but because the difference in
electronegativites is not great enough, the other atom pulls its electron back, until
the more electronegative atom takes it away again.
b)
The electron spends more time in the shells of the more electronegative atom
giving that atom (or that part of the molecule) a negative charge, and the atom
losing the electron a positive charge.
D.
If the difference in electronegativity between two atoms is less than 0.5, then a nonpolar covalent
bond is possible between compatible atoms.
1.
C-H bonds are nonpolar covalent.
2.
Carbon has an electronegativity of 2.5, Hydrogen has an electronegativity of 2.1 yielding
a difference of 0.4.
E.
In conclusion, two or more atoms may form ionic or covalent bonds (depending on their
electronegativities), if their valence shells are compatible.
Chemical reactions are chemical rearrangements, or changes in chemical bonds between atoms.
A.
Chemical reactions are represented by chemical equations.
B.
Chemical equations contain the following components.
1.
Chemical formulas represent compounds, atoms, and molecules, e.g. MgCl2.
2.
Coefficients are numbers in front of chemical formulas which state how many “moles” of
each kind of molecule or compound there are, e.g. 6CO2.
a)
The number, six, the coeffiecient, indicates there are six moles of carbon dioxide.
(1)
A mole is a standard unit of chemistry that represents a specific number of
molecules (6.02 x 1023 particles).
(2)
Just as the word “dozen” represents a specific number of items (12) so too
does the word “mole,” it is just a much larger number (6.02 x 1023).
b)
The coefficients in chemical reactions actually represent the number of moles of
atoms or molecules, rather than the actual number of atoms or molecules, e.g. if
the formula 6CO2 was part of a chemical equation it means six moles of carbon
dioxide rather than six molecules of carbon dioxide.
3.
Reactants are atoms and molecules that will be involved in the reaction.
37
C.
4.
A yield sign(arrow) indicates a chemical reaction has occurred.
5.
Products are the atoms and molecules generated(produced) by the chemical reaction.
Consider the following chemical equation representing a well known chemical reaction.
Reactants
Yield sign
C6H12O6 + 6O2
6H2O
Glucose
1.
2.
3.
Products
6CO2 +
Oxygen
Carbon dioxide
Coefficient(indicates
there
Water
are six water molecules)
In this equation there are two reactants, glucose and oxygen.
a)
The reaction requires one molecule (or moles) of glucose for every six molecules
or moles of the molecule, O2.
b)
The equation is like a recipe in this way, in that it gives the proportions of
reactants needed for the reaction to proceed.
The atoms of the reactants rearrange to form two products, carbon dioxide and water.
a)
The reaction will yield six molecules (or moles) carbon dioxide and six molecules
(or moles) of water.
b)
The chemical equation also demonstrates the relative quantities of the products.
Notice that the number of atoms of each element on the reactant side of the equation
equals the number of atoms of that element on the product side of the equation.
a)
Element
Number of reactant atoms
Number of product atoms
Carbon
6
6
Hydrogen
12
12
Oxygen
18
18
b)
D.
In chemical reactions atoms are neither created nor destroyed
(1)
This is consistent with Law of Conservation of Matter which states that
matter is neither created nor destroyed (in chemical reactions).
(2)
In chemical reactions atoms rearrange themselves, but maintain their
physical integrity, they do not “disappear” nor do they “appear” from
nothingness.
Some chemical reactions are reversible, and are represented by a yield sign that has one arrow in
each direction.
HCO3- + H+
H2CO3
E.
V.
There are a number of specific chemical reactions in nature, as well as general categories or
“types” of chemical reactions--these will be addressed, when relevant, throughout the course.
All elements in a vertical column in the periodic table are in the same chemical family.
A.
Members of a chemical family have similar chemical bonding characteristics.
B.
Members of a chemical family have similar chemical bonding characteristics because they have
the same number of electrons in their outer shell, called the valence shell.
C.
Refer back to the electron distributions of the elements diagrammed previously.
38
1.
D.
E.
F.
Hydrogen and Sodium each have one electron in their valence shell, and are in the same
chemical family.
2.
Calcium and Magnesium each have two electrons in their valence shell and are in the
same chemical family.
3.
Chlorine and Fluorine each have seven electrons in their valence shell and are in the same
chemical family.
4.
Neon and Argon each have eight electrons in their valence shell and are in the same
chemical family.
By comparing the valence shells of different families, one can determine if family members are
compatible for forming bonds with one another.
1.
H, Li, Na, K, etc each have one electron in their valence shell which they are likely to
lose, while F, Cl, Br, etc each have seven electrons in their valence shell, and need to gain
one electron to be chemically stable.
2.
Any member of the H, Li, Na family will form an ionic bond with any member of the F,
Cl, Br family.
The term valence describes the typical ion an atom will form, based on its valence shell.
1.
H, Li, Na, K all have a valence of +1.
2.
F, Cl, Br, etc have a valence of -1.
By combining valences to generate a compound of neutral charge, one can predict compatible
combinations.
1.
One H(+1) and one F(-1) may form HF(+1 -1 = 0).
2.
Two H(+1 each) and one O(-2) may form H2O (+1+1-2 = 0).
3.
One Ca(+2) and two Cl(-1 each) may form CaCl2(+2-1-1 =0).
39
40
Lesson 4: Biochemistry--Important Inorganics
I.
II.
Biochemistry is the study of the atoms and molecules of living things, as well as their interactions and
chemical reactions.
Biochemistry involves a variety of important molecules that can be divided into two major categories.
A.
Organic molecules are typically defined as having at least one C-H bond, and/or contain a chain
of carbon atoms (at least two carbons bonded together).
1.
Examples.
a)
CH4, C6H12O6, C2H5OH.
b)
Carbohydrates, proteins, lipids, and nucleic acids are all organic molecules.
2.
Some general characteristics of organic molecules are as follows.
a)
Organic molecules are typically (but not always) associated with living things
(organisms).
b)
Organic molecules contain the element carbon.
(1)
Carbon is an interesting element in that it has four electrons in its valence
shell, which means that it forms four (typically covalent) bonds.
(2)
Carbon readily forms double (more rarely triple or quadruple) bonds, and
also readily bonds with other carbons to form a “carbon chain,” including
rings.
c)
Because of carbon’s amazing traits, there is more molecular diversity among
organic molecules than inorganic molecules.
d)
Organic molecules tend to be larger and more complex than inorganic molecules.
e)
Most of the important organic biomolecules are composed of some basic unit
called a monomer(mono=one, mer=thing or part).
(1)
Think of a monomer as being like a single lego piece.
(2)
Legos can be put together to create structures of incredible diversity and
size.
(3)
In this way organic monomers are similar, in that they can be pieced
together to create incredibly complex molecules.
(4)
Complex organic molecules produced from organic monomers are called
polymers(poly=many, mer=thing or part).
(a)
Homopolymers are composed of identical monomers.
(i)
If you had blue lego pieces of a specific size and shape, you
could put them together to create a structure that would be
analogous to a homopolymer.
(ii)
You would not, however, have to assemble them the same
way each time, you could come up with many different
shapes and sizes--the same is also true in nature, different
homopolymers could be composed of the same monomer
but have different properties.
(iii)
Two examples from nature are starch, used as an energy
source, and cellulose, used as a structural component in
plant cell walls--both are homopolymers of the sugar
glucose, but they are constructed differently.
(b)
Heteropolymers are composed of monomers that are different from
one another.
(i)
Returning to our lego analogy, if you had lego pieces of
different colors, or even different shapes and sizes, and
41
f)
constructed something, this is the nature of a
heteropolymer.
(ii)
Proteins are classic heteropolymers in that they are
composed of amino acid monomers.
(a)
There are twenty different amino acids in nature.
(b)
At each position in a protein, which may be
constructed of hundreds of amino acids, any one of
the twenty amino acids could be used.
Monomers are joined together by means of a dehydration (condensation) reaction.
(1)
A dehydration reaction occurs when a Hydrogen from one monomer bonds
with an OH from another monomer forming a molecule of water, causing
the monomers to bond together.
(2)
Dehydration(Condensation) reaction.
Before:
H
Monomer
A
OH
H
OH
H2O
After:
Monomer
A
H
Monomer B
Monomer B
OH (Polymer)
+
H2O
(3)
g)
More dehydration reactions could add monomers to one end or the other,
extending the length of the polymer.
In the reverse reaction, polymers are split into monomers by a hydrolysis reaction.
(1)
An hydrolysis reaction occurs when a molecule of water is split
(hydro=water, lysis= split) into H and OH, each of which begins to share
electrons with adjacent monomers, breaking the chemical bond between
the monomers.
(2)
Hydrolysis reaction.
Before:
H
Monomer
A
Monomer B
OH (Polymer)
H2O
After:
H
Monomer
A
42
OH H
Monomer B
OH
h)
III.
Organic molecules have a “backbone” of carbon to which are attached so called
“functional groups” that give a particular molecule its chemical characteristics.
(1)
These groups include:
(a)
Methyl groups (CH2 or CH3)
(b)
Amino groups (NH2),
(c)
Carboxyl groups (also called organic acids (COOH)),
Please note: the carboxyl group has an error in the functional group
chart of the video.
(d)
Alcohol groups ((OH) called hydroxides in inorganic molecules),
(e)
Phosphate groups (PO4), and
(f)
Others.
(2)
All of the groups named above are polar, except for methyl groups, which
are nonpolar.
i)
Many organic molecules are isomers (iso=same, mer=thing or part).
(1)
Isomers are organic molecules that have the same chemical formula, but
different structures (be sure not to confuse isomers with isotopes).
(2)
Because of carbon’s ability to form several bonds, many molecular
structures can be constructed from the same atoms.
(3)
The sugars glucose, fructose, and galactose all have the same chemical
formula (C6H12O6) but different physical structures, and are isomers.
B.
Inorganic molecules lack a C-H bond, or lack carbon entirely.
1.
Examples: CO2, NH3, CaCl2.
2.
Water is an extremely important inorganic molecule.
Even though water is an inorganic molecule, it is required for life.
A.
There is no particular trait unique to water, but it has a combination of traits that make it unique
in nature, and it is required for life here on earth.
B.
Water makes up about 80% of the mass of most organisms, with organics making up about 19%,
and electrolytes less than 1%.
C.
Many of water’s characteristics are due to the fact that water is a polar molecule.
1.
The hydrogen-oxygen bond is a polar covalent bond.
2.
The oxygen has a higher electronegativity and tends to “hoard” shared electrons making
the oxygen slightly negative, and the hydrogens slightly positive.
3.
This leads to weak electromagnetic interactions between adjacent water molecules, as the
slightly positive hydrogens of one water molecule are attracted to slightly negative
oxygen atoms of a different molecule.
4.
These weak electromagnetic attractions involving hydrogen are called “Hydrogen bonds.”
a)
Hydrogen bonds are actually weak covalent bonds between atoms of adjacent
water molecules.
b)
The attraction is between atoms of different molecules, not atoms of the same
molecule.
5.
This polar nature of water molecules causes hydrogen bonds between adjacent molecules,
causing them to cling slightly to one another, as diagrammed below.
+H
+H
H+
O
+H
H+
O
O
Etc.
+H
H+
H+
O
43
D.
IV.
The major characteristics of water are as follows.
1.
Water is cohesive.
a)
Cohesion is defined as like substances sticking together.
b)
Water molecules cling to themselves.
c)
The cohesive nature of water leads to some of the other traits listed below.
2.
Water is adhesive.
a)
Adhesion is the sticking together of unlike substances.
b)
Water can stick a wet paper towel to a wall, because water will hydrogen bond
with the wall, hydrogen bond with itself, and hydrogen bond with the paper towel,
sticking the paper towel to the wall.
3.
Water exhibits capillary action, the tendency of a liquid to move up a tube (a narrow tube
is called a capillary).
a)
The narrower the tube , the higher the water climbs.
b)
Capillary action results from a combination of adhesion and cohesion.
c)
Water moves up plant stems, in part, from capillary action.
4.
Water has a high surface tension.
a)
Water will support objects, and form droplets.
b)
This trait is due to water’s cohesive nature.
5.
Water has a high specific heat.
a)
Water absorbs relatively large amounts of heat energy without changing
temperature.
b)
Heat is a measure of molecular or atomic movement, and cohesion limits
movement of water molecules, initially.
c)
Water will likewise retain heat energy once it is warm because once moving, the
water molecules tend to keep moving, so water cools less rapidly than other
substances.
d)
Coastal temperatures less variable because ocean heats and cools, slowly.
6.
Water has a relatively high heat of vaporization(boiling point).
7.
Water a good polar solvent.
a)
Solids (solutes) may dissolve in liquids (solvents) to create a solution.
b)
Water is the solvent of life, forming aqueous solutions.
c)
Many polar covalent molecules and ionic compounds dissolve in water.
8.
The crystalline solid form of water (ice) is less dense than the liquid (water is most dense
at 4 degrees Celsius).
a)
As ice forms, it traps air, making it float, and insulating the water underneath from
the colder temperatures.
b)
Large bodies of water will not freeze completely, even in very cold environments,
due to the insulating nature of ice, permitting the aquatic organisms to survive the
cold weather.
There are many other important inorganic substances.
A.
Acids are substances that when dissolved in water yield H+ (hydrogen ions).
1.
Acidic solutions are high in H+ and low in 0H-(hydroxide ions).
2.
Hydrogen ions are very reactive, and as a result, acids are very caustic or reactive.
3.
Example: Hydrochloric acid........HCl ----> H+ + Cl-.
4.
It is the abundance of H+ in solution that makes HCl a strong acid.
44
B.
C.
D.
E.
F.
A base(alkali) is a compound, that when placed in water, yields OH- (hydroxide ions), or
decreases the concentration of H+.
1.
Basic (alkaline) solutions are low in H+ (and high in OH-).
2.
Hydroxide ions are also very reactive, and bases are likewise very caustic or reactive.
3.
Example, Sodium hydroxide.........NaOH ----> Na+ + OH-.
4.
It is the abundance of OH- in solution that makes NaOH a strong base.
When combined, acids and bases neutralize one another.
1.
H+ is what makes an acid an acid.
2.
OH- is what makes a base a base.
3.
When H+ , the reactive component of an acid, is combined with OH- , the reactive
component of a base, they form HOH or H2O(water), a stable and harmless molecule.
4.
A salt(by definition) is formed as a result of a chemical reaction between an acid and a
base.
a)
Acid + Base ----> A Salt + Water
b)
A salt does not affect pH(pH will be discussed below).
c)
Example of a neutralization reaction: HCl + NaOH ---> NaCl (a salt) + HOH
(water).
pH is a numerical scale, from 0 to 14, that represents a measure of H+ concentration in a solution.
1.
It is a negative log scale, therefore, the lower the number, the greater the concentration of
H+ (and lower the OH- concentration) and the more acidic the solution.
2.
Acids have a pH of less than 7.
a)
The H+ concentration is greater than the OH- concentration, in acidic solutions.
b)
The closer to zero the pH value, the stronger the acid, i.e. the greater the H+
concentration and lower the hydroxide ion concentration.
3.
Bases have a pH greater than 7 (to 14).
a)
The H+ concentration is less than the OH- concentration, in basic (alkaline)
solutions.
b)
The closer to 14 the pH value, the stronger the base, i.e. the lower the H+
concentration and higher the hydroxide ion concentration.
4.
A solution with a pH of 7 is a neutral solution.
a)
A neutral solution has equal parts H+ and OH- , which combine to form water (H+
+ OH- ----> HOH).
b)
A neutral solution is neither acidic nor basic.
5.
The pH scale is based on the power of ten, so an acidic solution with a pH of 3 is ten
times stronger than an acidic solution of pH4.
pH is extremely important in both the external and internal environments of organisms.
1.
As mentioned, hydrogen and hydroxide ions will react with a variety of compounds,
including the cellular components of organisms.
2.
Organisms and their cells are adapted to specific environmental pH’s both externally and
internally.
3.
Changes in what is an optimal pH for an organism can lead to changes in the
biochemistry of that organism, with fatal results.
4.
The human internal pH is about 7.3, either an acidosis or alkalosis of our blood can be
fatal.
A buffer is a compound that acts to neutralize the effects of either an acid or a base, to maintain a
particular pH.
1.
It is usually difficult for an organism to regulate its external environmental pH, but
organisms utilize buffers to maintain their internal pH.
45
2.
3.
A typical buffer has the capacity to bind or release H+ to maintain pH when acid or base
is added to a solution.
Example: Carbonic acid.
H2CO3
a)
b)
c)
H+ + HCO3-
This equation represents a reversible reaction, as you can see.
If OH- is added to the solution, it will bind with H+ to form water.
(1)
This counteracts the effects of the OH-, but will decrease the concentration
of H+.
(2)
As H+ is removed from the solution in the formation of water molecules, it
causes the intact acid H2CO3 to dissociate releasing more H+ (the reaction
is driven to the right, as H+ is removed from the solution).
(3)
This keeps the solution at its original pH.
If H+ is added to the solution, it combines with free HCO3-, to form intact
carbonic acid (H2CO3).
(1)
Since H+ is bound as it is added to the solution, the pH does not change.
(2)
The addition of H+ drives the reaction to the left.
46
47
Lesson 5: Biochemistry--Carbohydrates and Protein Structure
I.
The first major class of biologically important organic molecules is the Carbohydrates.
A.
Carbohydrates(carbo=carbon, hydrate=water) are composed primarily of carbon, hydrogen, and
oxygen in a 1:2:1 ratio (CnH2nOn), or something very close to that ratio.
B.
Sugars, starch, and cellulose are examples of carbohydrates.
C.
Carbohydrates have several important functions.
1.
Carbohydrates serve as the primary source of energy in nature.
a)
The potential energy in carbon-carbon bonds can be exploited to do cellular work.
b)
Glucose, in particular, is a major source of cellular energy.
2.
Some carbohydrates are structural, e.g. cellulose, which is a major component of plant
cell walls.
3.
Some carbohydrates are involved in immune system recognition.
D.
The basic unit (monomer) of carbohydrates is the monosaccharide (mono=single,
saccharide=sugar), of which there are several types.
1.
Glucose is an extremely common monosaccharide found in virtually all organisms-animal, plant, fungi, protists, and bacteria.
a)
Glucose is the primary energy source in nature.
b)
Glucose, and many other organic molecules are often described with a structural
formula, rather than a chemical formula.
(1)
Several monosaccharides have the same chemical formula, but differ
structurally (isomers).
(2)
Structural formulas show the differences in their physical structure--the
structural formula for glucose is listed below.
CH2OH
c)
The structural formula listed above is a “shorthand” version of the structure of
glucose whose chemical formula is C6H12O6.
(1)
Shorthand formulas do not show all of the atoms of a molecule, as
repeatedly drawing these complex molecules is a tedious process.
(2)
Certain “conventions” of carbohydrate shorthand models are described
below.
(a)
A carbon atom exists at the intersection of two or more lines.
(b)
An alcohol group (OH), exists at the end of any line that extends
away from an intersection of two or more lines.
(c)
If a carbon shows less than four covalent bonds, in the shorthand
structural formula, then the bonds not shown are hydrogen atoms
bound to the carbon.
48
(3)
The complete structural formula is given below, and you should now be
able to find the six carbons, twelve hydrogens, and six oxygens described
by the chemical formula, C6H12O6.
CH2OH
H
C
C OH
HO
H
H
H C
C
C
OH
OH
H
(4)
2.
As you can see the complete structural formula is quite cluttered, hence the
need for a simplified version.
(5)
You also be aware that the structural formula is representing a 3
dimensional molecule on a two dimensional surface, so it is not a perfect
representation by any stretch of the imagination.
Fructose is another common monosaccharide that is an important source of cellular
energy.
a)
Fructose is sometimes called “fruit sugar” but it is found in honey, and in virtually
all plant and animal tissues.
b)
Fructose is an isomer with the same chemical formula as glucose, C6H12O6, but it
has a different structural formula.
HOH2C
49
O
CH2OH
c)
3.
This is the shorthand formula, so be sure you can determine where the six
carbons, twelve hydrogens, and six oxygens are located based on the discussion of
glucose.
Ribose and Deoxyribose are structural monosaccharides.
a)
They are components of the genetic materials, RNA and DNA, respectively.
b)
Structural formulas are listed below.
Ribose
O
HOH2C
HOH2C
Deoxyribose
O
c)
E.
Be sure you can complete the shorthand formulas, above, and determine their
chemical formulas from their structural formulas.
Two monosaccharides bonded together via a dehydration reaction form a disaccharide (di=two,
saccharide=sugar), of which there are several in nature.
1.
Maltose is composed of two glucose molecules.
a)
Before (two glucose)
CH2OH
CH2OH
Glucose
Glucose
OH
HO
H2O
b)
After(maltose)
CH2OH
CH2OH
Glucose
Glucose
O
50
+ H2 O
2.
II.
Sucrose.
a)
Sucrose is composed of one glucose and one fructose.
b)
Extremely common sugar, in fruits, vegetables, etc.
c)
Sucrose is our common table sugar.
3.
Lactose.
a)
Lactose is composed of glucose and a monosaccharide called galactose (also
C6H12O6).
b)
Lactose is found in dairy products.
c)
Many individuals cannot properly digest lactose into its monosaccharides leading
to “lactose intolerance” (intestinal discomfort).
F.
Several monosaccharides linked together (via dehydration reactions) form a polysaccharide
(poly=many, saccharide=sugar), of which there are many types.
1.
Starch.
a)
Starch is composed of repeating glucose sugars.
b)
Starch is produced by plants, and is how they store energy for a “rainy day,” as
starch can be hydrolyzed back into glucose, and glucose converted to cellular
energy.
2.
Glycogen.
a)
Glycogen is the animal (mammalian) equivalent of starch, composed of repeating
glucose sugars.
b)
Glycogen has more branching linkages between sugars, but like starch, is
compose of repeating glucose sugars.
c)
Humans store glycogen primarily in muscle tissue and in the liver, for quick
access to glucose for cellular energy.
3.
Cellulose.
a)
Cellulose is still another polysaccharide composed of repeating glucose subunits.
b)
Cellulose is a structural carbohydrate in plant cell walls, and is not produced by
animals.
c)
Humans cannot digest cellulose into glucose sugars, so it has no caloric value, and
is part of the “roughage” in our diets.
4.
Starch, glycogen, and cellulose are classic homopolymers of glucose.
Proteins are another important organic polymer.
A.
Proteins are heteropolymers composed primarily of the elements carbon, hydrogen, oxygen,
nitrogen and sulfur (CHONS).
B.
The basic unit or monomer of proteins is the amino acid, and proteins are heteropolymers of
amino acids bonded together via dehydration reactions.
C.
Basic amino acid structure is indicated below.
H
Amino
group
N
H
Carboxyl (organic acid)
group
H
O
C
C O H
R
Variable (radical) group
51
1.
2.
D.
NH2 is an amino group.
COOH is an organic acid (carboxyl group) which dissociates in water COOH ---> COO+ H+, yielding hydrogen ions(which makes it an acid).
3.
Hence the name amino acid.
4.
Notice that all amino acids have a carbon to which is bonded an amino group, a carboxyl
group and a Hydrogen atom.
5.
All twenty amino acids differ at the fourth bond to the carbon, represented by the letter R,
which stands for radical or variable group--each of the twenty amino acids will have a
different atom or group at that position.
Dehydration reactions bond amino acids together.
1.
Amino acids are also called peptides.
2.
The bond between amino acids (peptides) is called a peptide bond.
3.
A peptide bond forms between the carboxyl group of one amino acid, and the amino
group of another amino acid.
Before:
HO
2
H
N
H
H
H
O
C
C O H
N
H
R
H
O
C
C O H
R
After:
H
N-terminal end
H
H
N
C
O Peptide
bond
C
R
H
O
N
C
C O H
C-terminal end
R
4.
E.
H
The terms dipeptide(di=two), tripeptide(tri=three), and polypeptide(poly=many) refer to
length of the amino acid chain.
5.
Long chains of amino acids ( 50 or more) are considered proteins.
6.
No matter how long the peptide chain, one end will always have an amino group exposed
(and is called the N-terminal end), and the other will have a carboxyl group exposed (and
is called the C-terminal end).
7.
Knowing the N-terminal and C-terminal ends can be important when determining the
amino acid sequence of a protein or building a protein of known amino acid sequence.
Proteins are describes as being either complete or incomplete.
1.
There are 20 amino acids encoded by DNA.
2.
For humans, ten of the amino acids are essential amino acids, meaning they must be
ingested because the body cannot synthesize them.
52
3.
4.
F.
The other ten are non essential amino acids, that can be synthesized by the body.
Some proteins are considered “complete.”
a)
Complete proteins contain the ten essential amino acids.
b)
If a protein has the ten essential amino acids, it functionally has all twenty amino
acids because cells can make the other ten.
c)
Animal proteins (meat, dairy products, eggs, etc.) are complete proteins.
5.
Some proteins are considered “incomplete.”
a)
Incomplete proteins lack at least one of the ten essential amino acids.
b)
Plant proteins are typically incomplete proteins.
(1)
Some plants have been genetically engineered to make complete proteins.
(2)
None of the major crop plants have complete proteins.
c)
Lacto-ovo vegetarians can get complete proteins from eggs and dairy products.
d)
Strict vegetarians eat no animal products, and must combine “complementary”
plant proteins in their diet to ensure ingestion of the ten essential amino acids.
(1)
For example, one plant product with only six of the ten essential amino
acids, can be combined with a plant product that has the other four
essential amino acids to provide all ten essential amino acids.
(2)
A well known complement, is to combine grains(wheat, corn, etc.) with
legumes(beans, peas, lentils, etc.) to provide the ten essential amino acids
in the diet.
The structure of a protein determines its function.
1.
Primary structure refers to the sequence of amino acids(beginning from the N or C
terminal ends), not to a three dimensional structure, per se.
a)
The sequence of amino acids determines the three dimensional shape of the
protein, and ultimately, its function.
(1)
Changes in primary structure can severely alter function.
(2)
One form of diabetes is due to a single amino acid mutation in a
membrane protein.
(3)
Sickle cell anemia is caused by the protein sickle cell hemoglobin, which
differs from normal hemoglobin by a single amino acid substitution in a
chain of 300 amino acids.
b)
Interactions between the variable groups will shape the protein.
(1)
The variable groups of amino acids may be small, large, polar or nonpolar.
(2)
Interactions between variable groups include hydrogen bonds,
electromagnetic attractions and repulsions, nonpolar aggregations, and
covalent bonds.
2.
Primary structure of a protein will determine the secondary structure of a protein--there
are two secondary structures found in proteins.
a)
Some regions or sometimes entire proteins(e.g. collagen) will form an alpha helix
(a fancy name for a coil or spiral shape).
b)
Other regions or sometimes entire proteins (e.g. silk) will form beta pleated sheets
(like a sheet folded many times)
3.
Some proteins continue to fold up on themselves to form globular, 3-D proteins called
tertiary proteins, or proteins with tertiary structure.(e.g. antibodies).
53
4.
G.
H.
Sometimes two or more secondary or tertiary proteins may bond to form a massive
quaternary structure or protein (e.g. hemoglobin, which is composed of four tertiary
proteins).
Please note: in the video I inadvertently state that hemoglobin is composed of three
subunits, but is composed of four subunits as stated above in the outline.
5.
While it is true that amino acid sequence determines structure, many proteins also need
the help of “chaperone proteins” that help a protein fold into the appropriate shape, so it
is possible for two proteins to have the same sequence of amino acids, yet different three
dimensional shapes.
The amino acid sequence of proteins is determined by the genetic material, DNA, but we will
talk about this in great detail in future lessons.
Proteins are sensitive to environmental conditions.
1.
Proteins are particularly sensitive to pH and temperature.
2.
Changes in pH or temperature denatures (inactivates) a protein by changing the protein’s
shape.
54
Lesson 6: Biochemistry--Protein Function and Lipids
I.
Proteins have a variety of functions.
A.
Some proteins are enzymes.
1.
An enzyme is a protein catalyst.
2.
A catalyst speeds up the rate of a chemical reaction without being consumed in the
reaction.
3.
For chemical reactions to occur, they typically require an input of energy.
a)
The initial energy required to start a chemical reaction is called the energy of
activation.
b)
Many reactions once started, release enough energy to keep the reaction going.
4.
Enzymes function by lowering the energy of activation required to make a reaction occur.
a)
The energy of activation can be supplied by kinetic energy (the energy of
movement) of random collision, but the force and angle of collision are crucial,
and may rarely happen spontaneously.
b)
Enzymes function to lower the energy of activation in one or more of the
following ways.
(1)
By placing reactants in the optimal position for bond formation.
(2)
By creating a physical stress on bonds for rearrangements or breakage of
bonds.
(3)
By introducing chemical energy from an outside source (typically the
hydrolysis of a molecule called ATP).
c)
An example of a chemical reaction that can involve an enzyme is the hydrolysis
(break up) of a molecule called urea, into ammonia and carbon dioxide.
(1)
Urease is an enzyme that will hydrolyze 30,000 urea molecules to
ammonia and carbon dioxide each second, at 220 Celsius, and a pH of 8.
(2)
At the same conditions, without urease, it would take 3 million years for
those 30,000 urea molecules to hydrolyze spontaneously.
5.
Virtually all of the chemical reactions needed for life are enzyme mediated (accelerated
by enzymes), and will not occur fast enough spontaneously to keep us alive.
6.
Properly functioning enzymes are required to keep us alive.
a)
We expend a great deal of energy to maintain a uniform internal environment,
such as constant pH, temperature, nutrient dispersal, etc. so that enzymes will
function properly.
b)
Many enzymes need “cellular energy” to function which is why we need to take in
food every day--we need the energy in the food to run our chemical reactions.
7.
Although enzymes carry out a number of tasks, we will consider enzymatic function to
consist of three basic operations.
a)
Some enzymes break larger molecules into smaller molecules (catabolism).
b)
Some enzymes bond two or more smaller molecules into larger
molecules(anabolism).
c)
Some enzymes rearrange the atoms in a molecule, forming another molecule.
8.
The “induced fit” model (theory) describes how some enzymes work.
a)
Enzymes are large polar molecules with a particular shape determined by the
variable group (R group) interactions of its amino acids.
b)
The enzyme shape is crucial to its function, particularly a part of the enzyme
called the active site.
55
c)
II.
The active site binds one or more molecules that will chemically react, known as
the substrate molecules.
d)
The substrate molecules and the active site must have complementary shapes for
binding to occur, this makes enzymes extremely specific in what they will bind.
e)
The enzyme and substrate molecules will form an enzyme-substrate complex.
(1)
Binding the substrate molecules causes a redistribution of charge
(polarity), and changes the interactions between variable groups(R groups)
of amino acids.
(2)
The change in shape causes the substrate molecules to chemically react.
f)
The products will not have the proper shape to stay in the active site and will be
released.
g)
Once the product is released, the enzyme goes back to its original shape, and can
bind more substrate molecules.
h)
So while substrate is converted to product, the enzyme is unchanged and will
continue to carry out the reaction.
9.
An alternate theory, the “lock and key” model of enzyme function, describes a slightly
different way in which some enzymes function.
a)
In the lock and key model, substrates bind to complementary active sites, but do
not induce any shape change in the enzyme.
b)
The substrate molecules will be in an optimum position to chemically react when
bound to the enzyme, and they react with one another to form a product or
products.
c)
The enzyme merely holds the substrate molecules in position for bonding.
d)
Once the product(s) is released the enzyme can be reused.
10.
Because enzymes are very specific in what they do, several enzymes typically work
together in metabolic pathways to accomplish “big jobs.”
11.
Enzymes required for a metabolic pathway are often encapsulated in a cellular organelle
(like a mitochondrion), or embedded in membranes.
B.
Some proteins are transport molecules that move specific molecules across cell membranes.
C.
Some are "carriers" in cytoplasm, interstitial fluid, lymph, or blood, binding to molecules and
delivering them to specific organs or tissues.
D.
Proteins play a role in communication.
1.
Some proteins (and amino acids) act as neurotransmitters.
2.
Some proteins are receptors for neurotransmitters.
3.
Some proteins are hormones.
4.
Some proteins are receptors for hormones.
Please note: In the video animation of a hormone and a receptor, I inadvertently say in
reference to the release of the red molecule, that it “causes some kind of a change
inside the molecule;” when I should have said the red molecule “causes some kind of
a change inside the cell.”
E.
Some proteins play a role in the immune responses (antibodies).
F.
Proteins typically compose the matrix that connects tissues or body parts(tendons, ligaments,
etc).
G.
The mere presence of proteins in body fluids is important in maintaining or regulating fluid
balance (osmosis).
H.
Proteins can function as an energy source.
Lipids are a third class of important organic biomolecule.
A.
The term “lipid” is a descriptive term for greasy or oily substances.
56
B.
III.
Chemically, lipids are hydrocarbons, meaning they are composed primarily of carbon and
hydrogen, which makes them nonpolar and hydrophobic.
C.
Although there are features common to many different lipids, they do not always follow the
classic monomer-polymer theme we have seen in carbohydrates and proteins.
D.
There are many terms that are synonymous that are used in reference to lipids and other
molecules.
1.
Polar = water soluble = fat insoluble = hydrophilic (water loving) = lipophobic (fat
fearing).
2.
Non-polar = water insoluble = fat soluble = hydrophobic = lipophilic.
E.
Lipids, like other molecules we have discussed so far, have numerous functions.
1.
A major function of lipids is as an energy source, we package excess energy for long-term
storage as triglycerides in adipose tissue.
2.
Some lipids have structural functions, such as phospholipids and cholesterol in plasma
membranes.
3.
Some lipids are important in communication as hormones e.g. steroid hormones.
4.
Some lipids are important in immune system stimulation.
There are several important lipids.
A.
Fatty acids are long carbon chains with a terminal carboxyl group.
1.
Fatty acids have energy potential, and some are components of structural lipids.
2.
Fatty acids can be categorized as follows.
a)
Saturated fatty acids are so-called because carbon bonds are "saturated" with
hydrogens, and the carbon-carbon bonds are all single covalent bonds.
(1)
Example:
H
H
H
H
O
H C
C
C
(CH2)n
C
C OH
H
H
H
H
Carboxyl
Numerous methyl groups
group
(2)
b)
Saturated fatty acids tend to be more solid at room temperature than
unsaturated fatty acids(discussed below).
(3)
Animal fats are highly saturated.
(4)
Saturated fatty acids travel in the bloodstream as Low Density
Lipoproteins(LDL's) or "bad" cholesterol.
(5)
LDL’s significantly contribute to cardiovascular disease.
Unsaturated fatty acids have at least one double bond(or triple bond) between
carbons.
(1)
Example:
H
H C
H
H
C
H
C
57
(CH2)n
H
C
H
O
C
OH
(2)
c)
d)
If the fatty acid has only one double bond between carbons it is a
“monounsaturated” fatty acid.
(3)
If it has two or more double bonds between carbons it is a
“polyunsaturated” fatty acid.
(4)
Being unsaturated is a subtle, but significant, change in a fatty acid.
(a)
Unsaturated fatty acids tend to be oils at room temperature.
(b)
Plant oils are high in unsaturated fatty acids.
(c)
Unsaturated fatty acids travel in the bloodstream as HDL’s (High
Density Lipoproteins) or "good" cholesterol.
(d)
HDL’s are less likely to contribute to cardiovascular disease.
Essential fatty acids are unsaturated fatty acids that cannot be made by humans.
(1)
Linoleic acid and Linolenic acid are essential fatty acids.
(2)
They are found in the oils of many grains (flax seed oil is particularly
high), and fish oils.
Hydrogenated fatty acids are generated in a commercial process that transforms
polyunsaturated fatty acids to saturated fatty acids.
(1)
The purpose is to change the texture of cheap, but less palatable vegetable
oils, into a more palatable lipid with a long shelf life.
(2)
If hydrogenation is complete, you end up with a saturated fatty acid.
(3)
Partially hydrogenated fatty acids are not completely saturated, but they
have some negative characteristics.
(a)
Polyunsaturated fatty acids have a “kink” in the carbon chain,
known as a “cis” chain.
(b)
Partially hydrogenated fatty acids have the cis configuration
transformed to a “trans” configuration.
(c)
B.
The trans (partially hydrogenated) fatty acids contribute more
significantly to LDL’s in the blood than cis (polyunsaturated) fatty
acids.
(d)
Trans fatty acids are thought to have fewer biological uses than cis
fatty acids, and as a result, are more likely to be stored as fat than
cis or saturated fatty acids.
(e)
To improve cardiovascular health one should reduce total lipid
consumption, and replace saturated and partially hydrogenated
fatty acids with unsaturated fatty acids in diet--do not add
unsaturated fatty acids to whatever lipids you are already
consuming.
Triglycerides are commonly know as “fats.”
1.
Mammals store triglycerides in adipose tissue for future energy needs.
58
2.
Triglycerides are composed of three fatty acids bound to a three carbon molecule called
glycerol, by dehydration reactions.
G
L
Y
C
E
R
O
L
C.
Fatty acid 1
Fatty acid 2
Fatty acid 3
3.
The fatty acids may be saturated, unsaturated, or partially hydrogenated.
4.
High levels of triglycerides in the bloodstream contribute to cardiovascular disease.
Phospholipids are important structural lipids.
1.
Phospholipids usually have two fatty acids (saturated or unsaturated) bonded to a polar
complex that includes a phosphate group (PO4).
2.
The fatty acid “tails” are nonpolar and hydrophobic, and the polar phosphate “head” is
polar and hydrophilic.
3.
This makes for an interesting molecule that will spontaneously arrange its hydrophobic
tails together in an aqueous(water) solution.
4.
Phospholipids readily form bilayer sheets (with tails inwards and heads outwards) that act
as one of the major components of the plasma membranes of all cells in nature.
5.
The phospholipid bilayer is an effective barrier because hydrophilic molecules have
difficulty passing through the hydrophobic tails, and hydrophobic molecules have trouble
passing through the hydrophilic heads.
Phosphate head
Phospholipid
bilayer
Fatty acid
tails
IV.
V.
VI.
VII.
D.
Waxes, steroid hormones, and prostaglandins are also biologically important lipids.
There are many other important organic molecules such as nucleic acids, vitamins, glycoproteins,
glycolipids, and others, many of which we will discuss later in the course.
As mentioned previously, certain carbohydrates, proteins, and lipids have energetic value.
We normally refer to these energetic molecules as having “caloric value.”
A.
A calorie is a unit of energy (energy required to raise one ml of water, one degree Celsius).
B.
The Calorie listed on packaged foods is called the “nutritionist’s calorie,” and is equivalent to a
chemist’s Kilocalorie (Kcal), meaning that 1 Nutritionist’s Calorie = 1 Kilocalorie.
The caloric value of foods is as follows:
A.
Carbohydrates = approximately 4 Kcal/gram.
B.
Proteins = approximately 4 Kcal/gram.
59
C.
D.
Lipids = approximately 9 Kcal/gram.
Ethyl alcohol (in alcoholic beverages) = approximately 7 Kcal/ gram.
60
61
Lesson 7: Plasma Membranes and Solutions
I.
II.
III.
I have already mentioned that cells, and organisms themselves, are composed of highly complex
solutions, such as the cytoplasm of a cell, blood plasma, etc.
A.
Furthermore, especially in complex cells, the cytoplasm is not only contained within a complex
plasma membrane, but the cytoplasm is further compartmentalized by still more plasma
membranes.
B.
Before looking at cell structure, I would like us to first examine the nature of plasma membranes,
solutions, and the movement of particles in solution (including how particles move across plasma
membranes).
Plasma membranes not only form the outer boundary of the cell, but act like “cellular tupperware or
baggies,” to package enzymes, other chemicals, and structures needed for specific tasks by the cell, i.e.
they organize the cell.
A.
The cell membrane of a cell, is a plasma membrane.
1.
The cell membrane is the outermost boundary of a cell.
2.
The cell membrane, like all plasma membranes, is a selectively (or semi) permeable
barrier, which means some things pass through the membrane, and some things do not.
B.
Internal cellular membranes, which form various cellular organelles, are also composed of
plasma membranes, and include the following:
1.
Vacuoles, which are large membranous sacks that enclose solutions or objects within
cells.
2.
The nuclear envelope, which encapsulates a cell’s chromosomes.
3.
Peroxisomes, which are small saccules that contain specific enzymes.
4.
Etc.
Plasma membranes have a fairly universal structure in nature.
A.
Plasma membranes are composed of a bilayer sheet of phospholipids.
1.
The hydrophobic fatty acid tails are found in the interior of the membrane.
2.
The hydrophilic phosphate heads form the exterior of the membrane.
3.
The membrane, though pliable, is an effective barrier.
a)
Polar molecules do not easily pass through the nonpolar tails.
b)
Nonpolar molecules do not easily pass through the polar phosphate heads.
B.
Embedded in the biphospholipid membrane are proteins, often with carbohydrates or other
molecules associated with them.
1.
Some proteins are “pore” proteins or “channel” proteins that are like hollow cylinders
extending through the plasma membrane.
a)
Pore proteins give the plasma membrane its “semipermeable” or “selectively
permeable” characteristics.
b)
If particles are the right size and charge, they may pass through a particular pore
protein to the other side of the membrane.
c)
If particles are too big or have the wrong charge, they do not pass through the
membrane.
2.
Some membrane proteins are transport proteins.
a)
If a particle is too large to move through a pore protein it may be bound by a
transport protein which will move it across the membrane.
b)
Transport proteins work somewhat like enzymes, in that the binding site of the
protein and the molecule to be bound must be complementary, i.e. transport
proteins are specific in what they will move across a membrane.
c)
Some transport proteins require cellular energy to function.
62
3.
IV.
V.
Many other proteins are also embedded in the phospholipid bilayer, exhibiting a variety
of functions, such as: providing support or shape to the membrane, being enzymes, being
involved in immune response, etc.
C.
Lipids (other than phospholipids) and carbohydrates may also be associated with a plasma
membrane.
A solution is a homogenous mixture of ions or molecules of two or more substances.
A.
The solute of a solution is the relatively less abundant substance.
B.
The solvent of a solution is the relatively more abundant substance.
C.
In cells, cytoplasm is a very complex solution.
1.
Water is the solvent, making cytoplasm an aqueous solution.
2.
Ions, and a host of organic molecules form the solute.
D.
While solutions can involve any combination of solid, liquid, and gas, when we talk about
solutions in this class, we will assume them to be aqueous solutions, with a dissolved solid or gas
as the solute.
E.
Aqueous solutions are normally described by solute concentration, i.e. a 2.0% NaCl solution is
also 98.0% water.
The particles in an aqueous solution move, sometimes across membranes.
A.
Some particle movement is “passive” which means that no “cellular energy” is needed for the
movement to occur.
1.
Cellular energy is released by the hydrolysis of a molecule called ATP (we will learn
much more about ATP later in the course).
2.
Passive processes do not require energy from ATP.
B.
There are several types of passive processes.
1.
Diffusion is the movement of particles in solution from an area of high concentration to
an area of lesser concentration (they follow their concentration gradient) eventually
resulting in a relatively equal distribution of particles.
a)
Particles in aqueous solutions have an energy of movement called kinetic energy
that leads to collisions of particles.
b)
The kinetic energy and resultant collisions disperse particles from an area of high
concentration, to one of lesser concentration.
c)
A system left to diffuse naturally, in which no other mechanisms are active, will
eventually reach an equilibrium of distribution.
2.
Osmosis is a special case of diffusion: it is the diffusion of water across a semi-permeable
membrane (abbreviated spm).
a)
In nature, the spm is a plasma membrane.
b)
Tonicity is a term that describes the comparative osmotic pressure of two
solutions.
(1)
Osmotic pressure is a measure of the ability of a solution to draw water
towards itself by osmosis, from another solution.
(2)
Several factors affect osmotic pressure, but we are going to take the
simplistic approach that solute concentration, alone, determines osmotic
pressure.
(3)
In our simplistic view, the greater the solute concentration, the greater the
osmotic pressure of a solution.
(a)
Imagine that two solutions, a 5% glucose solution, and a 10%
glucose solution, are separated by an spm, and that the spm is
permeable to water, but not the sugar, glucose.
(b)
The 10% glucose solution has a greater osmotic pressure than the
5% glucose solution, for the reasons described below.
63
(i)
c)
The 10% glucose solution is also a 90% water solution,
whereas the 5% glucose solution is also a 95% water
solution.
(ii)
Water will follow its concentration gradient, diffusing
across the spm from where water is more concentrated (the
5% glucose solution) to where it is less concentrated (the
10% glucose solution).
(iii)
Since more water diffuses (osmoses) towards the 10%
glucose solution, the 10% glucose solution has a greater
osmotic pressure than the 5% glucose solution.
(4)
Three terms are used to describe the tonicity of solutions.
(a)
A hypertonic solution is defined as having a greater osmotic
pressure (higher solute concentration) than another solution to
which it is being compared, so in our example above, the 10%
glucose solution is hypertonic to the 5% glucose solution.
(b)
A hypotonic solution is defined as having a lesser osmotic pressure
(lower solute concentration) than another solution to which it is
being compared, so in our example above, the 5% glucose solution
is hypotonic to the 10% glucose solution (which is hypertonic).
(c)
Isotonic solutions have the same osmotic pressures (solute
concentrations) as one another, and there is no net osmosis in
either direction (two 5% glucose solutions would be isotonic).
(5)
The terms hypertonic and hypotonic are comparative terms, in that, if you
say one solution is hypertonic, it means another solution is hypotonic.
(6)
If conditions permit, the hypertonic solution and hypotonic solutions will
eventually become isotonic to one another, so in the example above, the
two solutions would eventually equilibrate at 7.5% glucose.
Consider another example showing a red blood cell (RBC) in a saline (salt)
solution:
SPM(plasma membrane)
RBC
0.7% NaCl
(Hypotonic
solution)
(1)
(2)
Net water
movement
(osmosis)
2.0% NaCl
(Hypertonic
solution)
The cytoplasm of the RBC has a solute concentration of 0.7%,and the
surrounding saline solution has a solute concentration of 2.0%.
(a)
The RBC cytoplasm has a lesser solute (NaCl) concentration and is
therefore the hypotonic solution.
(b)
The surrounding saline solution has a greater solute (NaCl)
concentration and is therefore the hypertonic solution.
Because the cytoplasm of the RBC is 0.7% NaCl, it is also 99.3% water,
and because the saline solution 2.0% NaCl, it is 98.0% water.
(a)
Particles inherently diffuse from areas where they are highly
concentrated to areas where they are less concentrated.
64
(b)
Water will diffuse(osmose) from the cytoplasm of the RBC into the
surrounding saline, through the cell membrane of the RBC,
following its concentration gradient (99.3% ---> 98.0%).
(3)
I hope you can see that the hypertonic solution has a greater osmotic
pressure, causing water to osmose (diffuse) into it, because of its lower
concentration of water compared to the hypotonic solution--water merely
follows its concentration gradient.
(4)
Tonicity and osmosis can be confusing because we define hypertonic and
hypotonic by solute concentration, yet osmosis occurs in response to
differences in water concentration.
Please note: In the video animation of tonicity involving a red blood cell, it shows the solute
(saline) concentration as .07 % NaCl, but it should read 0.7 %.
3.
Dialysis is another special case of diffusion: the diffusion of solute particles across an
spm.
a)
Plasma membranes are permeable to water.
b)
Plasma membranes are also permeable to some solute particles but not others.
c)
Particle size and particle charge influence whether a particle can diffuse through
membrane pores or channels.
d)
In dialysis, solute particles follow their concentration gradient.
e)
Dialysis of solute particles may occur in one direction across a membrane, while
osmosis of water goes in the other.
f)
Consider the example below, in which two solutions are separated by an spm
(semi-permeable membrane), that is permeable to water and the sugar, glucose.
5.0%
Glucose
(dialysis
)
Water
(1)
spm
6.0%
Glucose
Glucose
(osmosis)
The 6% glucose solution is hypertonic, so water will osmose into it from
the 5% glucose solution.
65
(2)
C.
If the flow of water is not too severe, glucose molecules from the 6%
glucose solution will follow their concentration gradient and diffuse (via
dialysis) through the spm into the 5% glucose solution.
(3)
The diffusion of the water across the spm is osmosis, the diffusion of the
glucose through the spm is dialysis.
4.
Facilitated transport (diffusion) occurs when a solute particle binds to a membrane
transport protein with a complementary binding site, and is moved across the membrane
by that protein and deposited on the other side.
a)
Facilitated transport follows the concentration gradient, meaning solute particles
are transported across the membrane from a solution of high concentration to one
of lesser concentration.
b)
No cellular energy (ATP) is required for facilitated transport, so it is a passive
process.
c)
An example of a solute that is transported across human plasma membranes is
fructose.
Active processes are those that require cellular energy (ATP), and because they utilize energy,
can oppose concentration gradients.
1.
Active transport (like facilitated transport) occurs when a solute particle binds to a
membrane transport protein with a complementary binding site, and is moved across the
membrane by that protein and deposited on the other side, using ATP in the process.
a)
Active transport can oppose the concentration gradient of solute, taking solute
particles from a solution where they are less concentrated, across the plasma
membrane, to a solution where they are more concentrated.
b)
Example-- “sodium/potassium pump” proteins move sodium and potassium ions
across the cell membranes of nerve cells, priming them for nerve impulse
conduction.
2.
Endocytosis is the invagination of the cell membrane of a cell, to bring in particles and
objects from the outside of the cell to the inside(especially particles too large for diffusion
or transport)--see diagram below.
a)
VI.
The diagram shows a large particle outside the cell membrane, which slowly
invaginates bringing the particle into the cell within a membranous vesicle.
b)
This process requires cellular energy (ATP), and can oppose concentration
gradients.
(1)
Pinocytosis (pino=drink, cyto=cell) is “cell drinking,” in which a cell
endocytoses small vesicles containing water and small solute particles.
(2)
Phagocytosis (phago=eat, cyto=cell) is “cell eating,” in which a cell
endocytoses large particles or objects forming vacuoles.
3.
Exocytosis is the reverse of endocytosis, where vesicles containing particles for export
from the cell, fuse to the cell membrane expelling their contents--this process also
requires ATP and can oppose concentration gradients.
As mentioned previously, we are taking a simplistic view of osmotic pressure and tonicity, in that we are
defining hypertonic, hypotonic, and isotonic solutions in terms of their solute concentrations.
A.
This will work when we are comparing solutions with identical solutes.
66
B.
VII.
It is not this easy, however, when comparing solutions with different solutes, for example, a 1%
NaCl solution is not isotonic to a 1% CaCl2 .
C.
Factors other than concentration that affect osmotic pressure (tonicity) include the following:
1.
The number of electrolytes into which it will dissociate in water.
a)
NaCl ---> Na + Cl, two electrolytes.
b)
CaCl2 ---> Ca + 2Cl, three electrolytes.
2.
The mass of the particles.
a)
Calcium has a greater mass than Sodium.
b)
The greater the mass of the particles the greater the osmotic pressure.
3.
The charge of the particles can also play a role.
Certain factors affect the speed or rate of diffusion, these include the following:
A.
Temperature
1.
The higher the temperature the faster the rate of diffusion.
2.
Particles at 370 C diffuse faster than the same particles at 270 C.
B.
Mass of particles.
1.
The smaller the mass of a particle the faster it diffuses.
2.
Sodium, with a mass of 11 Daltons diffuses faster than Calcium with a mass of 20
Daltons.
C.
Concentration gradient.
1.
The larger the concentration gradient, the faster particles diffuse.
2.
Glucose molecules will diffuse (dialysis) from a 10% glucose solution to a 1% glucose
solution (separated by a semi-permeable membrane) more rapidly, than they will from a
2% glucose solution to a 1% glucose solution (separated by a semi-permeable
membrane).
67
68
Lesson 8: Cells and Their Structures
I.
II.
III.
IV.
The surface area to volume ratio is the major factor limiting the size of cells.
A.
Nutrients and waste products must diffuse in and out of a cell, through the cell membrane.
B.
If there is not enough surface area compared to cell volume, then nutrients cannot diffuse in, nor
can waste products (which can be toxic in high concentrations) diffuse out, rapidly enough to
maintain life.
C.
Volume increases more rapidly than surface area, as a cell gets bigger.
D.
Imagine the cell as a cube--a cube has six sides.
1.
The surface area of the cube is computed by taking the length of one side (l), squaring it
(yielding the surface area for one side), and multiplying times 6 to get the surface area for
all six sides: 6 (l2) = surface area of a cube.
2.
The volume of cube is computed by cubing the length of one side: l3 = volume of a cube.
3.
The ratio is determined by dividing surface area by the volume.
4.
Consider the following:
a)
If one side = 1 mm, then surface area = 6 sq mm, and volume = 1 cu mm, which is
a surface area to volume ratio of six to one.
b)
If one side = 2 mm, then surface area = 24 sq mm, and volume = 8 cu mm, which
is a surface area to volume ratio of 3 to one.
c)
If one side = 4 mm, then surface area = 96 sq mm, and volume = 64 cu mm, which
is a surface area to volume ratio of 1.5 to one.
d)
If however instead of one cell of 4 mm, put four 1 mm cells together then: surface
area = 384 sq mm( 6 sq mm x 64 cubes), and volume = 64 ( 1 cu mm x 64 cubes),
which is a surface area to volume ratio of 6 to one.
E.
From this you can see organisms maintain a much higher surface area to volume ratio by
breaking up their volume into many smaller cells.
F.
Some cellular adaptations that increase surface to volume ratio, are as follows.
1.
Small cells.
2.
Highly folded cell membranes.
3.
Broad, flat cells.
There are two types of cells in nature, prokaryotic cells and eukaryotic cells.
Prokaryotic (pro=before, karyo=nut(nucleus)) cells lack a nuclear envelope, and therefore lack a “true
nucleus,” and they lack membrane bound organelles-- small structures with specific functions that are
bound by one or more plasma membranes.
A.
A nuclear envelope is a double plasma membrane that surrounds the genetic material in more
complex cells (called eukaryotic cells; eu=true, karyo=nut(nucleus)).
B.
Prokaryotes are sometimes described as having “naked” or unprotected DNA.
C.
Prokaryotic organisms are placed in the taxonomic Kingdom Monera, and include organisms like
bacteria, and blue-green algae(cyanobacteria).
D.
Prokaryotic cells tend to be very small, around 2 um (micrometers) is a typical size, compared to
eukaryotic cells with a typical size of 100 um.
There are several important prokaryotic cell structures with specific and important functions, and in the
list below, the structures will be described starting structures that are outermost, working towards the
interior of the cell.
A.
Pili extend from the cell wall.
1.
Pili are rather stiff tubelike structures that project from the cell wall, they are composed
of a protein, pilin.
2.
Pili attach a cell to other cells or surfaces.
69
3.
B.
C.
D.
E.
F.
G.
H.
I.
Specialized pili are used in bacterial sex (conjugation) to create a conjugation tube for
transfer or exchange of DNA.
Some bacteria secrete a "capsule", or "slime layer.”
1.
The capsule when present is composed primarily of carbohydrates that have a “slimy”
texture.
2.
The slime layer may be toxic, or it may prevent phagocytosis by white blood cells, it is
typical of many disease causing bacteria.
3.
The slime layer is highly porous and does not function as a semi-permeable barrier--it is
considered external to the cell, and “non-living”--much like hair and fingernails are to us.
Internal to the slime layer is the prokaryotic cell wall , which is typically composed, at least in
part, of a material called peptidoglycans, which is a complex of amino acids and sugars.
1.
The cell wall lends structural support and protection to the cell.
2.
The cell wall does not control movement of materials in and out of the cell.
3.
In some bacteria, the cell wall contains toxic materials, that can cause disease (called
endotoxins).
4.
Like the slime layer it is considered “outside” the cell.
Associated with the cell wall, many bacteria have flagella, whiplike appendages that are used for
locomotion through their aqueous environment.
1.
Prokaryotic flagella are embedded in the cell wall, and attach to the cell membrane, which
is internal to the cell wall.
2.
Prokaryotic flagella have a different structure, composition, and mechanism of operation
than eukaryotic flagella, though they accomplish the same thing.
3.
Prokaryotic flagella are thought to spin around at their base acting like a propeller, pulling
the prokaryote through the water.
Internal to the cell wall is the cell membrane.
1.
Like all plasma membranes in nature, it is composed of a phospholipid bilayer, with
embedded proteins.
2.
The cell membrane is the structure that controls movement of particles in and out of the
cell, i.e. it is selectively permeable.
3.
All fluid and structures within the cell membrane are part of the cytoplasm (cyto=cell,
plasm=material)
Within the cytoplasm is a dense mass of DNA called the nucleoid.
1.
The nucleoid is a large circular chromosome of DNA, tightly compacted into a dense
mass.
2.
The “nucleoid” of prokaryotic cells is not a “nucleus” because they lack a protective
nuclear envelope, composed of plasma membranes, surrounding the chromosome.
3.
An enfolding of the prokaryote's cell membrane called the mesosome, attaches to the
chromosome and plays a role in separating genetic material during cell division.
Prokaryotes typically have additional, small loops of DNA, free in the cytoplasm called plasmids.
1.
Plasmids have genes just like chromosomes.
2.
Plasmids may be transferred from one prokaryote to another (prokaryotic “sex”).
3.
This is how bacteria transmit antibiotic resistance to one another.
Prokaryotic ribosomes are also free in the cytoplasm.
1.
They function to build proteins from amino acids.
2.
Prokaryotic ribosomes are smaller than eukaryotic ribosomes, but function in similar way.
Some prokaryotes, like blue-green algae and some "photosynthetic" bacteria, have sheets of
plasma membranes called “photosynthetic membranes.”
1.
The molecules needed for photosynthesis, or similar processes, are embedded in these
membranes.
70
2.
V.
VI.
VII.
The sheets of internal plasma membrane provide a large surface area for the absorption of
light energy, and related chemical processes.
Eukaryotic cells possess a "true nucleus".
A.
The DNA(in the form of linear chromosomes) is enclosed by a double plasma membrane called
the nuclear envelope, forming a structure called a nucleus.
B.
The DNA is bonded to proteins in eukaryotic cells to form a complex called chromatin (which in
turn forms chromosomes).
C.
The envelope acts to protect the genetic material by acting as a selectively permeable barrier that
controls what enters and leaves the nucleus.
D.
Eukaryotic cells are much larger than prokaryotic cells.
Eukaryotic cell cytoplasm is much more complex than prokaryotic cells, with numerous organelles that
have specialized functions.
A.
Organelles are membranous (composed of plasma membranes) intracellular structures with
specialized functions.
B.
The abundance of one type of organelle or structure over another allows cells to develop
specialized functions.
C.
Not all cells have all of the organelles or structures discussed below, so be aware of which
structures are specific to certain types of cells.
Eukaryotic cell structures and their functions will be discussed below, beginning from the outermost part
of the cell, inwards.
A.
The eukaryotic cell membrane is a typical plasma membrane, i.e. a bilayer of phospholipids, with
proteins embedded.
1.
Everything internal to the cell membrane is part of the cytoplasm of the cell.
2.
The cell membrane is a selectively permeable barrier.
B.
The cytoskeleton is a system of support fibrils found in animal cells and some others.
1.
It is composed of proteins that lend shape, and flexible support and orientation to interior
of cell.
2.
The cytoskeleton is anchored to proteins in the cell membrane.
3.
Some fibrils run around the periphery of the cell giving it shape.
4.
Some fibrils run into the interior of the cell attaching to organelles.
a)
Some fibrils may act as “highways” along which molecules and vesicles may be
transported from one part of the cell to another.
b)
Others lend a “flexible stability” to the cytoplasm, giving it a gelatinous texture
and positioning organelles.
c)
Still other fibrils are involved in movement, allowing some cells to crawl along
surfaces (like amoebae and white blood cells), or contract as our muscle cells do.
C.
The nucleus is the most obvious structure in the cytoplasm of a eukaryotic cell.
1.
The nucleus is surrounded by the nuclear envelope, which is composed of two plasma
membranes, the outer one being continuous with the endoplasmic reticulum.
2.
The nuclear envelope contains the nuclear chromatin which is a complex of DNA and
proteins, which forms chromosomes.
a)
Think of the relationship between chromatin and chromosomes like that of cement
and a sidewalk--cement is the material from which the sidewalk (a structure) is
made, likewise a chromosome (a structure) is composed of material called
chromatin.
b)
DNA is the genetic component of a chromosome, with the proteins acting to
protect, stabilize, and control expression of the DNA.
c)
The DNA double helix wraps around certain proteins for stability, while still other
proteins bond directly to the strands of DNA.
71
d)
D.
E.
F.
Eukaryotic chromosomes are linear, not circular like the prokaryotic
chromosomes.
3.
The nucleus also contains one or more areas called nucleoli (nucleolus=singular).
a)
Nucleoli are concentrations of RNA, within the nucleus, that typically stain very
darkly.
b)
The nucleoli are sites where ribosomes are partially constructed.
4.
The nucleus is the "brain" of the cell controlling activities of the cell by determining the
proteins to be made--some of these will be enzymes that will in turn influence cell
activities.
5.
Nucleoplasm is the material within the nuclear envelope (including the chromatin and
nucleoli).
Endoplasmic reticulum (ER) is a sheetlike network of plasma membranes that are continuous
with the outer membrane of the nuclear envelope, extending throughout the cytoplasm, forming
membranous channels that run throughout the cytoplasm.
1.
Rough endoplasmic reticulum (RER) is one type of ER.
a)
It is called "rough" because ribosomes are embedded in its plasma membranes
giving it a grainy or rough appearance in electron micrographs.
b)
RER is continuous with the outer membrane of the nuclear envelope, and is a
primary site of protein synthesis (construction).
2.
Continuous with the RER is the smooth endoplasmic reticulum (SER).
a)
The SER lacks ribosomes.
b)
The SER are involved in steroid, lipid, and carbohydrate metabolism; and in some
cells is important in detoxification of chemicals.
3.
The ER act to concentrate proteins and the products of enzyme metabolism, as well as act
as a pathway by which the substances may travel through the cell.
Ribosomes work with RNA produced in the nucleus to make proteins.
1.
Ribosomes (ribo=RNA, soma=body) are composed of RNA and proteins, are partially
constructed in the nucleoli of the nucleus before being shipped to the cytoplasm.
2.
Ribosomes are found both free in cytoplasm or attached to membranes.
3.
Ribosomes may attach to one another to form polyribosomes.
4.
They are composed of two subunits that must come together to function, so-called large
and small subunits.
5.
Prokaryotic and eukaryotic ribosomes are structurally and functionally similar, but
eukaryotic ribosomes are much larger.
The golgi complex (also called golgi body, or golgi apparatus) is composed of a series of
flattened membranous sacs, stacked on one another, like coins.
1.
These sacs are called cisternae.
2.
The golgi complex acts to modify the products of the endoplasmic reticulum.
3.
These modified products are released into the cytoplasm as secretory vesicles (which will
be exported from the cell), or lysosomes (an organelle which will be discussed later).
4.
The golgi complex functions in the following way:
a)
A particular golgi complex works with a specific region of RER or SER.
b)
The cisternae closest to the ER accepts vesicles from the ER.
(1)
Vesicles are small membranous sacs.
(2)
The ER will pinch off small vesicles containing ER products.
(3)
These move through the cytoplasm (along fibrils of the cytoskeleton) to
the golgi complex where they fuse with the golgi plasma membrane.
(4)
The contents of the vesicle are now within the cisternae of the golgi
complex.
72
c)
G.
H.
I.
J.
K.
The material from the ER is modified within the cisternae, and may be shuttled
from one cisterna to another by vesicles.
d)
The golgi cisternae most distant from the ER eventually pinch off vesicles that
follow one of these scenarios.
(1)
Some vesicles are transported to the cell membrane where they fuse with
the cell membrane to release their products to the outside, in a process
known as secretion.
(a)
Secretion is way to move large quantities of a product out of the
cell, quickly and efficiently.
(b)
Secretory products include hormones, lubricants, and many other
products.
(2)
Some vesicles will be marked for delivery to some other cell location or
organelle.
(3)
Some vesicles will merge with one another to form lysosomes, organelles
which are discussed below.
Mitochondria produce large amounts of ATP (cellular energy), via a process known as cellular
respiration.
1.
Mitochondria are sometimes referred to as the "power houses" of the cell.
2.
The more energy a cell uses, the more mitochondria the cell will have.
3.
Mitochondria are composed of a double plasma membrane: an outer membrane, and an
inner membrane that is highly folded to increase surface area--this is where most of the
ATP are produced.
4.
There is substantial evidence that mitochondria evolved from prokaryotic ancestors.
a)
Mitochondria have physical similarities to some bacteria.
b)
Mitochondria have their own “prokaryotic” DNA (called mitochondrial DNA or
mtDNA), and share genes with bacteria.
c)
They undergo a bacterial form of division, independent of the cell.
d)
There are many examples of symbiotic bacteria living within eukaryotes in nature,
in symbiotic relationships.
Lysosomes (lyse=break or split, soma=body) are membranous vesicles that contain
digestive(catabolic) enzymes.
1.
They normally digest particles brought in through endocytosis (phagocytosis), or cellular
organelles that are defective.
2.
Immune cells that are phagocytic, like our white blood cells, have many lysosomes.
3.
Some diseases, like rheumatoid arthritis and gout, may be related to abnormal lysosome
activity.
Microsomes(microbodies) are small vesicles derived from rough endoplasmic reticulum that
contain crystals of enzymes that catalyze specific reactions, e.g. peroxisomes contain the enzyme,
catalase, that breaks up hydrogen peroxide to water and oxygen.
Vacuoles are large membranous sacs in the cytoplasm.
1.
Storage vacuoles typically store fluids, ingested food, or other chemicals.
2.
Contractile vacuoles are specialized vacuoles that pump excess water out of some single
celled organisms that live in aquatic environments.
a)
Aquatic protozoans, for example, are hypertonic to their environment, so water
osmoses into the cells.
b)
Water that diffuses into the cell is pumped out by the contractile vacuole.
Centrioles are two cylindrical masses of microtubules (micro=small) located near the nucleus of
animal cells.
73
1.
L.
M.
N.
O.
P.
Centrioles are identical in structure to basal bodies, but differ in location and general
function.
2.
Centrioles play a role in animal cell division.
3.
Each “cylinder” of the centriole is composed of 9 "triplets" of microtubules.
Cilia (latin for hair) are relatively short “hair-like” structures that extend from the cell membrane.
1.
They beat back and forth propelling a cell through an aqueous habitat, or move fluids past
a cell if the cell is in a fixed position (as are the cells that line our respiratory passages).
2.
Cilia are composed of a cylinder of microtubules in a “9 + 2” formation: nine
"doublets"(pairs) of microtubules, surrounding a central doublet (or pair).
3.
The cylinder of microtubules is covered by the cell membrane.
4.
At the base of each cylinder of microtubules is a basal body (discussed below) that
controls cilium activity--cilia beat in a coordinated fashion not randomly.
5.
Microtubules slide past one another to cause the back and forth motion of the cilia.
Flagella (Latin for whip) are identical to cilia, except they are longer (whip-like), and usually
fewer in number: all other statements made about cilia also apply to flagella.
Basal bodies.
1.
Found at base of all cilia and flagella, and are thought to be involved in the construction
and control of cilial and flagellar action.
2.
Basal bodies are identical in structure to centrioles.
Plants, algae (kelp is an example), and most fungi have a cell wall--ANIMAL CELLS DO NOT
HAVE A CELL WALL.
1.
Plant and some algae cell walls are composed primarily of cellulose(a structural
carbohydrate--remember?), and sometimes a substance called lignin which gives woody
plants their texture.
2.
Fungal cell walls composed mainly of chitin, a complex of glucose and nitrogen.
3.
Like the prokaryotic cell wall, the eukaryotic cell wall provides structure and support, but
is not considered a selectively permeable barrier.
Plastids are also double membrane organelles that are similar to mitochondria in many ways,
although they have different functions, and they are found only in plants and algae.
1.
Chloroplasts (chloro=green, plast=molded) contain the green pigment chlorophyll and
other light absorbing pigments.
a)
Chloroplasts are found in cells exposed to the sun.
b)
Chloroplasts are the site of photosynthesis: a process that converts light energy to
cellular energy (ATP), and then uses the energy in ATP to make organic
molecules from carbon dioxide and water.
(1)
Chloroplasts have a double plasma membrane structure not unlike
mitochondria.
(2)
The inner membrane is highly folded and increases surface area for
absorption of light energy and production of ATP.
2.
Leucoplasts (leuco=white) are converted chloroplasts that store starch, protein, or oils
(depending on the plant, and cell location).
a)
Leucoplasts are found in cells that are not exposed to light, like root cells.
b)
Since no light reaches the cells, photosynthesis cannot occur, and the organelle is
used for storage.
c)
Leucoplasts most typically store starch, which is white in color, hence the name
leucoplast.
3.
Chromoplasts (chromo=color) are converted chloroplasts that contain color pigments, and
are found in cells of flower petals, carrots, and other pigmented plant parts.
74
4.
VIII.
IX.
Plastids, like mitochondria, also evolved from symbiotic prokaryotes, showing physical,
genetic, and behavioral similarities to prokaryotes.
Q.
Tonoplasts are large central fluid filled vacuoles that lend internal support to plant cells.
R.
Plasmodesmata are cytoplasmic connections between plant cells, that extend through large pores
in the cell walls of adjacent cells, making the cytoplasm of one cell actually continuous with
another.
The organelles of cells and their functions have been determined by a technique known as cell
fractionation.
A.
It usually begins with a blender which ruptures the cells creating a homogenate.
B.
A centrifuge is a machine that spins rapidly.
1.
The homogenate is put in a centrifuge and spun at high revolutions.
2.
The homogenate will separate according to the density and other properties of the
organelles.
a)
The cell wall debris will form a band.
b)
Chloroplasts will form a band.
c)
Mitochondria will form a band, etc.
C.
A specific band can be further centrifuged to see if it is a mixture of structures or not.
D.
Once a band is considered to be composed of like organelles it can be chemically analyzed for
function and microscopically examined.
While there are many types of microscopes that are used to observe microscopic structures and
organisms, we are going to consider two major classes of microscopes: light microscopes and electron
microscopes.
A.
The strength and quality of a microscope is measured by its resolving power.
1.
Resolving power is a measure of the clarity of a microscope, or the minimum distance at
which one can still distinguish two points from one point.
2.
Magnification (a measure of how many times an image has been increased from its actual
size) is related to resolving power, but resolving power also is determined by the clarity
of the image, e.g. a highly magnified but blurred image would have a poor resolving
power (RP).
B.
The naked eye has a resolving power of .1 mm, and a magnification of 1x.
C.
Light microscopes work with light waves--usually visible light.
1.
Light microscopes either bounce light off a specimen (reflected light), or pass light
through a specimen (transmitted light), then passing the light through a system of lenses
and/or mirrors to the eye.
2.
The resolving power is limited by wavelengths of light used.
3.
Violet light has shortest wavelength and the greatest resolving power: approximately 0.2
um (micrometers=millionths of a meter) , and a magnification of 2,000x.
D.
Electron microscopes use electron beams instead of light waves.
1.
Electron beams are either reflected off, or transmitted through specimens impregnated
with heavy metals like silver, then the electron beam exposes film or creates an image on
a monitor.
2.
Electron microscopes are much more powerful than light microscopes, with a resolving
power of .0002 um, and magnifications of 1,000,000x.
75
Lesson 9: Energetics, An Overview
I.
II.
The flow of energy is crucial to how ecosystems function, and so a brief review of the two major laws of
energy is in order.
A.
The first law of energy states that energy is neither created nor destroyed, but merely changes
form (e.g. solar energy converts to chemical energy, which converts to heat energy).
B.
The second law of energy states that some energy dissipates to the surroundings as it flows or
changes form.
1.
There are many ways to state this law, e.g. some high quality energy degrades to low
quality energy (heat) as it flows or changes form, etc.
2.
Example: A certain amount of electrical energy produced at Hoover Dam is sent through
wires to Las Vegas.
a)
The amount of electrical energy received in Las Vegas will be less than that
produced at Hoover Dam.
b)
As the electricity moves through the wires, heat is generated, so some of the
electrical energy degrades to heat energy and is lost from the “system,” in this case
the electrical wires.
c)
This seems to be a contradiction of the first law of energy, but if you measured the
amount of electrical energy received in Las Vegas, and measured the amount of
energy lost as heat, it would equal the amount of energy produced by the
generators at Hoover Dam.
d)
The energy is not “lost” in the sense that it is destroyed, it is merely lost from the
system as a lesser quality (usually heat) energy.
C.
I would like to review two terms we will be using in our discussion of energy: mole and
kilocalorie.
1.
A mole of ATP molecules will yield approximately 7.3 Kcal of energy per high energy
bond ( remember a mole is a standard unit of chemistry that represents 6.02 x 1023
particles).
2.
A Kcal (Kilocalorie) is a unit of heat energy, and is listed on food packaging as a
“CALORIE,” i.e. a chemist’s Kilocalorie is a nutritionist’s CALORIE.
An understanding of ATP and the general nature of energetic reactions is essential to understanding how
energy flows in nature.
A.
Virtually all chemical reactions that require energy in an organism, use energy contained in the
high energy phosphate bonds of ATP.
1.
The initials, ATP, stand for Adenosine Triphosphate.
a)
The “adenosine” in ATP, is composed of a nitrogenous base called adenine,
bound to a ribose sugar.
b)
The “triphosphate” is three phosphate groups (PO4) bound to the ribose sugar of
the adenosine.
2.
The “third” phosphate group of the ATP is bound to the second phosphate group by a
“high energy bond.”
a)
The term "high energy bond" means that it is easily broken, and yields a form of
chemical energy, usually called "cellular energy,” that is usable by enzymes.
b)
ATP will react with a water molecule in a hydrolysis reaction to yield one
phosphate group (abbreviated Pi or PO4); one ADP(adenosine diphosphate, socalled because two phosphate groups are still attached to the adenosine unit); and
energy, which has the potential to drive enzyme mediated chemical reactions.
76
c)
B.
The hydrolysis of ATP is summarized below:
ATP + H2O -----> ADP + Pi , releasing 7.3 Kcal of energy
d)
Conversely, ATP is synthesized via a condensation (dehydration) reaction that
requires an input of energy:
ADP + Pi + 7.3 Kcal of energy -----> ATP + H2O
(1)
The molecules used to make ATP (ADP, and Pi,) are recycled, but the
energy required for its construction cannot be recycled (due to the second
law of energy), and must be constantly renewed(see food chains below).
(2)
There are only about two ounces of ATP in the body, at any given
moment, but you synthesize (and hydrolyze) a weight equivalent to your
body weight daily.
3.
ATP is synthesized in very specific locations and reactions.
a)
Substrate level phosphorulation is a relatively simple process in which an enzyme
transfers a phosphate group from one molecule to an ADP to form ATP--these
reactions are carried out by only a few enzymes and are therefore rare, producing
relatively little ATP.
b)
Chemiosmotic phosphorulation is the major producer of ATP in cells, occurring
within mitochondria and chloroplasts (this process will be described later).
Chemical reactions can be classified by whether they demonstrate a net release, or consumption
of energy, and ATP is best discussed in the context of these reactions.
1.
Exergonic (ex=out, erg=labor) reactions are those that release energy to the surroundings.
a)
The products of these reactions have less “free energy” than the reactants, which
means there is less energy in the bonds of the products than the reactants, because
energy has been released to the surroundings.
b)
The primary exergonic reaction of living things is the hydrolysis of ATP, which
contains “cellular energy.”
(1)
ATP + H2O ---> ADP + Pi :  G = - 7.3 Kcal per mole (1Kcal = 4.1855
Kjoule).
(2)
The  (delta) in G stands “change”, and G is the symbol for free energy.
(a)
G thus stands for the change in free energy in a chemical reaction.
(b)
In exergonic reactions the G has a negative value because the
products have less “free energy” than the reactants due to the
release of energy to the surroundings.
2.
Endergonic (endo=inward) reactions are those that require energy from an outside source.
a)
The products have more free energy than the reactants, due to a consumption of
energy.
b)
An important endergonic reaction is synthesis of ATP:
ADP + Pi ---> ATP + H2O G = +7.3 Kcal.
c)
Endergonic reactions have a positive G due to the fact that energy from an
outside source has been absorbed, so there is more energy in the chemical bonds
of the products than the reactants.
3.
Exergonic and endergonic reactions are normally “coupled” in biological systems.
a)
An enzyme that is catalyzing an endergonic reaction will simultaneously catalyze
an exergonic reaction (the hydrolysis of ATP) so that the energy needed for the
endergonic reaction is available.
b)
Example: Lets say that an enzyme catalyzes the binding of a molecule “A” with a
molecule, “B” to form a product “AB,” and the G is + 3.2 Kcal :
A + B -----> AB G = 3.2 Kcal
(1)
This is an endergonic reaction that requires energy.
77
(2)
(3)
Enzymes that require energy to function get their energy from the
hydrolysis of ATP.
If the enzyme “couples” the endergonic reaction, requiring 3.2 Kcal of
energy, with the exergonic hydrolysis of ATP, which releases 7.3 Kcal of
energy, so that both reactions are occurring together, there is enough
energy provided by the hydrolysis of ATP to enable the enzyme to form
the product, AB.
ATP + H2O
ADP + Pi
A + B ---------> AB
(4)
III.
G = - 7.3 Kcal
G = +3.2 Kcal
You may have noticed that hydrolysis of ATP yields more energy than is
required (7.3 Kcal to 3.2) for the formation of AB-- the excess energy (4.1
Kcal in this case) is lost to the surroundings as heat.
(a)
The hydrolysis of ATP and release of energy is an all or nothing
event, because the hydrolysis of ATP yields 7.3 Kcal regardless of
how much energy is required by the endergonic reaction to which it
is coupled.
(b)
If an endergonic reaction requires 0.1 Kcal of energy, then the
hydrolysis of ATP would still provide 7.3 Kcal of energy, meaning
that 7.2 Kcal of energy would be lost to the surroundings as heat.
c)
If the synthesis of ATP from ADP and Pi is an endergonic reaction, and
endergonic get their energy from the exergonic hydrolysis of ATP into ADP and
Pi , then how can ATP be synthesized without ATP--this is a question we will
address in detail shortly.
Before getting into the specifics of how ATP is produced, an overview of the flow of energy in nature is
in order.
A.
The primary source of energy that drives our ecosystems is solar energy (light energy) produced
by the sun.
1.
The energy is produced by a nuclear reaction called fusion.
a)
Nuclear reactions lead to a change in the nuclei of the atoms involved (rather than
a change in electron configurations as in chemical reactions).
(1)
In nuclear reactions some mass is converted to energy according to
Einstein’s famous equation E = mc2, where E=energy, m=mass, and
c=speed of light (approximately 186,000 miles per second).
(2)
According to this equation, even a tiny mass converts into a tremendous
amount of energy.
b)
In a fusion reaction, isotopes of Hydrogen nuclei (of which the sun is composed),
under conditions of extreme heat (meaning the nuclei are moving at remarkable
speeds) collide and fuse to form Helium nuclei.
(1)
The resulting Helium nucleus has less mass than the Hydrogen isotopes
that fused to form it, i.e. mass is converted to energy.
(2)
Humans have duplicated this fusion reaction here on earth, in the form of a
hydrogen bomb.
2.
Some of the energy produced by fusion is solar energy.
78
a)
B.
Solar energy is a type of electromagnetic energy which exhibits particle and wave
dynamics.
b)
Electromagnetic energy can be visualized as tiny particles called photons, moving
in a wave-like trajectory.
c)
The distance from the crest of one wave to the crest of another is a wavelength.
d)
The shorter the wavelength the greater the energy of the photon, and the higher the
frequency (wavelength and frequency are inversely related).
3.
The electromagnetic spectrum generated by the sun (solar energy) is composed of the
following (from shortest wavelength to longest) : gamma rays, x-rays, ultraviolet rays, the
visual spectrum (violet, blue, green, yellow, orange, red), infra-red rays, and radio waves.
4.
It is the solar energy of the visual spectrum that plays such an important role here on
earth.
Solar energy (particularly light in the violet-blue, and orange-red spectra) is absorbed by plants
(also algae and blue green algae) and converted to chemical energy in the process of
photosynthesis, becoming the foundation of virtually all of earth’s ecosystems.
1.
The “overall” equation for photosynthesis is as follows:
6CO2 +
Carbon dioxide
C.
Chlorophyll-a
6H2O + 686Kcal(solar E) -------------------> C6H12O6 + 6O2
water
glucose
oxygen
Please note: the video labels carbon dioxide as “carbon,” this is an error.
a)
The reaction requires a green pigment, chlorophyll-a, as a catalyst.
b)
You can see that the process of photosynthesis yields an organic molecule
(glucose) from inorganic ones (carbon dioxide and water), as well as the planet’s
oxygen supply.
2.
Photosynthesis is actually a complex, two-step process--the light dependent reaction, and
the light independent (dark) reaction.
a)
The light dependent reaction requires chlorophyll-a and is a process in which solar
energy is converted to cellular energy (in the form of ATP), and the hydrogen and
oxygen atoms of water are separated from one another.
b)
The light independent reaction uses enzymes to build glucose molecules from
carbon dioxide and hydrogen (from the light dependent reaction), using ATP (also
generated in the light reaction) as the energy source (remember enzymes get their
energy from ATP not from light energy).
3.
Once the plant has made glucose, the glucose can be converted into other biomolecules
(amino acids, lipids, nucleic acids, etc) needed to grow, etc.
4.
Through photosynthesis, solar energy is converted to chemical energy in the carboncarbon covalent bonds of organic molecules.
This chemical energy is then transferred through the ecosystem by means of a food chain.
1.
As mentioned previously, photosynthetic organisms (plants in most ecosystems), are the
producers of organic molecules (= food), and are at the bottom of all food chains.
2.
Organisms that cannot photosynthesize, must ingest organic molecules for the energy they
contain, and are called “consumers.”
3.
There may be several types of consumers in a food chain.
a)
Primary consumers feed on producers.
(1)
Consumers cannot make organic molecules from inorganic sources.
(2)
Consumers must ingest organic molecules.
(3)
Consumers can convert one type of organic molecule to another.
b)
Secondary consumers feed on primary consumers.
79
c)
d)
e)
f)
g)
D.
Tertiary consumers feed on primary consumers.
Quaternary (4th degree) consumers feed on tertiary consumers.
Etc.
Secondary consumers and above are called higher order consumers.
Solar energy is converted to chemical energy by the producers, and the chemical
energy is transferred from consumer to consumer via ingestion.
4.
Consider the following food chain: grass is eaten by a beetle, which is eaten by a lizard,
which is eaten by a snake, which is eaten by a coyote.
a)
The grass is the producer.
b)
The beetle is the primary consumer.
c)
The frog is the secondary consumer.
d)
The snake is the tertiary consumer.
e)
The coyote is the quaternary consumers.
f)
The frog, snake, and coyote are higher order consumers in this food chain.
Consumers exploit the potential energy in producer tissues by means of a process called the
cellular oxidation of glucose.
1.
Consumers (via the cellular oxidation of glucose) take apart the sugars (and other organic
molecules) produced by plants and other producers, capturing some of the energy released
by the process, in the form of ATP.
2.
Consumers cannot generate ATP from solar energy, so enzymes that require energy to
catalyze a chemical reaction use ATP generated by the cellular oxidation of glucose.
3.
The overall equation for the cellular oxidation of glucose is as follows:
C6H12O6 +
6O2 -------------------> 6CO2 +
6H2O + 686Kcal
glucose
oxygen
water
Carbon dioxide
(263Kcal as ATP, 423Kcal as heat)
Please note: I inadvertently state in the video that 262 Kcal of energy is used for the
production of ATP, I meant to say 263 Kcal as represented in the graphic.
a)
Like the equation for photosynthesis, this simple equation belies a complex, twopart process, which can yield 36 ATP per molecule of glucose.
(1)
Glycolysis takes place in the cytoplasm (cytosol) of a cell, and not within
any organelle.
(a)
Glycolysis is an enzymatic pathway that breaks the six-carbon
glucose sugar into two, three-carbon, pyruvate molecules.
(b)
Glycolysis generates two ATP molecules per glucose and does not
require oxygen.
(c)
All organisms practice glycolysis (both prokaryotes and
eukaryotes)--it is a universal process in nature.
(2)
Cellular respiration takes place in the mitochondria of the cell.
(a)
Pyruvates diffuse into the mitochondria from the cytoplasm.
(b)
For each glucose (or two pyruvates), 34 ATP are produced by
cellular respiration.
(c)
Cellular respiration requires oxygen.
(d)
Only organisms with mitochondria (eukaryotes) carry out cellular
respiration.
b)
As you can see, the overall equation in some ways reverses the work of
photosynthesis by returning carbon dioxide and water vapor to the atmosphere.
80
c)
E.
F.
G.
Most of the energy released by the cellular oxidation of glucose is lost as heat (the
second law of energy at work again), with some energy converted to the cellular
energy of ATP.
d)
Plants also have mitochondria and use the cellular oxidation of glucose to produce
ATP when photosynthesis cannot be carried out (like nighttime), but ATP is still
needed by enzymes.
4.
It may not seem very energy efficient to convert solar energy to the cellular energy of
ATP, to the chemical energy of carbon bonds in sugars (in the producer), only to break
the carbon bonds (in the consumer) for the purpose of making ATP again--especially
when this ATP might be used to remake sugars, or polysaccharides, proteins, etc, to build
the consumer’s tissues.
a)
You are right, it is an energy efficiency nightmare, because energy is lost in each
conversion of energy : solar to cellular to chemical to cellular to chemical etc.
b)
However inefficient it is, however, it works, because the sun provides a virtually
limitless supply of energy, some of which moves through the food chain, and most
of which is lost to the surroundings as heat.
c)
Only about 1% of the solar energy absorbed by a plant is converted to energy in
carbon-carbons bonds.
Individual food chains may overlap forming food webs.
1.
In the food chain above the grass, beetle, lizard, snake, and coyote could all be prey to
other organisms in the ecosystem.
2.
All of the above could likewise play other roles in different food chains, for examples
coyotes may feed directly on plant material, acting as primary consumers.
Some other terms are associated with food chains.
1.
Herbivores (herb=plant, vor=devour) are animals that primarily eat plants.
2.
Carnivores (carn=meat, vor=devour) are animals that primarily eat other animals.
3.
Omnivores (omni=all, vor=devour) are animals that have a relatively balanced diet of
plants and animals.
The energy distribution in an ecosystem always forms an energy pyramid.
1.
A pyramid is widest at the base and tapers towards the top.
2.
The energy in the tissues of the producers form the base of the energy pyramid, with
primary and higher order consumers forming subsequent levels, each with less energy.
3.
Only about 10% of the energy in one level of the pyramid (comparable with a level of a
food chain) goes to the next level, which means that about 90% of the energy dissipates to
the surroundings (due to the second law of energy)--this is called the “ten percent rule.”
4.
The higher an organism(or species) feeds in a food chain, the more at risk it is, because:
a)
It has less energy available to it for feeding, and
b)
The greater the chance for food chain disruption.
5.
Being omnivorous, humans can maximize food (energy) supply by moving down in the
food chain--each step down in the food chain increases our food supply by ten times.
6.
The energy pyramid.
10% of
energy to
next level,
90% lost to
surroundings
Etc.
81trophic level (quaternary
Fifth
cons.)
Fourth trophic level (tertiary consumers)
Third trophic level (secondary consumers)
H.
What is crucial to understand about food chains is this:
1.
Matter is recycled in food chains, e.g. Carbon in CO2 is absorbed in photosynthesis by
producers, and is passed to consumers via the food chain, which release the carbon back
to the atmosphere as CO2 allowing the cycle can start over.
a)
Carbon, oxygen, hydrogen, nitrogen, etc are passed from organism to organism
via the food chain.
b)
The atoms of which you are composed were in other animals or plants prior to
being incorporated into your tissues.
c)
These atoms are as old as the earth itself (approximately 4.6 billion years).
2.
Energy is not recycled in food chains it is a one way flow.
a)
Energy dissipates into the surroundings as heat, as it moves through the food
chain, according to the second law of energy.
b)
High quality energy must be continually introduced into food chains by
photosynthesis.
c)
The reason organisms must consume food regularly is that the high quality energy
consumed previously, dissipates, and must be replaced.
82
Module 10: Energetics, Pieces of the Energy Puzzle
I.
There are some important chemical reactions, and processes, that play essential roles in cellular
energetics, --their relevance will not become clear until we discuss photosynthesis and the cellular
oxidation of glucose, in detail.
A.
The first type of reactions to be discussed are oxidation and reduction reactions.
1.
Oxidation occurs when a substance loses electrons, and/or hydrogen atoms.
2.
Reduction occurs when a substance gains electrons, and/or hydrogen atoms.
3.
Oxidation and reduction typically occur in the same chemical reaction with one substance
being oxidized, the other reduced.
4.
Oxidation-reduction reactions are important in cellular energetics because they help move
electrons from one location to another, and electrons have potential energy.
5.
There are three molecules that play important roles in energetics by being oxidized or
reduced.
a)
NAD+ , NADP+ , and FAD are complex organic molecules that are readily
oxidized and reduced.
(1)
These molecules are coenzymes derived from B vitamins.
(a)
Coenzymes work with enzymes to complete an enzyme mediated
reaction.
(b)
If coenzymes are required, the reaction cannot occur without them.
(c)
NAD+ , NADP+ , FAD act as electron carriers, because they take
electrons (and hydrogen) physically away from an enzymesubstrate complex so that a reaction can proceed.
(d)
By picking up the electrons and hydrogens, the coenzymes are
reduced, and the substrate is oxidized (because it loses the
electrons and hydrogen).
(2)
The coenzymes are reduced according to the following equations:
NAD+ + 2e- + H+ -------> NADH ,
NADP+ + 2e- + H+ -------> NADPH .
FAD + 2e- + 2H+ -------> FADH2
b)
Once coenzymes have picked up electrons these electrons have potential energy
that can be converted to cellular energy in the form of ATP.
(1)
The coenzymes will “carry” the electrons (by diffusion) to a different place
in the cell, where they will give up their electrons for the production of
cellular energy (being oxidized in the process).
(2)
The coenzymes are oxidized as follows:
NADH -------> NAD+ + 2e- + H+,
NADPH -------> NADP+ + 2e- + H+ ,
FADH2 -------> FAD + 2e- + 2H+ .
(3)
Because of oxidation, the NAD+ , NADP+ , FAD are now available to act
as coenzymes again, and can be reduced and carry electrons once again.
c)
Although it will not be obvious until later, the oxidation and reduction of
coenzymes keeps essential chemical reactions going in a cell, and provides a
steady supply of electrons for the production of cellular energy (ATP).
B.
Enzymatic pathways, and enzymatic cycles, are two ways to carry out complex reactions in a
series of steps.
1.
An enzymatic pathway is a series of enzyme driven reactions, that work, in sequence, to
generate products different from the initial molecules involved.
83
a)
b)
2.
Each enzyme has a small part in the overall reaction.
The product of one enzymatic reaction is the substrate of the next enzyme, and its
product the substrate of another enzyme, and so on.
c)
It is considered a “pathway” because what you end up with (the product) is
different than what you start with, just as a pathway takes you to a different
location from where you started.
d)
An enzymatic pathway is like an assembly line in an automotive plant making car
engines.
(1)
Each worker on the assembly line is like an enzyme, doing one small part
of the overall job of making an engine.
(2)
Each worker adds a part and it moves on to the next worker.
(3)
You start with a pile of parts, you end up with an engine.
e)
An example from nature is a process called glycolysis which we will study later.
An enzymatic cycle is a series of enzymatic reactions that start and end with the same
substance, although atoms or molecules are added and then removed in a different form,
along the way.
a)
It is similar to a pathway, in that a series of enzymes each do a small part of the
overall reaction, and the product of one enzyme becomes the substrate of the next
enzyme, and so on.
b)
It differs from a pathway in that atoms or molecules added to the original
molecule(s) are removed--in a different form-- generating one or more products,
while leaving the original molecule to be reused at the end of the cycle.
c)
To visualize this process imagine an assembly line that forms a circle with
workers along the way--and their job is to fasten six items together to make some
product.
(1)
Each worker will find it difficult to fasten the items together by him or
herself, because each will need to use two hands for the fastening, yet the
pieces will need to be positioned while they are being fastened--they need
another hand for positioning.
(2)
Now imagine that the first worker has a “mold” to hold the first two pieces
in place while they are fastened--he/she puts the pieces in the mold, fastens
them together and the mold moves to the next worker, with the two pieces
now fastened together still in the mold.
(3)
The next worker adds the third piece to the mold and fastens it to the two
already fastened by worker number 1.
(4)
The mold then moves along the conveyor to workers 3, 4, and 5 who add
pieces 4, 5, and 6 to the mold, fastening each piece to the others previously
fastened together.
(5)
Worker number 5, after fastening the last piece, lifts the completed
product from the mold, and the mold continues to worker number one who
starts the whole process over again.
(6)
This is like an enzymatic cycle in that you start and end with the mold, but
you add pieces to the mold, and later remove them in a different form,
each worker functions like an enzyme, doing a small part of the total job-the workers generate a product along the way , but they also end up with
what they started with, the mold.
d)
Examples from nature include the Krebs cycle and the Calvin cycle, both of which
we will study later.
84
C.
Electron Transport Systems (ETS), and the Chemiosmotic Phosphorulation (COP) they support,
are at the heart of nature’s ATP production, and as a result, its flow of energy.
1.
An electron transport system (ETS), sometimes called an electron transport chain, is a
series of molecules, embedded in a plasma membrane, that move electrons from one
location to another.
a)
At one end of the ETS there must be a source of electrons.
(1)
In the case of photosynthesis, that source is chlorophyll-a (and indirectly,
water molecules).
(2)
In the cellular oxidation of glucose NADH and FADH2 provide the
electrons, and are oxidized back to NAD+ and FAD respectively (they can
then function again as coenzymes).
b)
Electrons are passed from one molecule to another until they get to end of the
chain.
(1)
The first molecule of the ETS gives up the electrons it has received to the
second molecule, which passes them to the third molecule, etc.
(2)
Each molecule in the chain can only accept electrons if it has already given
up its electrons.
c)
At the far end of an ETS there must be a terminal electron acceptor.
(1)
The terminal electron acceptor picks up the electrons at the end of the
ETS, and is reduced in the process.
(2)
The terminal electron acceptor is free of the membrane and will diffuse
away after picking up electrons (and usually hydrogen ions).
(3)
The terminal electron acceptor ensures that electrons are removed from the
system so that a flow of electrons is maintained in the ETS, the importance
of the electron flow is described below.
(4)
In photosynthesis, NADP+ is the terminal electron acceptor, and is reduced
to NADPH; in the cellular oxidation of glucose, oxygen is the terminal
electron acceptor and is reduced to form water.
d)
As long as there is a source of electrons and adequate quantities of the terminal
electron acceptor, there will be a flow of electrons through the ETS.
(1)
This flow of electrons creates a miniature electrical current, with the
potential to do work.
(2)
Some ETS’s use this electrical current to “pump” hydrogen ions across the
plasma membrane in which they are embedded.
(a)
Some of the molecules in an ETS are pore(or channel) proteins that
span the width of the plasma membrane, and are considered
“hydrogen pumps,” or “proton pumps” (hydrogen ions are single
protons).
(b)
As electrons are accepted by the hydrogen pump molecule, they
attract H+ ions on one side of the plasma membrane into a channel
that runs through the center of the molecule.
(c)
The electrons then move the length of the channel, dragging H+
ions with them (because of their opposite electromagnetic charges),
before being passed on to the next molecule in the ETS, leaving the
H+ ions deposited on the other side of the plasma membrane from
where they started.
(d)
This elegant process uses the flow of electrons to move H+ ions
across the plasma membrane.
e)
So what is the point of pumping the Hydrogen ions?
85
The pumping of H+ ions, by the ETS creates a concentration gradient, with
more H+ ions on one side of the plasma membrane than the other.
(2)
As we learned in our discussion of solutions, particles follow their
concentration gradient, i.e. they “want” to diffuse from where they are in
high concentration to lower concentration.
(3)
The highly concentrated H+ ions have potential energy which can be
released as kinetic energy (the energy of movement) as they diffuse across
the membrane--the problem is how do they diffuse back across the
membrane.
(a)
The protons cannot diffuse back through the hydrogen pump
molecules.
(b)
The protons cannot diffuse through the bilipid plasma membrane.
(c)
But they can diffuse through a channel in a large enzyme called
ATP synthase, that spans the plasma membrane--and this diffusion
of H+ ions through the ATP synthase powers ATP production via
chemiosmotic phosphorulation.
+
The H ions, pumped by the electrical current of an ETS, drive chemiosmotic
phosphorulation.
a)
H+ ions are concentrated on one side of a plasma membrane by the hydrogen
pumps of an ETS.
b)
The H+ ions will diffuse back across the membrane, through a large channel
protein called ATP synthase-- an enzyme that makes ATP.
c)
As H+ ions diffuse through the channel in the ATP synthase, they trigger shape
changes in the ATP synthase enzyme, causing it to bind ADP and free phosphate,
releasing them as ATP.
d)
The name “chemiosmotic phosphorulation” is a mouthful, but it describes the
process--the osmosis (diffusion across a plasma membrane) of chemicals
(Hydrogen ions), drives phosphorulation (the binding of phosphate to another
molecule).
e)
The importance of chemiosmotic phosphorulation cannot be overstated-Chemiosmotic phosphorulation produces vast quantities of ATP without using the
hydrolysis of ATP as an energy source, instead, exploiting the energy generated by
electron flow, hydrogen pumping, and the diffusion of charged particles to drive
ATP production.
An electron transport system with chemiosmotic phosphorulation is diagrammed below.
(1)
2.
3.
AN ELECTRON TRANSPORT SYSTEM WITH CHEMIOSMOTIC PHOSPHORULATION
2H+
2H+
+
2H
Plasma membrane
2e-
2e
Electron source
2H+
2H+
2eHydrogen
2ePump
2H+
2H+
+
2H
2H+
2H+
2H+
2H+
ATP
Synthase
2e2e-
ADP + Pi
2e
Terminal
electron acceptor
86
ATP
2H+
87
Lesson 11: Energetics, Photosynthesis
I.
Let us now reexamine the crucial process of photosynthesis, as practiced by plants, in more detail.
A.
The “overall” equation for photosynthesis is as follows:
6CO2 +
Carbon dioxide
B.
Chlorophyll-a
6H2O + 686Kcal (solar E) -------------------> C6H12O6 + 6O2
water
glucose
oxygen
Please note: the video shows carbon dioxide as “carbon,” this is an error, it should be
carbon dioxide.
1.
The reaction requires the green pigment, chlorophyll-a, as a catalyst.
2.
The process of photosynthesis yields an organic molecule (glucose) from inorganic ones
(carbon dioxide and water), and releases oxygen.
3.
The importance of photosynthesis is summarized as follows:
a)
It is the only process in nature that converts solar energy, which is useless to
enzymes, to cellular energy in ATP which is useful to enzymes.
b)
It produces organic molecules (glucose) from inorganic molecules (something
consumers cannot do).
c)
It produces the earth’s oxygen supply.
4.
While the process of photosynthesis can be described by this simple reaction, it is actually
a very complex process that is usually considered in two parts, the Light Reaction
(sometimes called the Light Dependent Reaction), and the Dark Reaction (also called the
Light Independent Reaction, the Calvin Cycle, and the 3-Carbon Pathway).
5.
Photosynthesis is verbally described in the sections that follow, but be sure to refer to the
diagrammatic representation of photosynthesis as well.
Photosynthesis takes place within the chloroplast of a plant cell, usually in the leaf of a plant.
1.
Leaves provide a large surface area for the absorption of light, and leaves are composed
of cells which contain chloroplasts.
2.
Chloroplasts have the following structures; be sure to review chloroplast structure using
pictures in your text.
a)
An outer plasma membrane called the outer membrane.
b)
An inner plasma membrane called the inner membrane.
c)
A fluid filled space separates the membranes and is called the outer compartment.
d)
The inner membrane is a highly folded, complex structure.
(1)
The membrane is folded into stacks of membrane called grana (so-called
because they were once thought to be crystals, or grains).
(2)
The highly folded stacks of membrane that form the grana are called
thylakoid membranes, but they are in fact, extensions of the inner
membrane.
(a)
Embedded in the thylakoid membrane are the photosystems (to be
described later), composed of chlorophyll and other molecules that
are involved in the light dependent reaction of photosynthesis.
(b)
The highly folded thylakoid membranes form a huge surface area
for the absorption of light, and an abundance of photosystems.
(3)
A tiny space is formed between the thylakoid membranes and is called the
thylakoid space, but it is actually an extension of the outer compartment.
88
e)
C.
Fluid fills the space surrounding the grana (thylakoid membranes) and is called
the stroma; the stroma is the site of the Dark Reaction of photosynthesis.
f)
The green color associated with chloroplasts and plants is due to the pigment
chlorophyll.
The light reaction of photosynthesis converts solar energy to cellular energy--it is light
dependent.
1.
Before discussing the mechanics of the light reaction, it is important to understand the
role of photosystems.
a)
Photosystems are aggregations of approximately 400 molecules, embedded in the
thylakoid membranes of the grana.
b)
Photosystems are composed, in part, of photosynthetic pigments that form
“antennae molecules” or an “antennae complex” that absorb and concentrate light
energy.
(1)
Electrons of photosynthetic pigments absorb photons of light energy of
specific wavelengths.
(2)
This boost of energy “excites” an electron to move from its normal orbital
into an orbital farther from the nucleus.
(3)
In most photosynthetic pigments, called accessory pigments, the electron
cannot maintain itself in its new orbital (as energy dissipates), and
transfers its energy to an adjacent pigment molecule, as the electron slips
back to its original orbital.
(a)
This transfer of energy is called “resonance transfer,” and the
energy absorbed by the adjacent molecule will excite one of its
electrons to a new orbital.
(b)
The original pigment molecule can now be re-excited and pass on
more energy.
(c)
The molecule that absorbed the energy via resonance transfer will
likewise pass on the energy to another molecule and so on.
(d)
In summary, accessory pigments absorb light energy and resonance
transfer energy which excites their electrons, this energy is
momentarily stored before it is passed on to adjacent pigment
molecules exciting them--it is a dynamic system of absorption and
transfer of energy.
(e)
Examples of accessory pigments include:
(i)
Chlorophyll-b
(ii)
Carotenes.
(iii)
Xanthophylls.
(iv)
Carotenes and xanthophylls account for the “fall colors” of
broadleaf trees.
Please note: in the video Chlorophyll-a, and Chlorophyll-b are
not spelled correctly.
(f)
Accessory pigments are considered “accessory” because, while
important, they are not absolutely required for photosynthesis.
(4)
Chlorophyll-a molecules are also part of the antenna complex of the
photosystem.
(a)
Chlorophyll-a molecules are essential pigments because they are
required for photosynthesis to occur.
89
(b)
2.
Chlorophyll-a molecules are similar to the accessory pigments in
that they will absorb light energy and energy from resonance
transfer, thereby exciting electrons to an outer orbital.
(c)
Chlorophyll-a molecules differ from accessory pigments in that
excited electrons can become so excited that they are lost from the
chlorophyll-a.
(i)
These electrons have potential energy and are the driving
force of photosynthesis.
(ii)
Chlorophyll-a initiates photosynthesis by the loss of
electrons.
(iii)
The loss of these electrons, is, in turn, driven by the
absorption and concentration of solar energy from the
antenna molecules of the photosystem.
(d)
There are two types of chlorophyll-a.
(i)
P-680 chlorophyll-a absorbs light energy with a wavelength
of 680 nanometers (billionths of a meter).
(ii)
P-700 chlorophyll-a absorbs light energy with a wavelength
of 700 nanometers (billionths of a meter).
(iii)
A photosystem is considered a “Photosystem I” if it has a p700 chlorophyll-a at its reaction center, and a “Photosystem
II” if it has a p-680 chlorophyll-a at its reaction center
(describe below).
(5)
The photosystems absorb solar energy most strongly in the violet-blue, and
orange-red wavelengths (green is reflected or transmitted, not absorbed).
c)
Electrons lost by one chlorophyll-a molecule are typically picked up by another
chlorophyll-a molecule that has lost an excited electron (allowing it to be reexcited)--except at the “reaction center” of a photosystem.
(1)
At the reaction center of a photosystem, there is an electron transport
system associated with a chlorophyll-a, and it takes electrons away from
the photosystem.
(2)
Electrons lost to an electron transport system are replaced by electrons
from the enzymatic break-up of water molecules at the reaction center.
Now we will discuss the mechanics of the light reaction.
a)
We will begin our discussion of the light reaction of photosynthesis with a
photosystem II, embedded in the thylakoid membrane of a grana of a chloroplast.
b)
The antennae molecules of the photosystem excite a p680 chlorophyll-a at the
reaction center to lose electrons.
c)
The electrons are picked up by the electron transport system, and carried away
from the p-680 chlorophyll-a.
d)
The electrons lost by the p-680 chlorophyll-a are replaced by electrons generated
from the enzymatic breakup of water molecules.
(1)
A large enzyme embedded in the thylakoid membrane breaks apart water
molecules.
(2)
The water molecules yield the following:
(a)
Electrons, which replace those lost by the p-680 chlorophyll-a
(biologists generally consider water as the “source” of electrons for
the electron transport systems of photosynthesis).
90
H+ ions which can be “pumped” for ATP production by
chemiosmotic phosphorulation, and which also passively follow
electrons that move through the electron transport systems.
(c)
Oxygen gas which diffuses out of the leaf and into the atmosphere.
e)
The electrons moving through the electron transport system of photosystem II
pump hydrogen ions across the thylakoid membrane, from the stroma to the
thylakoid space (outer compartment).
f)
The hydrogen ions pumped across the membrane, diffuse back across the
membrane through ATP synthase enzymes, providing the energy for them to
produce ATP (chemiosmotic phosphorulation).
g)
The electron transport system of a photosystem II ends at the reaction center of a
photosystem I, where p-700 chlorophyll-a molecules are excited by solar energy to
lose electrons to another electron transport system that take the electrons away
from the p-700 molecule.
h)
The electrons lost from p-700 chlorophyll-a are replaced by those from the
electron transport system of photosystem II (with the p-680 chlorophyll-a).
Please note: I inadvertently refer to the photosystem II, as a photosystem I, during
the discussion of electron flow. I apologize if it confuses you.
i)
The electrons flowing through the electron transport system of photosystem I do
not pump hydrogen ions, but do deliver electrons (and passively following H+
ions) to a location where NADP+ will pick them up, reducing the NADP+ to
NADPH.
j)
The NADP+ acts as the terminal electron acceptor, taking electrons away so more
electrons can flow through the electron transport systems of the photosystems.
k)
The importance of electron flow cannot be minimized.
(1)
The flow of electrons through the electron transport systems is driven by
the absorption of solar energy and resulting excitation of chlorophyll-a
molecules, causing them to lose electrons to the electron transport systems.
(2)
Electrons flow through the electron transport systems of the photosystems,
pumping hydrogens across the thylakoid membranes, so they can diffuse
back across the membrane through ATP synthase, triggering the
production of ATP (chemiosmotic phosphorulation).
(3)
Water molecules are considered the source of electrons for the electron
transport systems because electrons lost from chlorophyll-a molecules are
replaced by electrons from the enzymatic breakup of water.
(4)
NADP+ is considered the terminal electron acceptor because it removes
electrons at the very end of electron transport, thereby allowing electron
flow through the electron transport systems of photosystems II and I,
pumping H+ ions, kinetic energy for the production of ATP.
3.
ATP and NADPH produced by the light reaction are required for the Dark Reaction.
a)
The ATP will provide cellular energy to the enzymes of the Dark Reaction so they
can construct glucose--this conversion of solar energy to cellular energy is key to
life, enzymes cannot use solar energy to do work.
b)
The NADPH will provide hydrogen atoms (from the enzymatic break up of water
molecules) for the construction of glucose.
The Dark Reaction is the “construction phase” of photosynthesis, i.e. this is where glucose is
made.
1.
The dark reaction is an enzymatic cycle, that does not require light, although it needs
products of the light reaction.
(b)
D.
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E.
Please note: I refer to the dark reaction as a “chemical cycle,” in the video. Consider a
chemical cycle to be the same as an enzymatic cycle.
2.
The cycle can be divided into three basic operations.
a)
The first step is the fixation of carbon.
(1)
Adding carbon from an inorganic source (in this case, carbon dioxide), to
an organic molecule, is called fixation of carbon.
(2)
The cycle begins with six molecules of a sugar named Ribulose
Bisphosphate.
(a)
Ribulose Bisphosphate is abbreviated RuBP.
(b)
RuBP is a five carbon sugar.
(3)
An enzyme named RubisCO binds the six RuBP’s and adds carbon from
carbon dioxide to each RuBP.
(a)
This step requires six carbon dioxide molecules to make one
glucose (remember, glucose is a six carbon sugar).
(b)
The product of this step, is six molecules, each composed of six
carbons.
b)
The next part of the cycle is the production of glucose.
(1)
The six carbon molecules generated by the fixation of carbon dioxide,
undergo a series of reactions that eventually remove six carbons from the
cycle in the form of glucose.
(2)
This part of the cycle requires ATP and NADPH from the light reaction.
(a)
The ATP provides cellular energy to the enzymes.
(b)
The NADPH provides hydrogen atoms for the construction of
glucose (the original source of the hydrogen atoms is water
molecules).
c)
The last part of the cycle is the regeneration of the six RuBP molecules we started
with.
(1)
The molecules left over from the production of glucose undergo another
series of reactions, and are converted back to six molecules of RuBP.
(2)
This regeneration of RuBP sugars requires more ATP from the light
reaction.
(3)
The dark reaction (or Calvin Cycle) is an enzymatic cycle because we start
and end with six RuBP molecules.
3.
To produce one glucose molecule the dark reaction requires the following:
a)
Six carbon dioxide molecules from the atmosphere.
b)
Eighteen ATP molecules from the light reaction.
c)
Twelve NADPH molecules from the light reaction.
4.
Products of the Calvin Cycle are as follows:
a)
Glucose.
b)
ADP and phosphate (which can be reused by the light reaction to generate more
ATP).
c)
NADP+ which can pick up more electrons ( and hydrogen ions) from the light
reaction.
A diagrammatic representation of photosynthesis.
The Light Reaction
Thylakoid Space of Chloroplast
Reaction Center of Photosystem II
Reaction Center of Photosystem I
92
Solar Energy
2H+
Solar Energy
93
Please note: I strongly suggest you review the video to this point, one or more times, before
proceeding. I have gone relatively fast, and it is a complex process that deserves repeated
exposure.
F.
There are a few subtle nuances to photosynthesis we have not yet discussed.
1.
There is a process by which electrons can be recycled in the light reaction.
a)
If NADP+ is in short supply, another electron carrier can pick up electrons and
return them to the electron transport system at the reaction center of a
photosystem II.
b)
This allows the electrons to pump more hydrogens, leading to more ATP
production per electron.
c)
Not all plants are equally efficient at electron cycling.
d)
The typical efficiency for plants, is one ATP produced per electron released by the
lysis of water.
2.
In looking at the process you may wonder, “Why make glucose?”
a)
Once the plant makes glucose, the carbons in glucose can be used to make other
organic molecules, like polysaccharides, nucleic acids, lipids, proteins, etc.
(1)
These molecules are used to build the tissues of the plant, allowing it to
grow.
(2)
The plant will build these other molecules if it can absorb the required
elements from the soil (e.g. nitrogen, phosphorus, etc.)
b)
Glucose has potential cellular energy in its carbon-carbon bonds.
(1)
ATP is relatively unstable, its bonds are easily broken, and storing energy
in the form of ATP is wasteful.
(2)
Glucose can be disassembled in the process of cellular oxidation,
generating ATP for cellular work.
(3)
Glucose is a more stable form in which to store energy.
(4)
This stability is magnified when the plant binds glucose to form starch.
c)
When plant enzymes need ATP to do work they have two possible sources of
ATP.
(1)
When there is light, the plant can rely on the light reaction of
photosynthesis for ATP production.
(2)
When it is dark, the plant must turn to its stores of glucose for the
production of ATP.
3.
You may also have wondered how carbon dioxide gets into the leaf (and eventually the
cells and chloroplasts).
a)
The surface of most leaves is covered by a waxy substance called the cuticle.
(1)
The cuticle is impermeable to water, and prevents water loss by the plant.
(2)
The cuticle is also impermeable to gases, including carbon dioxide.
(3)
This creates a problem because carbon dioxide is required for
photosynthesis, but it cannot diffuse through the cuticle.
b)
Fortunately, the undersurface of most leaves has openings into the interior of the
leaf.
(1)
The openings are called stomata (the singular is stoma).
(2)
Each stoma is surrounded by two bean shaped cells, called guard cells, that
form the opening or stoma.
94
4.
5.
6.
(3)
The size of the stoma is controlled by the guard cells.
c)
The stomata can be opened or closed, and either situation creates problems for the
plant.
(1)
If the stomata are open, carbon dioxide can diffuse into the leaf interior for
photosynthesis, but water vapor will diffuse out, dehydrating the plant.
(2)
If the stomata are closed, the opposite problems result, i.e. water is
conserved, but there is no carbon dioxide for photosynthesis.
d)
How then, do the guard cells control the stomata.
(1)
The fluid (cytoplasm) within guard cells exerts pressure, internally, on the
cell walls of a plant cell—it is known as turgor pressure.
(2)
When turgor pressure is low within the guard cells (due to dehydration),
they sag together, closing the opening (stoma) between them.
(3)
When turgor pressure is high within the guard cells, the guard cells
become more curved, creating an opening between them, i.e., the stoma
will open.
(4)
Potassium ions are actively transported in or out of guard cells to increase
or decrease turgor pressure, respectively.
e)
Plants in arid habitats tend to grow slowly because they cannot afford to have the
stomata open for long periods of time--this saves water, but limits photosynthesis,
and growth.
The RubisCO enzyme, which fixes carbon dioxide in the dark reaction, is a terribly
inefficient enzyme.
a)
While most enzymes carry out 25,000 to 35,000 reactions per second, RubisCO
carries out only 2 to 3 reactions per second (10,000X less efficient).
b)
Furthermore, RubisCO will bind oxygen as readily as it binds carbon dioxide,
triggering a process known as photorespiration.
(1)
In photorespiration, RubisCO binds oxygen causing it to break up RuBP,
consuming ATP and releasing carbon dioxide in the process.
(2)
This is exactly the opposite of what photosynthesis will do.
(3)
It is thought that photosynthesis evolved when there was an abundance of
carbon dioxide, and little atmospheric oxygen, so RubisCO could afford
the inefficiency.
Some plants, called C4 plants, have evolved techniques to reduce photorespiration and
water loss.
a)
C4 plants are typically found in arid regions.
b)
They have evolved techniques to move carbon dioxide into cells quickly, so the
stomata do not have to remain open for long periods.
(1)
C4 plant cells bind carbon dioxide to a three carbon molecule, forming a
four carbon molecule (hence, C4) as it diffuses into a cell.
(2)
The carbon dioxide is slowly released to RubisCO throughout the day,
minimizing the effects of photorespiration, while conserving water.
CAM plants are similar to C4 plants but with an added adaptation to conserve water--they
close their stomata in the day, and open their stomata at night--many desert cacti are
CAMs.
95
96
Lesson 12: Energetics, The Cellular Oxidation of Glucose
I.
It is now time for a detailed look at the other major energetic process in nature, the cellular oxidation of
glucose.
A.
The overall equation for the cellular oxidation of glucose is:
C6H12O6
glucose
B.
C.
D.
+
6O2 ---->
oxygen
6H2O +
water
6CO2
+
carbon dioxide
36ATP +
263 Kcal
heat
423Kcal
As you can see this process in some ways reverses the outcome of photosynthesis.
1.
Glucose and oxygen, the products of photosynthesis, yield water and carbon dioxide, the
reactants of photosynthesis.
2.
The processes are interdependent, recycling carbon, hydrogen, and oxygen.
The importance of the cellular oxidation of glucose is that it yields ATP for cellular work.
1.
Consumers cannot generate ATP from solar energy.
2.
Consumers generate their ATP from the cellular oxidation of glucose.
3.
Consumers cannot make glucose from inorganic sources, so they must ingest glucose or
other organic molecules that can be converted to glucose.
4.
Plants also generate ATP by this process, when it is dark, or when they need more ATP
than can be produced via the light reaction.
The first phase of the cellular oxidation of glucose is glycolysis (glyco = sugar, lysis=break
apart).
1.
Glycolysis is an enzymatic pathway, that takes place in the cytosol or cytoplasm (not
within an organelle) of all cells.
a)
Glycolysis does not require oxygen.
b)
Glycolysis occurs in all cells be they eukaryotic or prokaryotic--it is a ubiquitous
process in nature.
2.
Glycolysis begins with the molecule, glucose.
3.
Glucose, a six carbon sugar, goes through a series of enzymatic reactions, eventually
yielding two molecules of pyruvate.
a)
Pyruvate is a three carbon sugar.
b)
Two pyruvates are generated per glucose.
4.
Glycolysis yields a net gain of 2 ATP per glucose molecule.
a)
Glycolysis consumes 2 ATP early in the enzymatic pathway.
b)
Late in the enzymatic pathway 4 ATP are produced (in what is called a substrate
level phosphorulation, a simple, but relatively rare type of ATP production).
c)
The net gain is 2 ATP per glucose.
Please note: In the video graphic of glycolysis it shows 2ATP in the IN colum, and 4
AT in the OUT column. It should read 4 ATP in the OUT column.
5.
Glycolysis also requires the coenzyme NAD+ , which is reduced to NADH.
6.
Glycolysis can be aerobic or anaerobic.
a)
The term “aerobic,” means oxygen is present, and yields two pyruvates per
glucose, as already described.
b)
The term “anaerobic,” means without oxygen, and anaerobic glycolysis is
sometimes described as fermentation.
(1)
In mammals, anaerobic glycolysis is called lactate fermentation.
(a)
Pyruvate is bound by an enzyme that converts it to lactate.
(b)
The conversion of pyruvate to lactate requires the oxidation of
NADH back to NAD+ (reducing pyruvate to lactate).
97
(i)
E.
F.
The importance of lactate fermentation is that it regenerates
NAD+.
(ii)
NAD+ is required for glycolysis, and as long as glycolysis
can occur, ATP is produced (albeit, two ATP per glucose is
not very much).
(iii)
Even when oxygen is missing, ATP can be made--this may
be the difference between life and death in some situations.
(c)
Fortunately, lactate fermentation is reversible.
(i)
When oxygen is present, lactate is converted back to
pyruvate, reducing NAD+ back to NADH.
(ii)
We typically experience lactate fermentation in our muscles
when exercising.
(iii)
The muscles get oxygen deprived, producing lactate which
causes the muscles to “burn.”
(iv)
When we stop exercising oxygen levels rise in the muscle
tissue, and lactate is converted back to pyruvate.
(2)
Yeast, and many bacteria practice alcohol fermentation.
(a)
In alcohol fermentation, a carbon dioxide molecule is removed
from the pyruvate, while oxidizing NADH back to NAD+ .
(b)
This changes the three-carbon pyruvate sugar, into a two carbon
alcohol (ethyl alcohol).
(c)
This process is not reversible.
(d)
This process produces our alcoholic beverages.
After glycolysis, the next step in the cellular oxidation of glucose, is cellular respiration.
1.
Cellular respiration occurs within the mitochondria.
a)
Mitochondrial structure is not dissimilar to a chloroplast.
b)
There is an outer plasma membrane and an inner plasma membrane, with a fluid
filled outer compartment between the two.
c)
Like the chloroplast, the inner membrane is highly folded.
(1)
The folds are called cristae, and increase surface area.
(2)
Imbedded in the inner membrane are mitochondrial electron transport
systems (with hydrogen pumps), and ATP synthase enzymes, for the
chemiosmotic phosphorulation of ATP.
d)
Within the inner membrane is a fluid filled compartment known as the matrix of
the mitochondrion.
e)
The matrix contains enzymes for the conversion of pyruvate to acetyl coenzyme
A, and the Kreb’s cycle (to be discussed below).
2.
Cellular respiration requires oxygen.
3.
Prokaryotes cannot carry out cellular respiration because they lack mitochondria.
4.
Prokaryotes and some unicellular eukaryotes can survive on glycolysis alone (2 ATP per
glucose), but multicellular eukaryotes need more ATP per glucose, than glycolysis can
provide.
5.
Cellular respiration adds another 34 ATP to the 2 ATP produced in glycolysis for a total
of 36 ATP per glucose.
Let’s now consider the sequence of events involved in cellular respiration.
1.
If oxygen is present, pyruvate (2 per glucose) will diffuse from the cytosol, through the
outer and inner mitochondrial membranes, and into the matrix of the mitochondria.
2.
Here pyruvate is converted to acetyl coenzyme A (Acetyl CoA).
98
a)
3.
4.
In this conversion pyruvate is bound by an enzyme that removes CO2 from the
three carbon pyruvate, making it a two carbon chain, also called an acetyl group.
b)
The enzyme then attaches the two carbons to a coenzyme, coenzyme A, to form
Acetyl Coenzyme A (abbreviated Acetyl CoA).
c)
This step also reduces NAD+ to NADH.
d)
The acetyl coenzyme A, shuttles carbons into the next step in cellular respiration,
the Krebs Cycle.
e)
Two Acetyl CoA’s are produced per glucose.
The Krebs Cycle is an enzymatic cycle named for its discoverer, Krebs.
a)
The Krebs Cycle begins and ends with a four carbon molecule named
oxaloacetate.
b)
An enzyme binds oxaloacetate and Acetyl CoA, and transfers the acetyl group (the
two carbons) from Acetyl CoA to the oxaloacetate to produce a six carbon
molecule called citrate.
(1)
The Coenzyme A is released by the enzyme, and it can be reused in the
conversion of pyruvate to Acetyl CoA.
(2)
For each glucose we begin with, two Acetyl CoA’s are produced, which
will cause two “turns” of the Krebs Cycle.
c)
The Krebs cycle then subjects the citrate to a series of reactions that will
eventually regenerate oxlaoacetate so the cycle can begin anew--several enzymes
require NAD+ or FAD to remove electrons.
d)
The major actions of the Krebs cycle are as follows:
(1)
Two carbons are removed from citrate in the form of carbon dioxide to
regenerate oxaloacetate (four carbon dioxide per glucose).
(2)
NAD+ is reduced to NADH (six per glucose).
(3)
FAD is reduced to FADH2 (two per glucose).
(4)
ATP is produced via substrate level phosphorulation (two per glucose).
e)
The Krebs cycle as you can see, does not produce any more ATP than glycolysis
(2 ATP per glucose), but it does produce an avalanche of NADH and FADH2
,whose electrons have the potential to produce volumes of ATP via chemiosmotic
phosphorulation.
The last part of cellular respiration is the mitochondrial electron transport system (ETS),
and the chemiosmotic phosphorulation it supports.
a)
Mitochondrial electron transport systems and ATP synthase enzymes are
embedded in the inner membrane of the mitochondrion.
b)
The mitochondrial ETS will pump hydrogens at three locations.
c)
The source of electrons for the mitochondrial ETS is NADH and FADH2
produced by glycolysis, the conversion of pyruvate to Acetyl CoA, and the Krebs
Cycle.
d)
NADH and FADH2 give up their electrons to the mitochondrial ETS and are
oxidized back to NAD+ and FAD respectively--they are now free to go pick up
more electrons in the reactions mentioned above, and will be recycled over and
over.
e)
As the electrons flow through the mitochondrial ETS they pump H+ from the
matrix into the outer compartment.
f)
The H+ diffuse through ATP synthase back into the matrix, causing the ATP
synthase to make ATP as we have discussed previously.
g)
Chemiosmotic phosphorulation accounts for 32 ATP per glucose, out of a total of
36 ATP produced by the entire process of cellular oxidation of glucose.
99
h)
G.
There is one factor left to consider in the mitochondrial ETS, and it is the terminal
electron acceptor-- oxygen.
(1)
NADH and FADH2 are the source of electrons for the ETS, and to
maintain flow, oxygen gas must be at the end of the ETS to pick up
electrons.
(2)
Oxygen will bind electrons and hydrogen protons, which will passively
follow electrons, to form water.
(3)
This is why we need oxygen--to pick up electrons at the end of the
mitochondrial ETS, and in the process it is chemically reduced to form
water.
5.
Why do we die when we do not get oxygen?
a)
Without oxygen, there is no terminal electron acceptor to pick up electrons at the
end of the mitochondrial ETS.
b)
Without the terminal electron acceptor, electron flow through the ETS stops.
c)
Without electron flow, H+ pumping stops, and chemiosmotic phosphorulation
stops, thus eliminating the production of 32 ATP per glucose.
d)
Without electron flow, NADH and FADH2 cannot be oxidized back to NAD+ and
FAD, creating a shortage of NAD+ and FAD.
e)
Without NAD+ and FAD, the enzymes in the Krebs Cycle that require them,
cannot function, so the Krebs Cycle shuts down, eliminating another 2 ATP per
glucose.
f)
Without NAD+, pyruvate cannot be converted to Acetyl CoA, so pyruvate
accumulates in the mitochondrion, and will no longer diffuse into the
mitochondrion from the cytoplasm (because there is no concentration gradient).
g)
Pyruvate accumulates in the cytosol, where it is converted to lactate, regenerating
a tiny amount of NAD+, which keeps glycolysis going until toxic levels of lactate
accumulate.
h)
Cells are forced to survive on the 2 ATP per glucose, generated by glycolysis.
i)
Within minutes, available ATP is consumed, enzymatic reactions stop, and death
follows.
Carbohydrates, proteins, and lipids are linked by cellular respiration.
1.
I have previously mentioned that carbohydrates, proteins, and lipids have energy value-this means they can be used to generate ATP.
a)
Carbohydrates and proteins yield 4 Kcal/gram, and lipids yield 9 Kcal/gram.
b)
The carbons in proteins and lipids can be enzymatically broken into two-carbon
chains and attached to Coenzyme A (CoA) to form Acetyl Coenzyme A (Acetyl
CoA), just like carbohydrates.
c)
These AcCoA then enter the Kreb’s cycle generating ATP in the Kreb’s cycle, or
via chemiosmotic phosphorulation.
2.
Since all foodstuffs can be converted to Acetyl CoA, one type of food can be converted to
another.
a)
For example, if you consume more energy than you burn, carbohydrates or
proteins can be broken down to Acetyl CoA, and the carbons can be diverted to a
pathway to make lipids--which can be stored as fat.
b)
This is why a person may reduce fat intake yet still gain weight and body fat.
c)
Lipids, on the other hand, can contribute carbons via Acetyl CoA for the
production of carbohydrates and non-essential amino acids.
100
H.
I.
It is also important to know that plant cells have mitochondria and will consume some of the
glucose and oxygen generated by photosynthesis, to produce ATP by the cellular oxidation of
glucose.
1.
Photosynthetic cells may need ATP at night.
2.
Some plant tissues, root tissues for example, never photosynthesize and are dependent on
the cellular oxidation of glucose for their ATP.
The cellular oxidation of glucose is highly conserved in nature, with glycolysis practiced by all
organisms and cellular respiration practiced by all eukaryotes, and it is one of many lines of
evidence that suggest a common ancestry of organisms.
101
J.
A diagrammatic representation of the cellular oxidation of glucose.
GLYCOLYSIS
IN
2 ATP
2 NAD+
2NAD+
1 Glucose
2NADH
2 Ethanol + 2CO2
Glycolysis yields a net gain
of 2 ATP per glucose
OUT
4 ATP
2 NADH (electrons to ETS, oxidized back to NAD+ )
2NADH
2NAD+
2 Lactate
2 Pyruvate
Alcohol Fermentation (yeasts)
Lactate Fermentation (mammals)
Oxygen from atmosphere
Outer membrane of mitochondrion
+
2H+ + 2H
2H
Electron
Transport
System
Outer compartment
Inner membrane of mitochondrion
2e-
CELLULAR
RESPIRATION
2H+
2H+
2H
Matrix of
mitochondrion
+
ATP (32 per
2H glucose)
+
2H+
OUT
NADH
FADH2
CO2
ATP (2 per
glucose)
To ETS
From ETS
IN
H2O (formed when oxygen, the
terminal electron acceptor, binds
electrons from the ETS and
Hydrogen ions, that are following
electrons in the matrix)
NAD+
FAD
ADP
Pi
Pyuvate
NAD+
NADH
ATP Synthase
ADP + Pi
Coenzyme A
Acetyl CoA
+
CO2
OUT
Citrate (6 C)
Krebs
Cycle
(2 ATP per
glucose)
IN
Oxyloacetate (4C)
102
103
Lesson 16, Mitosis
I.
II.
Cell division plays a central role in the processes of growth and reproduction.
A.
Growth is an increase in size.
1.
Unicellular (one celled) organisms, like an amoeba, grow because the individual cell gets
larger.
2.
Multicellular organisms grow because we produce more cells.
B.
Reproduction is the process of generating new individuals.
1.
Asexual reproduction does not involve a mate.
a)
A unicellular organism may divide to form a population of two individuals.
b)
A cutting from a plant may grow into a new plant.
c)
Asexual reproduction produces clones, i.e. individuals that are genetically
identical to one another.
2.
Sexual reproduction requires a mate.
a)
Each partner in sexual reproduction contributes DNA to the next generation.
b)
Offspring are not genetically identical to the parents.
C.
We will be considering cell division and reproduction of eukaryotic organisms.
Before getting into the mechanics of cell divisions and their relationship to reproduction, we need to
discuss the topic of ploidy.
A.
Ploidy refers to the number of sets of chromosomes in a cell nucleus.
B.
The term haploid (1n) means a cell has one complete set of chromosomes.
1.
A cell may have many different chromosomes, each with different genes.
2.
Humans have 23 chromosomes, each with different characteristics encoded in the DNA.
3.
A haploid cell in humans, then, has one set of the 23 chromosomes.
4.
In humans, the only haploid cells in our bodies are the sex cells, also called gametes
(sperm and ovum).
C.
The term diploid (2n) means that a cell has two complete sets of chromosomes.
1.
In a diploid cell, the chromosomes occur in what are called "homologous pairs."
2.
In humans, a diploid cell has two sets of 23 chromosomes, or 23 homologous pairs.
a)
The homologous pairs of chromosomes are similar but not identical.
b)
Homologous chromosomes carry genes for the same general traits, but not
necessarily exactly the same gene.
(1)
Chromosome number 1, for example, may carry a gene for eye pigment.
(2)
That gene comes in two versions: one that produces brown pigment, and
one that does not produce a brown pigment.
(3)
The different versions of a gene are called alleles.
(4)
A diploid cell, then, would have two of the chromosome number 1, but
there could be three possibilities if you looked at the eye pigment gene
pair:
(a)
Each chromosome could have the brown pigment allele.
(b)
Each chromosome could have the allele for no pigment.
(c)
One chromosome could have an allele for brown pigment, and the
other chromosome an allele for no pigment.
3.
In humans, and most organisms, all cells except for the sex cells are diploid cells.
4.
The diploid cells in humans are called somatic(body) cells.
D.
The term polyploidy means many sets of chromosomes.
1.
Many organisms will have multiple sets of chromosomes: triploid (3n), tetraploid(4n),
etc.
2.
Many domesticated plants are polyploid.
104
III.
Mitosis is the process of eukaryotic cell division, and it is part of a larger process called the cell cycle.
A.
The mitotic cell cycle has the following characteristics:
1.
It begins with a parent cell and yields two daughter cells.
2.
The daughter cells will be genetically identical to the parent cell and one another.
3.
The ploidy of cells does not change from parent to daughter cells.
4.
In humans, one diploid parent cell yields two diploid, genetically identical daughter cells.
B.
The major steps in a mitotic cell division include the following:
1.
The chromosomes are replicated.
2.
The chromosomes are separated into two identical groups within the cell (karyokinesis).
3.
The cytoplasm is divided yielding two genetically identical cells (cytokinesis).
C.
The cell cycle is a reproductive cycle of a cell, rather than a reproductive cycle of an organism.
D.
We are going to begin discussion of the cycle with a stage called interphase.
Please note: In the cell cycle graphic Metaphase is misspelled, “metaphase.”
1.
An interphase cell is a relatively “normal” looking cell when viewed under a light
microscope.
a)
The nucleus is intact.
b)
The chromatin is not condensed into visible chromosomes or chromatids.
2.
Interphase is sometimes called the “resting phase,” although this is not an accurate
representation of the phase.
a)
An interphase cell is typically a very active cell metabolically.
b)
Events in the interphase cell determine whether the cell will divide or not.
3.
There are three subphases to interphase--keep in mind that in all of these subphases of
interphase, the cell looks “normal.”
a)
Gap Phase 1, or G1 immediately follows the previous cell division.
(1)
Each new daughter cell is half the size of the original cell.
(2)
The cell must grow in size.
(3)
New organelles must be produced (mitochondria, golgi, ER, etc.).
b)
G1 is followed by Gap Phase 0, or G0 .
(1)
G0 is a non-dividing period for a cell.
(2)
In fetal cells or bone marrow cells, G0 will be absent, or last for only a few
minutes, because the business of these cells is to divide and make more
cells.
(3)
In some adult cells, an adult brain cell for example, G0 may last for
decades.
(4)
A G0 cell is not dividing, but it does carry out its normal “business” of
being a brain cell, or liver cell, or muscle cell, etc.
(5)
The cell is active, it is just not in the business of dividing while in G0.
c)
G0 is followed by the Synthesis Phase, or S phase.
(1)
The S-phase is the time of chromosome replication.
(2)
The importance of chromosomal replication is this: once chromosomes
begin DNA replication, the cell will divide (within minutes to hours).
(3)
Remember this process of replication is taking place in the nucleus of the
interphase cell, so the cell looks normal.
(4)
Each of the 46 chromosomes (23 homologous pairs) in a diploid somatic
cell nucleus are replicated.
(5)
The replicates are attached to one another by a mass of protein known as
the centromere.
(6)
While replicates are attached to one another they are called sister
chromatids.
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(7)
E.
Your book will describe the relationship between sister chromatids and
chromosomes as “a chromosome composed of two sister chromatids,” but
the reality is that each sister chromatid is a chromosome attached to its
replicate by a centromere.
(8)
Your book will also describe the cell as diploid after replication, but
functionally, the cell is now 4n.
(a)
Remember the 2n cell has two copies of each type of chromosome.
(b)
After chromosomal replication there are four copies of each type of
chromosome.
(c)
Eventually these four copies will be separated into two equal
halves, in separate cells, to reestablish a ploidy of 2n for each cell.
(9)
As mentioned previously, the S phase is extremely important because it is
the critical phase in cell division.
(a)
Once DNA replication begins it triggers cell division.
(b)
It is a time of intense interest to scientists.
(i)
Cancers are uncontrolled cell divisions, so understanding
controls of chromosomal replication is important.
(ii)
Regeneration of tissues is another area of interest.
(a)
Mature cells do no divide readily in most human
tissues.
(b)
If division could be stimulated, tissues killed in
heart attack, stroke, disease, and injury could
potentially be regenerated.
d)
The S phase is followed by Gap Phase 2 or G2 .
(1)
The cell begins to accumulate proteins, ATP and other molecules needed
for mitosis.
(2)
The pair of centrioles found in animal cells also replicate during G2 .
Interphase is followed by mitosis in the cell cycle--this is the actual cell division, and the cell will
not look “normal” in mitosis.
1.
Prophase is the first phase of mitosis,
a)
In prophase the chromosomes begin to hypercoil becoming visible.
(1)
Chromosomes (sister chromatids) are extremely long and thin.
(2)
They must be hypercoiled so they can be moved efficiently.
(3)
If you had 100 yards of thread and it needed to be moved from your
bedroom to your living room, you would spool or coil the thread before
moving it--same thing in the cell with DNA.
b)
The nucleus looks granular at the beginning of prophase, then expands, and one
finally sees chromosomes.
c)
The nuclear envelope disintegrates in early prophase, and nucleolus disappears.
d)
The chromosomes (sister chromatids) coil in a very precise manner, determined
by the nitrogenous base sequence-- not arbitrary coiling/folding.
e)
The sister chromatids are joined by a centromere.
f)
The centrioles replicated in interphase, begin to move towards either pole (end)of
the cell constructing a spindle of microtubules between them.
(1)
Only animal cells have centrioles.
(2)
Plant cells will also form a spindle even without centrioles.
(3)
The spindle is "football" shaped, and extends from one pole to other.
(4)
The mitotic spindle is composed of small protein tubules, called
microtubules, which are composed of the protein, tubulin.
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(5)
2.
3.
4.
Some microtubules align with others to run the length of the spindle
(continuous tubules), while others run from the pole to the chromatids
(discontinuous tubules), where they attach to the kinetochore of the
centromere.
g)
Astral rays radiate from the centrioles in animal cells (absent in plants) forming
the aster, and attaching to the cell membrane.
(1)
The aster is composed of microtubules.
(2)
The aster provides leverage for chromosomal movement, and plays a role
in animal cell cytokinesis (dividing the cell in half).
Please note: I inadvertently call a plant cell an animal cell in the micrographs
of prophase.
Metaphase follows prophase--in metaphase the chromatids are arranged at the equator of
the spindle.
a)
Spindle fibers from each pole of the spindle attach to a portion of the centromere
called the kinetochore, of the sister chromatids.
b)
In metaphase the sister chromatids are arranged at the equator of the spindle.
Anaphase follows metaphase.
a)
In anaphase, the kinetochore of the centromere "replicates"--this allows the
microtubules from each end of the spindle to pull the chromatids apart forming
sister chromosomes.
(1)
One sister chromosome gets pulled to one end (pole) of the cell, and the
other sister chromosome to the other end.
(2)
Discontinuous microtubules slide past continuous microtubules providing
the power to separate the chromatids.
b)
When viewed with the light microscope, the tails of the sister chromosomes will
trail behind the centrioles as they are pulled to the poles, distinguishing anaphase
from prophase or metaphase.
c)
This separation of sister chromosomes, one to each end of the cell, occurs for all
46 paired sister chromatids.
(1)
Remember that sister chromatids and the subsequent sister chomosomes
are genetically identical to one another.
(2)
If one set of the 46 sister chomosomes goes to one end of the cell, and the
other identical set of 46 sister chromosomes goes to the other end of the
cell, then the cell has divided its genetic material into two identical sets.
(3)
Furthermore, the 46 chromosomes at each end of the cell constitute 23
homologous pairs of chromosomes--all that remains is to divide the
cytoplasm in two, and we will have two genetically identical diploid cells.
(4)
This separation of the genetic material is called cytokinesis.
Telophase follows anaphase, and is the last phase of mitosis, and the cell cycle.
a)
Telophase begins when sister chromosomes have been pulled to either end of the
cell.
b)
The mitotic spindle and asters deteriorate, and a nuclear envelope will begin to
form around each mass of 46 sister chromosomes.
c)
The most visible sign of telophase is the beginning of cytoplasmic division, called
cytokinesis.
(1)
In animals, cytokinesis begins with invagination of the cell membrane,
from the outside in.
(a)
Astral rays cross at the equator of the cell, and pull in the cell
membrane.
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(b)
F.
G.
Eventually the cell membrane will pinch the cytoplasm (and
spindle) in half.
(2)
In plants, cytokinesis begins at the equator of the spindle, with the
formation of a cell plate which develops into a cell wall, partitioning the
cytoplasm in two.
d)
Telophase ends with the formation of two daughter cells, complete with newly
formed nuclei.
(1)
The daughter cells are genetically identical to one another.
(2)
The daughter cells are genetically identicallidentical to the parent cell.
(3)
Following mitosis the parent cell no longer exists.
(4)
Ploidy does not change from parent to daughter cell in mitosis.
The daughter cells are now in G1 of interphase, and the cell cycle continues.
Mitosis has the following uses in nature.
1.
It is the process by which multicellular organisms produce new cells in growth and
development (from feriltization to mature organism).
2.
It is a method of asexual reproduction.
a)
Unicellular organisms mitotically divide to produce more individuals.
b)
Plants, and a few animals (sponges), can generate new individuals through
mitosis.
c)
Mitotic asexual reproduction generates genetically identical individuals.
108
Lesson 17, Meiosis and Sexual Reproduction
I.
II.
Sexual reproduction requires a different kind of division.
A.
In sexual reproduction, sex cells (gametes) will fuse to form a new cell composed of genetic
material from each parent.
1.
If two diploid cells fused, it would yield a 4n offspring.
2.
If two 4n offspring mated they would produce an 8n offspring, etc. Etc.
3.
As you can see, there would soon be no room in a cell for all of the DNA, if this were the
case.
4.
The reality is that in sexual reproduction the ploidy does not change from generation to
generation, diploid humans mate and produce diploid babies.
B.
The evolution of a different kind of cell division permitted later evolution of sexual
reproduction--this “other” cell division is meiosis, also called reduction division.
1.
Meiosis begins with a “germ” cell that will yield gametes.
2.
A diploid germ cell will undergo meiosis to yield four haploid cells (gametes).
a)
Gametes are obviously genetically different from the parent or germ cell because
they have half the ploidy.
b)
The four haploid cells are also genetically different from one another due to two
processes we will discuss later: crossing over and independent assortment.
3.
Haploid gametes fuse to form a diploid cell, that will mitotically grow into a new
individual.
4.
Meiosis occurs within the sex organs (gonads), the testes of males, and ovaries of
females.
5.
Sexual reproduction cannot occur without meiosis.
C.
Sexual reproduction is summarized below.
1.
Within the gonads of each mate meiosis occurs.
a)
The cells of the gonad are diploid, but specific cells, called germ cells, will
undergo meiosis.
b)
Each diploid germ cell divides meiotically to yield four haploid (n) cells, which
may develop into sex cells called gametes.
c)
Males produce sperm, females produce ova.
2.
A single haploid sperm will fuse with a single haploid ovum in a process called
fertilization or syngamy.
3.
Syngamy yields a fertilized ovum called a zygote, which is diploid (2n).
4.
The zygote divides mitotically into a multicellular organism.
5.
All cells of the multicelluar organism are genetically identical due to the mitotic division
that produced them.
D.
The major events of meiosis are summarized below.
1.
The chromosomes replicate.
2.
Sister chromatids of homologous pairs mix genetic traits in a process called crossing
over.
3.
The cell at this point has four replicates of each type of the 23 chromosomes.
4.
The cell undergoes two rounds of karyokinesis and cytokinesis, without further
replication of chromosomes.
5.
This yields four cells, each with one set of chromosomes (haploid or n).
6.
The four haploid cells are genetically dissimilar to one another and the germ cell that
produced them.
Let’s now look at the process of meiosis in detail.
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A.
Meiosis I consists of chromosomal replication, and one round of division (karyokinesis and
cytokinesis).
1.
Meiosis I begins with Interphase I.
a)
The 46 chromosomes replicate as in mitosis, forming sister chromatids.
b)
The sister chromatids are held together by the centromere.
2.
Prophase I follows interphase I, and consists of the following events:
a)
Prophase I is similar to mitotic prophase in the following ways:
(1)
The nuclear envelope and nucleolus dissolve.
(2)
The sister chromatids condense and become visible.
(3)
The (meiotic) spindle forms.
b)
Prophase I is dissimilar to mitosis in that crossing over occurs.
(1)
The sister chromatids of homologous pairs form an association in which
they intertangle with one another, while matching up gene for gene.
(2)
This group of four sister chromatids of a homologous pair is called a
"tetrad.”
(3)
Enzymes will randomly cut and splice the DNA double helix of the
chromatids.
(a)
Most of the time a chromatid is cut and spliced back to itself.
(b)
Sometimes the cut ends of two different chromatids are spliced
together after being cut, this is called crossing over.
(4)
Crossing over yields chromatids that combine genes from different
homologous chromosomes.
(a)
This combines unique combinations of traits on a single
chromosome.
(b)
After crossing over the sister chromatids are no longer genetically
identical.
(5)
Consider the following situation.
(a)
Let’s say that chromosome number 1 (of the 23 distinct
chromosomes) has a gene for hair color at one end of the
chromosome and a gene for eye color at the other end.
(b)
The hair gene codes for brown hair (brh) or blonde hair (blh), and
the eye gene codes for brown eyes (bre) or blue eyes (ble).
(c)
Lets say Dad gives a chromosome 1 with genes for brown hair and
brown eyes to his daughter, and that Mom gives the daughter a
chromosome 1 with genes for blonde hair and blue eyes.
(d)
The daughter has a homologous pair of chromosome 1, composed
of one chromosome with genes for brown hair and brown eyes, and
one chromosome with genes for blonde hair and blue eyes.
(e)
The daughter grows up, reaches puberty, and begins to make ova of
her own via meiosis.
(i)
In Interphase I, the homologous chromosome number 1’s
replicate forming one set of sister chromatids with genes
for brown hair and brown eyes, and another set of sister
chromatids with genes for blonde hair and blue eyes.
brh
brh
blh
blh
bre
bre
ble
ble
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(ii)
In Prophase I the sister chromatids of the homologous pair
form a tetrad and a cross over occurs.
Before crossing over
After crossing over
(f)
3.
After crossing over, you can see that two of the chromatids have a
combination of genes that are new to the daughter and different
from the chromosomes contributed by her parents: her tetrad now
consists of the following sister chromatids:
(i)
One chromatid with genes for brown hair and brown eyes
(like the chromosome 1 Dad gave her).
(ii)
One chromatid with genes for blonde hair and blue eyes
(like the chromosome 1 that Mom gave her).
(iii)
One chromatid with genes for brown hair and blue eyes
(this chromatid is mixing genes inherited from Dad and
Mom onto a single chromosome).
(iv)
One chromatid with genes for blonde hair and brown eyes
(this chromatid is also mixing genes inherited form Dad
and Mom onto a single chromosome).
(g)
As you will see, the daughter no longer has to pass on to her child a
chromosome number 1 that has only genes inherited from her dad,
or a chromosome number 1 that has only genes inherited from her
mother, but she can now pass on a chromosome number 1 that has
a combination of genes from her mom and dad--a chromosome
unlike any in her parents’ cells, or in any of her own cells.
Metaphase I follows prophase I.
a)
In prophase I the tetrads line up at the spindle equator, the microtubules attached
to the centomeres.
b)
The tetrads arrange themselves independently of one another, i.e. the orientation
of one tetrad has nothing to do with the orientation of another tetrad.
c)
This is known as independent assortment.
(1)
Independent assortment contributes to genetic diversity of gametes.
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(2)
III.
When the tetrads are separated in anaphase I, independent assortment
helps ensure that neither a paternal set, nor maternal set of chromatids go
to one end of a cell.
(3)
The chromatids pulled to either end of the cell in anaphase I will be a
mixture of paternal and maternal chromosomes, even if crossing over did
not occur.
4.
Anaphase I follows metaphase I.
a)
The homologous pairs are separated, and pulled to the spindle poles.
b)
The centromere remains intact, holding the sister chromatids together.
c)
The sister chromatids are not genetically identical because of crossing over.
d)
Anaphase I completes the first round of karyokinesis.
5.
Telophase I follows anaphase I, and concludes meiosis I.
a)
The meiotic spindle will deteriorate.
b)
Telophase I will conclude meiosis I with the first round of cytokinesis.
c)
Telophase I concludes with two cells, each with 23 “pairs” of genetically
dissimilar sister chromatids attached to one another--in other words each cell has
two sets of 23 chromosomes, its just that the chromosomes are sister chromatids,
and the sister chromatids are genetically dissimilar due to crossing over.
B.
In Meiosis II a second round of karyokinesis and cytokinesis will occur, in each of the cells
produced in meiosis I, without chromosomal replication or crossing over.
1.
Interphase II begins meiosis II.
a)
The interphase II cell chemically prepares for division.
b)
No replication of chromosomes occurs in interphase II--if it did cell division could
not reduce ploidy.
2.
Prophase II follows interphase II.
a)
The meiotic spindles form, chromatids recondense, etc.
b)
Crossing over does not occur in prophase II.
3.
Metaphase II follows prophase II.
a)
The sister chromatids line up at spindle equators.
b)
Independent assortment is again at work--the orientation of one sister chromatid
pair, will not affect the orientation of another.
4.
Anaphase II follows metaphase II.
a)
The centromeres replicate and spindle fibers pull one set of 23 sister
chromosomes to each end of each cell.
b)
This completes the second round of karyokinesis.
5.
Telophase II is the final phase of meiosis.
a)
Cytokinesis yields four haploid cells.
b)
The four haploid cells are genetically dissimilar to one another due to crossing
over in prophase I, and independent assortment in metaphase I and II.
c)
It is virtually impossible for any two gametes to be genetically identical.
6.
All four haploid cells do not necessarily develop into gametes.
a)
In human females, only one of the four cells develops into an ovum, the other
three will deteriorate.
b)
In human males, all four develop into spermatozoa.
While meiosis is vital to sexual reproduction, it is a risky process, fraught with the potential for
mistakes.
A.
Translocations can occur.
1.
During crossover in prophase 1 a section of a chromatid can come loose from one
chromatid and splice to another.
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2.
IV.
Following meiosis this leaves one gamete with some genes missing, and another gamete
with duplicate genes.
3.
These excesses and shortages of genes can have drastic effects leading to developmental
disorders.
B.
Inversions.
1.
During crossover in prophase 1 a section of a chromatid can come loose and be reattached
in the opposite direction such that genes 123456789, might be reconfigured as
123456987.
2.
This may affect control of expression of genes and therefore affect the developmental
biology of the individual, following fertilization.
C.
Nondisjunction events.
1.
Sometimes chromatids do not dissociate properly in metaphase 1 or metaphase 2--for
example the centromere may not replicate in metaphase 2 so both chromatids go to the
same end of a cell rather than one to each end.
2.
This produces gametes with an excess chromosome, and gametes with a missing
chromosome.
3.
After fertilization the “diploid” cell may have a trisomy (three copies) of a chromosome
or be missing one of its chromosomes.
4.
Extra or missing chromosomes can have developmental consequences--Down’s
syndrome is caused by a trisomy of chromosome 21.
The great advantage of sexual reproduction over asexual reproduction is the genetic diversity it
generates.
A.
As we have seen, meiosis will mix parental genes on a single chromosome (crossing over) as
well as mix parental chromosomes (independent assortment).
B.
There is tremendous genetic diversity within gametes.
C.
Fertilization also mixes genes from two different individuals creating still more gene
combinations.
D.
The chance for mistakes in meiosis is significant--chromosomal breakages during anaphase for
example, and chromosomal duplications and omissions in crossing over, and would seem to put
species at risk.
E.
The great genetic diversity generated in the sexual process, however, protects the species against
a variety of challenges--there is direct evidence that genetic diversity reduces a population’s
parasite load, for example.
113
Lesson 18, Mendel and Single Trait Inheritance
I.
Before discussing the history and mechanism of inheritance, there are some important terms and
concepts I would like to discuss first.
A.
Offspring is a generic term for children, e.g. baby pea plants are not easily described as children.
B.
A chromosome is composed, in part, of DNA and DNA codes for proteins that account for
physical traits.
C.
A gene is a segment of chromosome that codes for a particular protein, or a particular physical
trait.
D.
Each gene has a locus, i.e. a particular place, on a particular chromosome where it is found.
E.
Diploid cells have two sets of chromosomes, one set inherited from each parent.
F.
Haploid gametes have one set of chromosomes, and are produced by meiosis from diploid cells.
G.
Paternal means coming from the father.
H.
Maternal means coming from the mother.
I.
A gene pair is the gene combination found at a particular locus.
1.
One chromosome of an homologous pair is inherited from each parent.
2.
A particular gene, for a particular trait, is found at a particular locus, on each
chromosome.
3.
These genes, one inherited from each parent, constitute a gene pair.
4.
As we discussed in mitosis and meiosis, a gene for any given trait may occur in two or
more alternate forms; these alternate forms are called alleles.
Please note: In the video graphic, the chromosome labeled “D”(for Dad) should be the same
length as the one labeled “M” (for Mom). The gene pair is at a particular locus on a
homologous pair of chromosomes.
a)
Alleles are either dominant or recessive.
(1)
Dominant alleles always show the physical trait for which they code.
(2)
Recessive alleles code for physical traits that can be hidden by dominant
alleles.
b)
For example, let’s say that a gene for flower color has two alleles (comes in two
possible versions).
(1)
One allele codes for an enzyme that makes a purple flower pigment, and
the allele will be represented by the upper case P.
(2)
The other allele codes for an enzyme that makes a defective pigment that
does not absorb light, so the flower is white, and that allele will be
represented by the lower case p.
c)
There are three possible genotypes (gene pairs), and two possible phenotypes
(physical appearances).
(1)
An individual with the genotype, Pp, has a phenotype of purple flowers.
(a)
It has one allele that makes the enzyme for purple pigment, hiding
the allele that does not produce pigment.
(b)
This gene combination shows that the P allele is dominant, and the
p allele is recessive, because the p allele is hidden by the P allele.
(c)
This gene pair is described as heterozygous ( hetero=different,
zygous= seed).
(2)
An individual with the genotype, PP, has a phenotype of purple flowers.
(a)
Both alleles make enzymes that make purple pigment.
(b)
This gene pair is described as homozygous (homo = same)
dominant.
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(3)
II.
III.
An individual with the genotype, pp, will have a phenotype of white
flowers.
(a)
It does not have a gene that will code for the enzyme that makes
purple pigment.
(b)
This gene pair is described as homozygous recessive.
(c)
The only way an individual can show a recessive phenotype is to
have a genotype of homozygous recessive.
We now have enough information to talk about the work of Gregor Mendel, a man described as the
father of classical genetics inheritance).
A.
Mendel was a German monk who did most of his work in a monastery in Brunn, Czechoslovakia
(in the mid 1800’s).
B.
Mendel came from a common background, but was fortunate enough to receive a superior
education in clerical studies at the University of Vienna, showing particular talent in science and
mathematics.
C.
Mendel had a keen eye for observation, an inquisitive nature, strong will, and the imagination to
create novel trials and tests of his ideas about inheritance.
D.
Mendel had no knowledge of chromosomes, genes, meiosis, or molecular genetics, yet he
inferred much about them, via mathematics.
E.
Mendel’s genius was not appreciated in his own lifetime, this may have been due to several
factors.
1.
It may have been overlooked--Mendel published his genetic results in 1860, a year after
Darwin published Origin of Species (1859), Darwin’s book on evolutionary theory , and
European science was preoccupied with Darwin and evolution.
2.
Lack of understanding.
a)
Mendel’s conclusions were drawn from probability theory and required close
scrutiny and detailed analysis to understand.
b)
Mendel’s genius is reinforced by Darwin himself.
(1)
Darwin duplicated Mendel’s work using snapdragons, but did not realize
the significance of his results.
(2)
Darwin certainly read Mendel’s work, but again did not realize its
significance, even though it was vital to Darwin’s own theory of evolution
(descent with modification).
3.
Class prejudice.
a)
Science was more hobby than profession, so scientists of the time period were
from wealthy families, and social status was equated with intelligence.
b)
Mendel was from a working class family, and not well known, so his work may
not have been seriously considered by those in scientific circles.
Mendel’s tool for testing inheritance theory, was the pea plant.
A.
Mendel was interested in testing the “blending” theory of inheritance, which was accepted in
Mendel’s time.
B.
Blending predicted that physical traits of parents mixed or “blended” in offspring to yield traits
intermediate to those found in the parents, e.g. a tall parent and a short parent would yield an
offspring of medium height.
C.
Mendel identified seven distinct phenotypes (physical characteristics) that would test the
blending hypothesis-- all seemed to occur in an “either-or” phenotype.
1.
Seed shapes were round or wrinkled.
2.
Seed colors were yellow or green.
3.
Flower colors were purple or white.
4.
Unripe seed pod colors were green or yellow.
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IV.
5.
Pod shapes were inflated or constricted.
6.
Stem lengths were long( 6-7 feet) or short (9-18 in).
7.
Flower positions were axial (flowers at nodes), or terminal (at stem tips).
Let’s look at seed color to see how Mendel developed his theories of inheritance.
A.
Mendel developed “pure breeding” or pure strains of pea plants.
1.
One pure breeding population had only yellow seeds, and when bred with one another
produced only offspring with yellow seeds.
2.
The other pure breeding population had only green seeds, and when bred with one
another produced only offspring with green seeds.
B.
He took pure strains for yellow and green (called the p generation, for “parental generation”) and
crossed them to get an F1 of all yellow seeded offspring.
1.
The term, F1 stands for “first filial” generation, or the first generation of offspring from a
parental cross.
2.
The fact that all offspring had only yellow seeds spelled doom for the blending
hypothesis.
C.
Mendel then allowed the F1 plants to self pollinate and produce an F2 generation.
1.
Remember, all of the F1 plants were yellow seeded.
2.
The F2 plants were both yellow and green seeds.
3.
Yellow seeds appeared in a 3:1 ratio, or 3/4 were yellow seeds, and 1/4 green seeds.
D.
Mendel observed similar results with the other traits he had studied:
1.
Pure breeding round and wrinkled seeds yielded an F1 of only round seeds, and F2 of 3/4
round to 1/4 wrinkled.
2.
Pure breeding purple flowered and white flowered plants yielded an F1 of only purple
flowers, and F2 or 3/4 purple to 1/4 white.
3.
And so on for the other traits.
4.
In each situation one trait disappeared in the F1 only to reappear in the F2 , and the traits
were always in a 3:1 ratio, something the mathematician in Mendel found to be
important.
E.
Mendel came to the following conclusions (modern terminology is used to describe his
conclusions):
1.
There were two versions of each gene: a dominant allele and recessive allele.
a)
This was true for the traits studied by Mendel.
b)
There can be more than two alleles, but in such cases there are more than one
dominant allele and/or more than one recessive allele.
2.
The mathematical ratios Mendel consistently observed in his research, led him to
conclude that parents carried two alleles, in a gene pair, i.e. that the parents were diploid.
3.
Mendel further reasoned, mathematically, that each parent plant could pass on only one of
its alleles (Mendel’s Law of Segregation) to their offspring, and that parents did not
choose which allele to pass on.
a)
Outcomes followed random laws of probability.
b)
Because of this, Mendel reasoned parents did not determine which alleles to pass
on to their offspring.
c)
We now call this process of packaging one set of chromosomes, which is one set
of alleles, into a gamete as meiosis.
(1)
Mendel knew nothing of chromosomes.
(2)
Mendel knew nothing of meiosis.
d)
Mendel further reasoned that by receiving only one allele from each parent
(actually one set of chromosomes from each parent), the offspring would duplicate
the condition of the parents, having two alleles (two sets of chromosomes).
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e)
V.
Mendel’s segregation of alleles (meiosis) ensured consistent genetic make-up
from generation to generation.
4.
When comparing multiple traits simultaneously, one could likewise predict outcomes due
to fact that separate traits assort themselves independently (Mendel’s Law of Independent
Assortment).
Let’s now consider how Mendel’s law of segregation can be used to predict the outcomes of genetic
crosses.
A.
Let’s examine crosses involving seed shape: R = round seed , r = wrinkled seed.
1.
What would be the expected genotypic and phenotypic ratios if a pure breeding round
seeded plant was crossed with a pure strain of wrinkled seeded plant?
a)
Pure strain or pure breeding means homozygous.
b)
A pure breeding (strain) round seeded plant has the genotype of RR.
c)
The pure strain wrinkled seeded plant has a genotype of rr.
d)
The RR plant has only R (dominant) alleles to pass to the next generation--all
gametes will carry one dominant (R) allele.
e)
The rr plant has only r (recessive) alleles to pass to the next generation--all
gametes will carry one recessive (r) allele.
f)
The cross is RR x rr
g)
We can set up a Punnett square to see possible offspring outcomes:
Possible gametes of the rr parent on
this axis (all gametes carry the r allele)
r
Possible gametes of
the RR parent on one
axis (all gametes
carry the R allele in
this case)
R
Rr
Possible outcomes of
offspring (100% of the
offspring will be Rr,
and be round seeded)
h)
2.
Answer: 100% of the offspring will have a genotype of RR, and 100% of the
offspring will have a phenotype of round seeds.
i)
The important thing to realize from this cross is that whenever a homozygous
dominant parent is crossed with a homozygous recessive parent, all the offspring
will be heterozygous, and show the dominant trait.
What would be the expected genotypic and phenotypic ratios in the next generation (F2 )
if two offspring from the previous problem were crossed with one another?
a)
The offspring were all heterozygous, Rr.
b)
The cross then would be Rr x Rr.
R
r
Possible gametes of the other Rr parent on
this axis (1/2 R, 1/2 r)
Possible gametes of
RR Rr
R
one Rr parent on this
axis (1/2 the gametes
Rr rr
r
carry the R allele, 1/2
the r allele)
There are four possible outcomes, with two of them having the same genotype. Expected
genotypic ratios are 1:2:1 (RR to Rr to rr) and phenotypically, 3:1 (round to wrinkled seeds).
We can also express the outcomes fractionally, 1/4 RR, 1/2 Rr, and 1/4 rr, and 3/4 round to
1/4 wrinkled. Regardless this is what we should expect anytime we cross heterozygotes.
117
c)
We can modify the Punnett square to show fractional outcomes.
1/2 R 1/2 r
Possible gametes of
one Rr parent on this
axis (1/2 the gametes
carry the R allele, 1/2
the r allele)
d)
e)
3.
Possible gametes of the other Rr parent on
this axis (1/2 R, 1/2 r)
1/2 R 1/4RR 1/4Rr
1/2 r 1/4Rr
1/4 rr
In this technique we use the multiplicative property of probability to determine
expected outcomes.
(1)
In this case if 1/2 the gametes of one parent are R and 1/2 the gametes of
the other parent are R, then the probability of an RR offspring is 1/2 x 1/2
= 1/4, and so on for the other possibilities.
(2)
There are two heterozygous outcomes, Rr and rR.
(a)
Both are considered heterozygous and written Rr.
(b)
The dominant and recessive alleles are coming from different
parents in each case however, so they must be counted as different
outcomes.
(3)
When events are random and independent of one another, we can multiply
the probability of each event to get the probability of the two occurring
together.
(4)
It is similar to the probability of flipping two coins and determining the
probability of getting two heads.
(a)
The probability of getting a heads on a coin is 1/2.
(b)
The probability of getting a heads on the other coin is 1/2.
(c)
The outcome of one coin has nothing to do with the outcome of the
other coin.
(d)
The probability of two heads is 1/2 x 1/2 = 1/4.
The most important thing to realize is this: when heterozygotes (also called
hybrids) are crossed 1/4 of the offspring are expected to be homozygous
dominant, 1/2 heterozygous, and 1/4 homozygous recessive; and 3/4 will show the
dominant phenotype, and 1/4 will show the recessive phenotype (in complete
dominance--explained later).
What would be the expected frequency of traits in a cross between a heterozygous plant,
Rr, and a homozygous dominant plant, RR?
1R
Possible gametes of
the Rr parent on this
axis (1/2 the gametes
carry the R allele, 1/2
1/2 R 1/2 RR118
1/2 r 1/2 Rr
Possible gametes of the RR parent on this
axis (1 R--100% R)
Genotypically, 1/2 will be RR, 1/2
Rr; phenotypically, 100% will be
When a heterozygous individual is crossed with a homozygous dominant individual you
expect 1/2 the offspring to be homozygous dominant, 1/2 to be heterozygous, and all to
show the dominant phenotype.
4.
How about a cross between a hybrid (heterozygote), Rr, and a homozygous recessive, rr,
individual.
Possible gametes of the rr parent on this
Possible gametes of
1r
axis (1 r--100% R)
the Rr parent on this
Genotypically, 1/2 will be Rr, 1/2
axis (1/2 the gametes 1/2 R 1/2 Rr
rr; phenotypically, 1/2 will be
carry the R allele, 1/2
1/2
rr
round seeded, 1/2 wrinkled seeded
1/2 r
the r allele)
B.
C.
D.
When you cross a heterozygous individual with a homozygous recessive individual you
expect 1/2 the offspring to be heterozygous, and 1/2 to be homozygous recessive; and 1/2
to show the dominant phenotype and 1/2 to show the recessive phenotype.
5.
Hopefully you can determine the possible outcomes from the last two possible crosses: a
homozygous dominant x homozygous dominant, and homozygous recessive x
homozygous recessive.
A summary of ratios generated by the different crosses is listed below.
1.
Homozygous dominant x homozygous dominant = 100% homozygous dominant; 100%
dominant phenotype.
2.
Homozygous recessive x homozygous recessive = 100% homozygous recessive; 100%
recessive phenotype.
3.
Homozygous dominant x homozygous recessive = 100% heterozygous; 100% dominant
phenotype.
4.
Heterozygous x heterozygous = 1/4 homozygous dominant, 1/2 heterozygous, 1/4
homozygous recessive; 3/4 dominant phenotype, 1/4 recessive phenotype.
5.
Heterozygous x homozygous dominant = 1/2 homozygous dominant, 1/2 heterozygous;
100% dominant phenotype.
6.
Heterozygous x homozygous recessive = 1/2 heterozygous, 1/2 homozygous recessive;
1/2 dominant phenotype, 1/2 recessive phenotype.
Mendel found these ratios to be consistent regardless of the trait he examined.
It is also important to note that the only way to show a recessive phenotype, is to have a genotype
of homozygous recessive.
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120
Lesson 19, Solving Problems and Multiple Trait Inheritance
I.
II.
III.
Mendel was fortunate that he happened to choose traits that exhibited complete dominance.
A.
Complete dominance means that there is no phenotypic difference between individuals that are
homozygous dominant and those that are heterozygous.
B.
For example, in pea plants if P = purple flowers and p = white flowers, a purple flowered plant
may be either PP or Pp--there is no way to tell by looking at the phenotype.
C.
Traits that exhibit complete dominance have only two phenotypes: a dominant phenotype and a
recessive phenotype.
D.
A testcross can determine the genotype of an organism showing a dominant phenotype.
1.
In a testcross, an organism of unknown genotype, but showing the dominant phenotype
(meaning they are either homozygous dominant, or heterozygous), is crossed with an
individual showing the recessive phenotype (meaning it has a genotype of homozygous
recessive).
2.
If all F1 offspring show the dominant phenotype then the unknown parent is homozygous
dominant, if (half) the offspring show the recessive phenotype, then the unknown parent
is heterozygous (see ratios above).
There is, however, another type of dominance at work in nature, and is probably responsible for the
theory of blending discussed earlier.
A.
A red flowered four o’clock is another type of plant.
B.
Its gene for flower color has two alleles: R for red flowers (and is dominant) and r for white
flowers (and is recessive).
C.
However, there is something unexpected in phenotypes, in that red flowered four o’clocks can
have red flowers, pink flowers, or white flowers.
1.
Individuals who are homozygous dominant, RR, have red flowers.
2.
Individuals who are heterozygous, Rr, have pink flowers.
3.
Individuals who are homozygous recessive, rr, have white flowers.
D.
Traits that exhibit incomplete dominance have three phenotypes: a dominant phenotype, a
recessive phenotype, and an incompletely dominant phenotype (an intermediate).
E.
If Mendel had worked with incompletely dominant traits, his ratios would have been more
complex and perhaps he would not have figured out inheritance.
Consider the following problems.
A.
What are the genotypes of all individuals if a hornless bull is mated to a horned cow and produce
a hornless offspring.
1.
H = hornless, h = horned.
2.
Cow is hh, offspring is Hh, and bull is either HH or Hh.
B.
Eye color pedigree.
1.
Males are squares.
2.
Females are circles.
3.
A specific phenotype will be shaded in the pedigree--in this case the shaded trait is for
brown eyes (B), and the non-shaded trait is for non-brown eyes(b).
4.
Lines connect related individuals, mates to children for example, and may show several
generations.
Bb
bb
Bb
bb
Bb
bb
Bb
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Bb
Bb
These three BB or Bb
bb
bb
C.
Mammal fur pigment pedigree.
1.
The shaded trait is red fur, the unshaded trait is black fur.
2.
If you did not know which trait was carried by the dominant allele, it could be
determined as follows.
a)
Determine whether the shaded trait is hidden in a generation.
b)
If so, then it is encoded by a recessive allele, because only recessive alleles ( and
the traits they encode) can be hidden.
c)
In this example, red fur is recessive because it is hidden in the second generation,
so black fur is dominant (B=black, b= red).
d)
List all genotypes.
BB or Bb
bb
Bb
Bb
Bb
Bb
bb
D.
BB or Bb
Co-dominance--when there is more than one dominant allele for a trait, each is completely
dominant over a recessive allele, but they are equally dominant with each other; the Landsteiner
blood types is an example.
1.
There are three different alleles which code for proteins in the membranes of red blood
cells, two of them are dominant, and are considered co-dominant.
a)
Allele A codes for protein A and is dominant.
b)
Allele B codes for protein B and is dominant.
c)
Allele O (or i) does not code for a protein and is recessive.
2.
This generates more than three gene pairs, and more than two phenotypes.
a)
Gene pair AA yields blood type A
b)
Gene pair AO (or Ai) yields blood type A.
c)
Gene pair BB yields blood type B
d)
Gene pair BO (or Bi)yields blood type B.
e)
Gene pair AB yields blood type AB(codominance).
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3.
f)
Gene pair OO (or ii) yields blood type O.
Consider the pedigree below and determine as many genotypes as possible.
AAor Ai A
B Bi
Ai A
AB, Ai, Bi, or ii
O ii
Ai
O ii
Ai or ii
A Ai
Please note: In the video I did not include “ii” as a possible genotype for the individual
with the arrow--it is a possible genotype.
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Lesson 20, Multiple Trait Inheritance and Epistasis
I.
Next Mendel began to consider genetic outcomes of offspring looking at combinations of traits.
A.
Mendel wondered if one trait influences how another trait is inherited, or whether they are
inherited independently of one another.--for example, does inheritance of hair color influence the
inheritance of eye color.
1.
Lets consider the following traits.
a)
R for round seed (complete dominance) , r for wrinkled seed.
b)
Y for yellow seed(complete dominance), y for green seed.
2.
Mendel bred a population of pea plants that were pure breeding (homozygous) for round,
yellow seeds (RRYY), and another population that was pure breeding (homozygous) for
wrinkled, green seeds (rryy) --these were the parental populations.
3.
A cross of members from each population yielded an F1 with all individuals round and
yellow seeded.
a)
These individuals are heterozygous for seed shape and seed color.
b)
These individuals are called dihybrids, because they are heterozygous for two
traits.
4.
Members of the F1 were allowed to cross and yielded an F2 with the following
characteristics:
a)
9/16 round, yellow; 3/16 round, green; 3/16 wrinkled, yellow; 1/16 wrinkled,
green.
b)
It can also be described more generically as: 9/16 with both dominant phenotypes;
3/16 with one dominant phenotype, and the other recessive phenotype; 3/16 with
the dominant and recessive phenotypes reversed; 1/16 showing both recessive
phenotypes.
5.
Mendel saw similar results regardless of which two traits he considered in combination.
B.
Mendel then compared predicted outcomes to the results he saw in his garden.
1.
In looking at the ratios, Mendel saw that 3/4 of the F1 generation had round seeds, and 1/4
had wrinkled.
a)
9/16 round, yellow + 3/16 round, green = 12/16 round = 3/4 round.
b)
3/16 wrinkled, yellow + 1/16 wrinkled, green = 4/16 wrinkled = 1/4 wrinkled.
c)
This is exactly as one would predict if considering the seed shape gene by itself.
(1)
Crossing parental homozygotes would yield 100% heterozygotes in the F1.
(2)
Crossing heterozygotes would yield an F2 of 3/4 dominant phenotype to
1/4 recessive phenotype.
2.
In looking at the ratios Mendel saw that 3/4 of the F1 generation had yellow seeds, and
1/4 had green.
a)
9/16 round, yellow + 3/16 wrinkled, yellow = 12/16 yellow = 3/4 yellow.
b)
3/16 round, green + 1/16 wrinkled, green = 4/16 green = 1/4 green.
c)
This is exactly as one would predict if considering the seed color gene by itself.
(1)
Crossing parental homozygotes would yield 100% heterozygotes in the F1.
(2)
Crossing heterozygotes would yield an F2 of 3/4 dominant phenotype to
1/4 recessive phenotype.
3.
If seed shape and seed color were acting randomly and independently of one another,
Mendel reasoned, he should be able to multiply the probabilities of each independent
outcome to get a combined probablilty to match his results.
a)
If 3/4 of the F1 are round, and 3/4 of them are yellow, then 9/16 should be round
and yellow.
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b)
C.
If 3/4 of the F1 are round, and 1/4 of them are green, then 3/16 should be round
and green.
c)
If 1/4 of the F1 are wrinkled, and 3/4 of them are yellow, then 3/16 should be
wrinkled and yellow.
d)
If 1/4 of the F1 are wrinkled, and 1/4 of them are green, then 1/16 should be
wrinkled and green.
e)
All of these probabilities matched Mendel’s results.
Mendel concluded that different traits are inherited independently of one another--this is
Mendel’s Law of Independent Assortment.
1.
If dihybrids are crossed and yield the 9:3:3:1 ratio discussed above, we will see that as
proof of independent assortment.
2.
What Mendel did not know about the traits he was studying, was that they were on
different chromosomes.
a)
As we learned in studying meiosis, tetrads in metaphase I, and sister chromatids in
metaphase II line up and assort independently.
b)
This allows traits on different chromosomes to assort independently.
c)
If traits are linked on the same chromosome, however, they do not assort
independently, and crossing dihybrids will not yield a 9:3:3:1 ratio seen with traits
that do assort independently.
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d)
Mendel’s dihybrid crosses involved parents with genes on different types of
chromosomes.
Homologous pair
chromosome number 1
Gene pair for
seed shape
R
Homologous pair
chromosome number 2
r
Y
R
R
Y
Possible gametes:
1/4
r
r
y
1/4
Gene pair for seed
color
y
Y
1/4
y
1/4
Each gamete gets one of each type of chromosome. The two traits assort
independently, so gametes with different combinations of alleles are equally likely,
i.e. 1/4 of the gametes will combine RY, 1/4 Ry, 1/4 rY, and 1/4 ry. When these are
put on each axis of a Punnett square and outcomes completed, careful analysis will
show the prediction of 9/16 round, yellow; 3/16 round, green; 3/16 wrinkled, yellow;
1/16 wrinkled, green. This type of square is relatively inefficient as there is overlap of
genotypes, and ratio determination is rather tedious.
RY
Ry
rY
ry
RY
RRYY RRYy RrYY RrYy
Ry
RRYy RRyy
RrYy Rryy
rY
RrYY RrYy
rrYY
rrYy
ry
RrYy Rryy
rrYy
rryy
If we know, however, that the traits of interest are on different chromosomes, and will assort
independently, we can use the multiplicative property of probability to modify our square to more
efficiently predict offspring and ratios. Remember that in our example of a dihybrid cross, we are
considering two different events: the cross of Rr x Rr, and Yy x Yy. Rr x Rr will yield 1/4 RR, 1/2
Rr and 1/4 rr. Yy x Yy will yield 1/4 YY, 1/2 Yy and 1/4 yy. If we put these outcomes on each axis
of a square, we will not only determine the possible genotypes, but also their predicted frequency
without any overlap of genotypes in the boxes. The modified square can also be done with
phenotypic ratios, which is the most efficient, if you are looking for phenotypic outcomes.
1/4 YY
1/2 Yy
1/4 yy
3/4 yellow
1/4 RR 1/16RRYY 1/8RRYy 1/16RRyy
1/2 Rr
1/8RrYY 1/4RrYy
1/8Rryy
1/4 rr
1/16rrYY
1/16 rryy
1/8rrYy
3/4 round
1/4 green
9/16 round,yellow 3/16 round,green
1/4 wrinkled 3/16 wrinkled,yell 1/16 wrinkled,green
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(1)
(2)
If traits assort independently, one can multiply to determine probabilities
of traits occurring in specific combinations.
For example:
(a)
What would be the expected genotypic ratio of offspring from the
following cross: RRYy x Rryy
(i)
The cross RR x Rr yields 1/2 RR and 1/2 Rr.
(ii)
The cross Yy x yy yields 1/2 Yy and 1/2 yy.
(iii)
If you put the genotypic outcomes for seed shape on one
axis of a square and the genotypic outcomes for seed color
on the other axis you will come up with the following
answer: 1/4 RRYy, 1/4 RRyy, 1/4 RrYy, and 1/4 Rryy.
|
1/2RR |
1/2 Rr |
(b)
(c)
1/2 Yy
1/4RRYy
1/4RrYy
1/2 yy
1/4RRyy
1/4Rryy
In a cross involving three traits, all on different types of
chromosomes, the answer must be done in two steps--let’s add the
trait for tallness to the traits we are already considering, such that T
= tall plants, t = short plants.
What would be the expected genotypic ratio of offspring from the
following cross: RRYyTT x Rryytt.
(i)
The cross RR x Rr yields 1/2 RR and 1/2 Rr.
(ii)
The cross Yy x yy yields 1/2 Yy and 1/2 yy.
(iii)
The cross TT x tt yields 100% Tt.
(iv)
First, determine the genotypic ratio for seed shape and seed
color as described above.
|
1/2RR |
1/2 Rr |
(v)
1/2 Yy
1/4RRYy
1/4RrYy
1/2 yy
1/4RRyy
1/4Rryy
Take the results for seed shape and color and put them on
one axis of a new square, and the expected ratios for
tallness on the other axis to get the answer : 1/4 RRYyTt,
1/4 RRyyTt, 1/4 RrYyTt, and 1/4 RryyTt.
1/4RRYy |
127
1 Tt
1/4RRYyTt
1/4Rryy |
1/4RrYy |
1/4Rryy |
128
1/4RryyTt
1/4RrYyTt
1/4RryyTt
e)
Why won’t different traits on the same chromosome assort independently?
Homologous pair
chromosome number 1
Gene pair for
seed shape
Possible
gametes
R
Y
R
r
Y
y
R
y
r
Y
r
y
As you can see the traits are linked on the same chromosome. More importantly, in the scenario pictured above,
the dominant alleles for each trait are linked on the same chromosome, and the recessive alleles for each trait are
linked on the same chromosome. The RY combination and ry combination are certain to appear in gametes
together. But what about the Ry and rY combinations? Are they possible, and if so how often do they occur?
The answers: they are possible due to crossing over, however, their frequency is not a random event and cannot
be predicted via simple laws of probability. The frequency of crossover is determined by the distance between
the genes: the closer the genes are together, the fewer crossovers occur; and the further apart they are, the more
crossovers occur. Because of this the four types of gametes will not occur in equal frequencies, and their
frequencies are not predictable. The RY and ry combinations would appear much more frequently than Ry or rY.
In summary, we would not see the classic 9:3:3:1 ratio because the traits do not assort independently since they
are on the same chromosome, and crossover frequencies are dependent on the distance between genes.
II.
Let’s consider problems, and assume independent assortment.
A.
What are the possible genotypic and phenotypic outcomes in the following cross: RRYy x Rryy
1.
Treat it initially as if it is two different problems.
a)
If you cross RR x Rr then the predicted outcome is 1/2 RR and 1/2 Rr offspring.
b)
If you cross Yy x yy then the predicted outcome is 1/2 Yy and 1/2 yy offspring.
2.
Then set up a grid with the R outcomes on one side, and the Y outcomes on the other, and
multiply to get the combined probabilities:
a)
The genotypic ratio is 1/4 RRYy, 1/4 RRyy, 1/4 RrYy, 1/4 Rryy.
b)
The phenotypic ratio is 1/2 round, yellow; 1/2 round, green.
B.
Lets consider a cross involving three traits--seed shape (round vs wrinkled), seed color (yellow vs
green) and plant length (tall(T) is dominant over short(t))-- RRYyTT x Rryytt.
1.
Treat it as if it three different problems.
a)
If you cross RR x Rr predicted outcome is 1/2 RR and 1/2 Rr offspring.
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III.
b)
If you cross Yy x yy then the predicted outcome is 1/2 Yy and 1/2 yy offspring.
c)
If you cross TT x tt then the predicted outcome is 100% Tt offspring.
2.
Then set up a grid with the R outcomes on one side, and the Y outcomes on the other, and
multiply to get the combined probabilities:
a)
The genotypic ratio is 1/4 RRYy, 1/4 RRyy, 1/4 RrYy, 1/4 Rryy.
b)
The phenotypic ratio is 1/2 round, yellow; 1/2 round, green.
3.
Then set up a grid with the R-Y outcomes on one side, and the T outcome(s) on the other.
a)
The genotypic ratio is 1/4 RRYyTt, 1/4 RRyyTt, 1/4 RrYyTt, 1/4 RryyTt.
b)
The phenotypic ratio is 1/2 round, yellow, tall; 1/2 round, green, tall.
C.
What would be the probability of getting an offspring with the following genotype, RRYyTt from
the following parental cross: RRYyTT x Rryytt.
1.
Treat it as if it three different problems.
a)
If you cross RR x Rr predicted outcome is 1/2 RR and 1/2 Rr offspring.
b)
If you cross Yy x yy then the predicted outcome is 1/2 Yy and 1/2 yy offspring.
c)
If you cross TT x tt then the predicted outcome is 100% Tt offspring.
2.
Multiply the probability of each genotypic outcome to get the combined probability.
a)
The probability of getting an RR offspring is 1/2.
b)
The probability of getting a Yy offspring is 1/2.
c)
The probability of getting a Tt offspring is 100%.
d)
1/2 x 1/2 x 1 = 1/4
Epistasis occurs when one gene interferes with expression of another gene--horse color involves
epistasis.
A.
Most horses have genes for black and red pigments.
B.
Horses have an “Agouti” gene which controls distribution of black pigment.
1.
The dominant allele for the Agouti locus (A) restricts expression of the black pigment to
the “points,” i.e. legs, mane, and tail, producing a chestnut horse with black points (called
a bay).
2.
The recessive allele for the Agouti locus (a) allows uniform expression of the black
pigment (masking any red pigment), producing a black horse.
3.
The Agouti genotypes and phenotypes are listed below.
a)
AA = bay horse (black restricted to points)
b)
Aa = bay horse (black restricted to points)
c)
aa = black horse (black uniformly distributed)
C.
Epistatic to the Agouti gene is one called the “Extension” gene, which regulates the Agouti gene.
1.
The dominant allele for the Extension (E) locus permits expression of the Agouti gene,
which will result in bay horses, if the horse has a dominant Agouti allele.
2.
The recessive Extension (e) allele inhibits expression of the Agouti locus, meaning that a
horse that is homozygous recessive for the Extension locus produces a chestnut (reddishbrown) horse, regardless of the gene pair at the Agouti locus.
3.
The Extension genotypes and phenotypes are listed below.
a)
EE = agouti expressed (bay or black horse).
b)
Ee = agouti expressed (bay or black horse).
c)
ee = agouti inhibited (chestnut horse).
D.
The possible combinations of Agouti and Extension gene pairs yield the basic horse colors.
a)
The following combinations yield a bay horse (dark points).
(1)
AAEE
(2)
AAEe
(3)
AaEE
(4)
AaEe
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b)
E.
The following combinations yield a black horse.
(1)
aaEE
(2)
aaEe
c)
The following combinations yield a chestnut horse.
(1)
AAee
(2)
Aaee
(3)
aaee
The Extension gene is considered epistatic to the Agouti gene, which is hypostatic to the
Extension gene.
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Lesson 21, X-Linked Traits, Imprinting, and More
I.
II.
III.
IV.
Mendel’s work was unappreciated and forgotten over the last part of the nineteenth century.
In the early twentieth century genetic research began to blossom.
A.
Darwin’s theory of evolution had stimulated interest in genetics to explain how traits, passed
from generation to generation, could be modified.
B.
Bold advances were also occurring in the medical field and questions existed about diseases that
“run in families.”
C.
Mendel’s work was rediscovered and duplicated, and then applied to other organisms.
Thomas Hunt Morgan was one of the pioneers of animal genetics research following the rediscovery of
Mendel's work in the early 1900's.
A.
Morgan chose to work with the common fruit fly, Drosophila melanogaster, an excellent choice
for the reasons listed below, although Morgan didn't realize some of them at the time.
1.
Drosophila is easy to keep, reproduces quickly, and in large numbers.
2.
It has only four pairs of chromosomes.
3.
Being an animal, it has sex chromosomes.
4.
There are several physical traits that show complete dominance, which are therefore easy
to follow.
B.
Drosophila became the primary tool for animal studies of genetic traits, and still is today,
although a nematode worm, Caenorhabditis elegans, is growing in popularity.
C.
Morgan made important discoveries that added to Mendel’s initial work.
1.
Morgan discovered that not all traits assort independently, and that they are linked on the
same chromosome.
2.
Morgan and his associates also determined the presence of sex chromosomes and Xlinked traits.
The sex chromosomes determine the sex of an animal; all other chromosomes are called autosomes.
A.
There are two types of sex chromosomes: X chromosomes and Y chromosomes.
B.
The X chromosome has many genes, most of which have nothing to do with sex determination.
C.
The Y chromosome has fewer genes, and Y genes do not match genes on the X chromosome.
1.
The XX chromosomal pair is homologous, and defines an individual as female.
2.
The XY chromosomal pair is non-homologous (because they do not have the same
genes), and defines an individual as male.
3.
The Y chromosome determines “maleness” in the offspring.
D.
Morgan and his associates determined the following about X linked genes (those found on the X
chromosome):
1.
A female will only show an X linked recessive trait if she is homozygous recessive
(assuming complete dominance) for a trait (just as in autosomal inheritance).
2.
A male will show an X linked recessive trait when he inherits a single X linked recessive
allele.
3.
Because there is no corresponding allele on the Y chromosome of a male, a recessive
allele on the X chromosome cannot be hidden and the male will show the recessive
phenotype.
4.
As a result, X-linked recessive phenotypes are expressed much more often in males, than
females, because they need inherit only one X linked recessive allele to show the
phenotype whereas females must inherit two X linked recessive alleles to show the
phenotype.
5.
Males pass Y chromosomes to their sons, and do not pass X linked genes to their sons as
a result-- mothers pass an X chromosome to sons.
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E.
F.
G.
Some well known X linked recessive phenotypes include hemophilia, color blindness, and
Duchenne’s muscular dystrophy.
Y linked traits certainly exist, but we will focus on X linked traits.
Let’s consider a few examples of X linked inheritance.
1.
If a woman is a carrier for muscular dystrophy and her husband is normal, can any of their
children have muscular dystrophy (N = normal, and n = muscular dystrophy)?
a)
A carrier is a woman who is heterozygous for an X linked trait, so in this case the
mother’s genotype is XNXn .
b)
The father is normal so his genotype is XNY.
c)
Make a Punnett square setting the mother’s and father’s chromosomal types on
different axes, to determine the possible genotypes of the children:
N
X
Y
XN
| XNXN
| XN Y
Xn
XNXn
Xn Y
d)
e)
f)
2.
As you can see, half of the sons would be expected to have muscular dystrophy.
None of the daughters would have the disease, but half would be carriers.
Notice that the male children get the Y chromosome from Dad and the X
chromosome from Mom.
g)
The female children get one X chromosome from each parent.
If a colorblind man marries a woman who is homozygous for normal color vision can any
of their children be colorblind (N = normal vision, n=colorblind)?
a)
The mother’s genotype is XNXN .
b)
The father is colorblind so his genotype is XnY.
c)
Make a Punnett square setting the mother’s and father’s chromosomal types on
different axes, to determine the possible genotypes of the children:
Xn
Y
XN
(all her gametes will carry only the XN chromosome)
| XNXn
| XN Y
d)
3.
As you can see, all the sons will have normal vision, because the woman only has
normal X’s to pass on to the children.
e)
All of the daughters will be carriers, because the man only has a colorblind X
chromosome to pass on to his daughters.
Label the genotypes of the individuals in the pedigree below.
XNY
XNY
XnY
XnY
XNXn
XNXn
XnY
XNY
XNX?
XNXn XNY
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XNY
XnY
XnY
XNX?
XNY
V.
X linked recessive phenotypes are dominated by males, whereas autosomal recessive
phenotypes will be equally shared by males and females.
H.
Mary Lyon determined that in mammals, only one X chromosome functions in a particular cell,
the other is turned off.
1.
In males, this is not a problem because the cells of a male have only one X chromosome.
2.
Females, however, have two X chromosomes, and so, one or the other must be
inactivated.
3.
This process of "turning off" one of the X chromosomes is called Lyonization.
a)
The inactive X condenses (hypercoils), as in mitosis.
b)
It is often visible in the nucleus as a dark granule, and is called a Barr body
(named for their discoverer).
4.
X inactivation appears to be completely random, but occurs during fetal development.
5.
During fetal development one cell may turn off one X, while another cell may turn off the
other.
a)
All subsequent daughter cells will also have the same X inactivated.
b)
This creates, in effect, a mosaic pattern of X chromosomal activity--one group of
cells with one functioning X, another group of cells with the other X functioning.
c)
This mosaic effect is visible in calico cats, where fur color is encoded on the X
chromosome.
(1)
Cats have two alleles for fur color on the X chromosome, B= black fur,
and Y = yellow fur.
(2)
The alleles are both dominant, or codominant.
(3)
There are several possible gene pairs: XB XB = female black cat, XY XY =
female yellow cat, XB Y = male black cat, XY Y = male yellow cat (do not
let the Y for yellow get confused for the Y for maleness), and XB XY
which produces a calico cat.
(4)
Why is the XB XY female cat a calico and not muddy brown or some color
formed by mixing yellow and black?
(a)
In some cells XY is active and that patch of fur is yellow.
(b)
Other cells have active XB and that patch of fur will be black.
(c)
Males cats are not calicos, because they have only one X.
Some other important factors can also affect inheritance and gene expression.
A.
Imprinting is the inactivation of either maternal or paternal copies of certain genes, so that only
one of the inherited copies function.
1.
Some genes need to exist in only one copy for normal development, this is true in some
developmental genes, for example.
a)
Genes are inherited, of course, in gene pairs because organisms are diploid and
inherit one set of chromosomes from each parent.
b)
In some gene pairs, if both genes are active, it causes problems so imprinting
inactivates one of the alleles or genes of a gene pair.
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2.
B.
C.
D.
E.
F.
Imprinting is an inactivation, thought to be induced by hypercoiling of DNA, by addtion
of methyl groups.
3.
If it is a gene in which the paternal copy is imprinted, then the sperm carry a copy of the
gene that is hypercoiled and it will not be active in the offspring, regardless of whether it
is dominant or recessive.
a)
When the offspring is an adult, the imprint is removed by the gamete producing
cells of meiosis within the gonads (ovaries or testes) of the offspring.
b)
If the offspring is female, she will make ova in which the gene is not imprinted.
c)
If the offspring is male, the imprint is reimposed, and he makes sperm with
imprinted genes, which he passes on to his own offspring.
4.
If it is a gene in which the maternal copy is imprinted, then the ova carry a copy of the
gene that is hypercoiled and it will not be active in the offspring, regardless of whether it
is dominant or recessive.
a)
When the offspring is an adult, the imprint is removed by the gamete producing
cells of meiosis within the gonads (ovaries or testes) of the offspring.
b)
If the offspring is female, she will make ova in which the gene is imprinted, which
she passes on to her own offspring.
c)
If the offspring is male, he makes sperm in which the gene is not imprinted.
Pleiotropy occurs when one gene effects many phenotypes-- an example is an allele for an
abnormal cartilage protein in rats will effect size of head, shape of snout, breathing rate, size, etc.
Polygenic inheritance occurs when many genes are responsible for a single physical trait.
1.
Height in humans is an example.
2.
Polygenic inheritance produces many phenotypes that generally cover a range of
intermediates.
Conditional gene expression occurs when environmental factors (typically temperature) influence
expression of a gene) e.g. heat inhibits expression of fur pigment in Siamese cats, so they are
dark on the legs, ears, and other cool places.
A sex limited trait is one that shows itself only in one sex, e.g. a gene for uterine cancer.
Sex influenced traits are expressed more frequently in one sex or another even though the genes
are inherited by both, e.g. male pattern baldness influenced by testosterone levels, males show it,
females usually just thinning of hair after menopause.
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136
EDUCATIONAL TELEVISION
GENERAL BIOLOGY
STUDY GUIDE
137
138
Lesson 0, Biology and the Characteristics of Organisms
I.
II.
Describe the characteristics of life shared by all organisms.
Should viruses be considered organisms? Why or Why not?
Lesson 1, The Nature of Science
1) Describe the major steps in the scientific method (process)
2) Describe, in detail, the components and characteristics of an experiment.
3) You wish to determine whether a fertilizer actually improves the yield of corn in your vegetable garden, so
you decide to do an experiment. 50 plants get fertilizer, 50 plants do not get fertilizer, and you weigh the
ears of corn as you harvest.
A. What is the independent variable?
B. What is the dependent variable?
C. Which is the experimental group and control group?
D. What variables must be controlled in the two groups?
E. Why must you have a control group?
4) What does it mean when scientists have a “95% confidence” in experimental results?
5) What do scientists mean when they say there is a “significant difference” between two groups?
6) How do scientists communicate results, and why is it important to do so?
7) Compare inductive and deductive logic.
8) Consider the following scenario: You are not doing well in a class and notice that some of the “good”
students study frequently. You poll fifty students and determine that “A” and “B” students study more
hours per week than “C through F” students. You conclude that the more one studies, the more likely that
person is to get an “A” or “B” grade. What type of logic have you used to reach this conclusion?
9) How is hypothesis different from theory?
10) How is the public’s use of the word “theory” different from the biologist’s use of the word?
11) What are the characteristics of good science?
12) Why does the public have such a difficult time recognizing good science from bad science?
13) Consider the following scenarios and determine the scientific problems with each:
a) You have a pain in your knee. Your doctor says you have either a broken patella (kneecap) or damaged
cartilage. He does an X-ray on your knee and the kneecap is not broken. He tells you it is cartilage
damage and needs to do surgery. Has he made a reasonable scientific conclusion?
b) A back injury has left you are paralyzed from the waist down and suffering from intense pain. You are
considering a treatment in another state that consists of a series of injections of a substance under the
skin along your back, which is purported to reduce pain, and improve mobility. A friend of yours had
the treatment, and feels he is improved. In your consultation with the practitioner, he tells you the
ingredients for the “medication” were revealed to him in a dream by God, and he is convinced that it
has helped dozens of people. He has not done a controlled study. You are impressed with his passion
and faith and decide to have the treatment. Is your decision a scientifically valid one? Explain.
Lesson 2, Basic Chemistry--Atomic Structure
1.
2.
3.
4.
5.
What is matter, and in what three states can it exist?
What is the relationship between an atom and an element?
Describe the three primary subatomic particles, their location, and approximate mass.
What is the unit of mass in atomic measurements?
What forces hold the nucleus of the atom together, and the electrons in their shells?
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6. Use the atomic number and atomic mass from the periodic table of elements to determine the number of
protons and neutrons in the nucleus of the most common atom of any three elements of your choosing.
7. Determine the number of protons, and neutrons in the nuclei of U-235 and U-238, two isotopes of
Uranium.
8. What are isotopes, and what makes some isotopes radioactive?
9. Describe the different types of radiation?
10. What is the maximum number of electrons that can be held by each of the first six electron shells of an
atom?
11. Diagram electron distributions for the following atoms: He, Li, C, O, Zn.
12. What is the valence shell of an atom?
13. Under what conditions is an atom chemically stable, and under what conditions is an atom chemically
reactive?
14. What are cations and anions and why do they form?
15. Determine which of the following atoms are chemically stable and which are chemically reactive
(unstable): He, Li, C, O, Zn.
Lesson 3, Basic Chemistry--Chemical Bonding and Reactions
1. What are the two chemical bonds that can form between atoms; which involves a transfer of electrons and
which a sharing of electrons; and which forms a stronger bond?
2. Make before and after diagrams of the following ionically bonded compounds: MgCl2, CaCl2, HCl, NaF.
3. Define the terms ion, cation, and anion.
4. What are characteristics of ionically bonded compounds?
5. Make before and after diagrams of the following covalently bonded molecules: NH3, H2, H2S.
6. How are polar and nonpolar covalent bonds different?
7. What actually holds the atoms together in an ionic bond?
8. What holds the atoms together in a covalent bond?
9. What are characteristics of covalently bonded molecules.
10. What are compounds and molecules?
11. What is electronegativity and how does it determine whether two or more atoms will form ionic, polar
covalent, or nonpolar covalent bonds?
12. Given the electronegativities below, determine whether the following compounds and molecules would be
composed of ionic, polar covalent, or nonpolar covalent bonds. Electronegativities: Li=1.0, H=2.2,
Ca=1.0, K=0.8, S=2.5, N=3.0, F=4.0, Cl=3.0, Zn=1.6. Compounds and molecules: HCl, H2S, LiF, KCl,
CaCl2, NH3., ZnCl2, H2.
13. Why is a bond between Ca and Li unlikely?
14. What is the relationship between the following terms: polar, nonpolar, water soluble, fat soluble,
hydrophilic, hydrophobic, lipophilic, and lipophobic?
15. What is a chemical reaction, and how is it related to a chemical equation?
16. In the following chemical equations identify the following and describe what they represent: atomic
symbols, chemical formulas, coefficients, yield signs, product(s), reactant(s).
2HF ---> H2 + 2F
H2CO3 ---> H+ + HCO3-
17. Does the equation in the previous question obey the Law of Conservation of Matter, i.e. do products and
reactants have the same number of each type of atom? Explain.
18. Why are elements arranged in vertical columns on the periodic table considered to be in the same chemical
family, i.e. why do they have similar chemical behaviors?
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Lesson 4, Biochemistry: General Concepts and Important Inorganics
1. Describe how organic molecules are different from inorganic molecules.
2. Which of the following are organic and which are inorganic: H2S, LiF, C5H10O5, KCl, CaCl2, CH2(NH2)2
3. What is the relationship between a monomer and polymer?
4. How is homopolymer different from a heteropolymer?
5. Describe a hydrolysis reaction.
6. What is the role of a hydrolysis reaction in biochemistry?
7. Describe a dehydration reaction.
8. What is the role of a dehydration reaction in biochemistry?
9. What are some examples of functional groups of organic molecules, and why are they important?
10. What are hydrogen bonds?
11. What are the major characteristics/properties of water?
12. In which properties of water is hydrogen bonding important?
13. What characteristic/property is unique to water, and why is water so important to life?
14. What is an acid, and give an example?
15. What is a base (alkali), and give an example?
16. Describe the chemical reaction that occurs when you combine equal parts acid and base?
17. What is the pH scale and describe the pH range of acid, base, and neutral solutions?
18. Assume the pH’s of several solutions are as follows: 2.0, 7.1, 5.4, 14.3, and 9.7. Answer the following
questions:
Which solution is the strongest acid?
Which solution is the strongest base?
Which solution is the weakest acid?
Which solution is the closest to neutral?
In which solutions is the H+ concentration greater than the OH- ?
19. What are buffers, how do they work, and why are they important?
Lesson 5, Biochemistry: Carbohydrates and Proteins
1. If a monosaccharide has five carbons, what is the molecular formula (i.e. how many hydrogens and
oxygens does it have)?
2. What is the function of carbohydrates in nature?
3. What is the monomer when talking about carbohydrates?
4. Glucose and fructose have the same chemical formula but different structural formulas--what is the term
for this situation?
5. Indicate where carbons, oxygens, and hydrogens would go on the (imaginary) sugar diagrammed below.
6. What chemical reaction bonds monosaccharides together to form monosaccharides and disaccharides?
7. Define monosaccharide, disaccharide, and polysaccharide, and give examples of each.
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8. Which carbohydrates are energetic and which are structural?
9. Describe the relationship of the following: carbohydrates, polysaccharides, glucose, fructose, sucrose,
maltose, ribose, deoxyribose, lactose, cellulose, glycogen, disaccharide, starch, monosaccharide.
10. Describe the relationship between proteins and amino acids.
11. How does the term “amino acid” relate to its structure?
12. Describe a peptide bond.
13. How are all twenty amino acids chemically similar, and how are they different?
14. What are essential and nonessential amino acids?
15. How are complete and incomplete proteins different, and how would you get each in your diet?
16. How do strict vegetarians ensure they get all essential amino acids?
17. What is the primary structure of a protein and why is it so important to the three dimensional shape of the
protein?
18. What kinds of interactions occur between side chains of amino acids in a protein?
19. Describe the three dimensional shapes of secondary, tertiary, and quaternary proteins.
20. What is the relationship between proteins, pH, and temperature?
21. What substance determines the primary structure of a protein? What effect might it have if the substance
made mistakes in the primary structure of a protein?
Lesson 6, Biochemistry: Proteins and Lipids
1.
2.
3.
4.
5.
What are enzymes and what do they do?
What is energy of activation and how is it related to enzyme function?
What is a catalyst, and how does an enzyme meet the definition?
What kinds of reactions do enzymes carry out?
Describe how enzymes do their work using the terms active site, substrate(s), enzyme-substrate complex,
complementary shapes, and product(s).
6. Describe the other major functions of enzymes?
7. What is the relationship between lipids and hydrocarbons?
8. Describe the structural similarities and differences in saturated, unsaturated (mono and poly), and partially
hydrogenated fatty acids.
9. Which fatty acids are associated with the following: HDL’s, packaged foods, cardiovascular disease, trans
chain, cis chain, LDL’s, good cholesterol, animal fats, plant oils, double bonds between carbons.
10. What are some other examples of lipids?
11. What is the approximate caloric value of carbohydrates, proteins, lipids, and ethyl alcohol.
Lesson 7, Plasma Membranes and Solutions
1)
2)
3)
4)
5)
6)
7)
8)
9)
What is the structure of a plasma membrane?
What are functions of plasma membranes?
Why are plasma membranes described as semi-permeable or selectively permeable?
What is the role of pore proteins and transport proteins in a plasma membrane?
Describe the relationship between the following: solute, solvent, solution.
What is an aqueous solution?
Identify and name the relative concentrations of the solute and solvent in a 10% glucose aqueous solution.
Describe diffusion, and how is diffusion and how is concentration gradient related to diffusion?
Do particles in solution diffuse from areas of high concentration to low concentration or vice versa?
142
10) Imagine a 5% starch solution is separated from a 7% starch solution by a semi-permeable membrane that is
permeable to water, but not the starch. Answer the following questions:
a) Which solution is hypertonic?
b) Which solution is hypotonic?
c) Which solution has a greater solute concentration?
d) Which solution has a greater solvent (water) concentration?
e) Which solution has a greater osmotic pressure?
f) Will osmosis occur, and if so towards which solution?
g) Define osmosis.
h) If the two solutions were to equilibrate at 6% starch, how would you describe the solutions in terms of
their tonicity (hypertonic, hypotonic, isotonic)?
11) Define dialysis.
12) Imagine a 5% glucose solution is separated from a 7% glucose solution by a semi-permeable membrane that
is permeable to both water and glucose. Answer the following questions:
a) Which solution is hypertonic?
b) Which solution is hypotonic?
c) Which solution has a greater solute concentration?
d) Which solution has a greater solvent (water) concentration?
e) Which solution has a greater osmotic pressure?
f) Will osmosis occur, and if so towards which solution?
g) Will dialysis occur, and if so how will glucose diffuse?
13) Compare and contrast facilitated transport with active transport.
14) Why can active transport oppose a concentration gradient, whereas facilitated transport cannot?
15) Describe endocytosis and exocytosis, and how are they related to pinocytosis and phagocytosis?
16) How do the following influence the osmotic pressure of a solution: the number of electrolytes it forms in
solution, and mass of the particles?
17) How do temperature, mass of a particle, and concentration gradient influence the rate of diffusion?
Lesson 8, Cells and Their Structures
1) Describe the relationship between surface area and volume as an object gets larger in size.
2) Why are cells so small?
3) Why do cells need a high surface area to volume ratio?
4) How are prokaryotic cells different from eukaryotic cells?
5) How are bacterial and plant cell walls different?
6) How are prokaryotic and eukaryotic chromosomes different?
7) What is the structure of the nuclear envelope, and what is its relationship with endoplasmic reticulum?
8) What is the relationship between chromosomes, chromatin, DNA, and protein?
9) What is the cytoskeleton of a cell?
10) Describe the relationship between endoplasmic reticulum, golgi bodies, and the cell membrane in the
process of secretion.
11) Why are mitochondria and plastids thought to be descended from bacterial ancestors?
12) What is a centrifuge and what is its contribution to the study of cell biology in fractionation?
13) Be sure to complete the cell worksheet on worksheet number 1.
Lesson 9, Energetics: An Overview
1. Describe the first and second laws of energy.
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2. What is a mole?
3. What is a Kilocalorie (Kcal)?
4. What is the function of ATP in a cell and why is it required for life?
5. Of what components if ATP composed?
6. Write the equation for the hydrolysis of ATP.
7. How much energy is released in the hydrolysis of mole of ATP?
8. Is this an endergonic or exergonic reaction, explain your answer?
9. What is meant by the term G?
10. Why is the G a negative number for an exergonic reaction?
11. What is an endergonic reaction and why is the G a positive number?
12. From where do enzymes that carry out endergonic reactions get the energy needed to complete the
endergonic reaction?
13. If an enzyme conducted the following reaction: C + D -----> CD,
G = + 0.5 Kcal, it would need to be coupled with an energy releasing reaction. Name the reaction it
would need to be coupled with, and what would happen to any extra energy not needed by the enzyme?
14. Describe a fusion reaction.
15. Why is it considered a nuclear reaction rather than a chemical reaction?
16. What happens to some of the mass of the particles in a nuclear reaction?
17. Where is fusion occurring in our solar system?
18. Place the following parts of the electromagetic spectrum in order from smallest wavelength to longest
wavelength: blue, xray, radio, orange, yellow, infra red, ultraviolet, red, violet, radio, yellow, green,
gamma.
19. What is the overall equation for photosynthesis?
20. What can plants do with solar energy that animals cannot?
21. What can plants do with carbon dioxide and water that animals cannot?
22. Why are plants considered producers in food chains?
23. What are consumers, and why do they “consume?”
24. Label the members of the following food chain as producers, and the appropriate consumer, i.e. primary,
secondary, tertiary, etc : flower nectar is eaten by a butterfly, which is eaten by a frog, which is eaten by a
snake, which is eaten by a red tailed hawk.
25. What are higher order consumers in the example above?
26. What is a food web?
27. Describe what is meant by the terms herbivore, carnivore, and omnivore, and give examples of each.
28. Describe an energy pyramid, and what is a trophic level.
29. What is the “10% rule” of nature?
30. How much energy in an ecosytem goes from one trophic level to the next, and what happens to the rest of
the energy?
31. What fraction of the energy originally in grain is held in human tissues if we do the following: grain is fed
to chickens, when chickens die they are ground up in cattle feed and fed to cattle, and the cattle are eaten by
humans.
32. What fraction of the energy originally in grain is held in human tissue if we do the following: grain eaten
by humans.
33. Explain the statement: matter recycles in ecosystems.
34. Explain the statement: energy does not recycle in ecosystems.
Lesson 10, Energetics: Pieces of the Energy Puzzle
1. What is oxidation?
2. What is reduction?
144
3. Is NAD+ oxidized or reduced in the following equation, explain:
NAD+ + 2e- + H+ -----> NADH
4. Is NADPH oxidized or reduced in the following equation, explain:
NADPH -----> NADP+ + 2e- + H+
5. What is a coenzyme and give three examples?
6. Why are coenzymes called electron carriers?
7. Describe how the coenzymes are reduced and oxidized.
8. What is an enzymatic pathway, can you think of an analogy?
9. What is an enzymatic cycle, can you think of an analogy?
10. What is an electron transport chain, or electron transport system (ETS)?
11. What is required at either end of an electron transport chain, to ensure a flow of electrons?
12. What is the hydrogen pump in an ETS, what is its function and how does it work?
13. What is ATP synthase, and how does it work, i.e. explain chemiosmotic phosphorulation.
14. Summarize, explain how an electron source, terminal electron acceptors, the electron transport system
(with its hydrogen pump), and ATP synthase work together to make ATP from ADP and phosphate in the
process of chemiosmotic phosphorulation.
15. What provides the energy for the production of ATP in chemiosmotic phosphorulation.
Lesson 11, Energetics: Photosynthesis
1. What is the overall equation for photosynthesis?
2. Be familiar with the following parts of a chloroplast: outer membrane, inner membrane, thylakoid
membrane, outer compartment, stroma, thylakoid space, grana.
3. What is the relationship between the inner membrane, thylakoid membrane, and grana.
4. What is the relationship between the outer compartment and the thylakoid space.
5. What are photosytems and where are they located?
6. What types of molecules gather energy in a photosystem, and why are they sometimes referred to as
antenna molecules?
7. What happens to photosystem molecules when they absorb a photon of light energy?
8. What is resonance transfer of energy?
9. Why is the cholorphyll-a at the reaction center required for the initiation of photosynthesis.
10. Describe the light reaction of photosynthesis.
11. Electrons lost by the p680 chlorophyll-a are replaced by electrons released from the breakup of what
molecule?
12. What is the role of the ETS of photosystem II, and what happens to the electrons at the end of the ETS of
photosystem II?
13. What replaces the electrons lost by the p700 chlorophyll-a of photosystem I?
14. What is the terminal electron acceptor for the ETS of photosystem I?
15. The light reaction has two ETS’s, what is considered the source of electrons and what is considered the
terminal electron acceptor?
16. Why is it called the light reaction, why is it limited to daylight in nature?
17. How is ATP produced in the light reaction?
18. What are the products of the light reaction?
19. What is cyclic flow of electrons in the light reaction?
20. Including cyclic flow, how many ATP are produced for each electron generated by the lysis (breakup) of
water?
21. Where does the light reaction take place within the chloroplast?
22. Explain the following statement: the light reaction converts solar energy to chemical energy in the form of
ATP.
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23. Describe the fixation of carbon dioxide in the Dark reaction (Calvin Cycle).
24. Describe the production of glucose in the Dark reaction (Calvin Cycle).
25. Describe the regeneration of RuBP in the Dark reaction (Calvin Cycle).
26. Is this an enzymatic cycle or pathway? Explain.
27. Where does the dark reaction occur?
28. What are the products of the dark reaction?
29. What is the role of RuBP?
30. What is the role of rubisco?
31. Where are ATP and NADPH required in the dark reaction?
32. What is the source of the ATP and NADPH required for the dark reaction?
33. How many ATP are required to make one glucose?
34. How many NADPH are required to make one glucose?
35. What happens to ADP, Pi, and NADP+ generated by the dark reaction?
36. Why does the plant make glucose?
37. What molecules does the dark reaction start and end with?
38. What is the source of carbon and oxygen for the construction of glucose in the dark reaction?
39. What is the source of hydrogen for the construction of glucose in the dark reaction?
40. Is light required for the dark reaction? Explain.
41. Summarize--talk yourself through the events of the light and dark reactions of photosynthesis.
42. Describe how carbon dioxide gets into the chloroplast from the ouside of a leaf, using the following in your
description: cuticle, guard cells, stomata, air spaces, mesophyll (parenchyma) cells, chloroplasts.
43. Under what conditions are the stomata of a leaf open, and under what conditions are the stomata closed?
44. When during the day are stomata typically open and closed?
45. What are C4 plants, and how are they adapted to drier climates?
46. What are CAM plants and how are they different from C4 plants?
Lesson 12, Energetics: The Cellular Oxidation of Glucose
1. Describe the structure of a mitochondrion using the follwing terms: outer membrane, inner membrane,
outer compartment, matrix, crista (cristae), and describe the relationship betweent the inner membrane and
cristae.
2. What is the equation for the cellular oxidation of glucose?
3. How many ATP are produced for each glucose, via the cellular oxidation of glucose?
4. Describe the process of glycolysis.
5. Where in the cell does glycolysis occur?
6. Is glycolysis an enzymatic pathway or cycle? Explain.
7. What types of organisms practice glycolysis, is it limited to animals?
8. With what molecule does glycolysis begin?
9. With what molecules does aerobic glycolysis end?
10. For each glucose, how many ATP are consumed in glycoylsis?
11. For each glucose, how many ATP are produced in glycolysis? What is the net gain in ATP from
glycoylysis?
12. Describe lactate fermentation and what organisms practice lactate fermentation?
13. Under what conditions does lactate fermentation occur, and what are the products of lactate fermentation?
14. Is lactate fermentation a reversible process? Explain.
15. Describe alcohol fermentation and what organisms practice alcohol fermentation?
16. Under what conditions does alcohol fermentation occur, and what are the products of alcohol fermentation?
17. Is alcohol fermentation a reversible process? Explain.
18. What types of organisms practice cellular respiration, and within what organelle does it occur?
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19. How does pyruvate get into the mitochondrion?
20. Describe the conversion of pyruvate to acetyl CoA?
21. How many pyruvates and acetylCoA molecules are produced for each glucose molecule?
22. Describe the Kreb’s Cycle, why is it an enzymatic cycle?
23. With what molecule does the Krebs Cycle start and end?
24. What are the products of the Krebs cycle?
25. How many ATP are generated for each glucose by the Kreb’s cycle?
26. What is the source of electrons for the mitochondrial electron transport system (ETS)?
27. What is the terminal electron acceptor for the mitochondrial ETS?
28. Describe how hydrogen ions are pumped by the ETS and how ATP is made by chemiosmotic
phosphorulation.
29. How many ATP per glucose are produced by the ETS and chemiosmotic phosphorulation?
30. What is the role of oxygen in ATP production?
31. What is the role of NADH, and FADH2 in ATP production?
32. How are NAD+ and FAD important in the cellular oxidation of glucose?
33. Describe the chain of events that occurs if oxygen is deprived, i.e. why will you die if you do not get
oxygen to your cells? Go step by step, the effect on the ETS and chemiosmotic phosphorulation, the Krebs
cycle, the conversion of pyruvate to acetyl coA, and glycolysis.
34. How many ATP are produced per glucose in the cellular oxidation of glucose by glycolysis, the Krebs
cycle, and chemiosmotic phosphorulation, and what is the total number of ATP produced per glucose?
35. How many ATP are produced per glucose by prokaryotes? Explain your answer.
36. How many ATP are produced per glucose by eukaryotes under anaerobic conditions?
37. How are proteins and lipids linked to cellular respiration?
38. How is it that one can eat excess carbohydrates and protein, yet still store excess Calories as fat?
Lesson 16, Mitosis
1. What is the difference between growth and reproduction?
2. How are sexual and asexual reproduction different?
3. Describe the relationship of the following terms: ploidy, haploid, diploid, polyploid, somatic cells, gametes,
sex cells, homologous pairs of chromosomes.
4. Describe the relationship between the following terms: mitosis, interphase, cell cycle, G1, G2, G0, S,
prophase, metaphase, anaphase, telophase.
5. Describe the events that occur in each of the subphases of interphase.
6. What does an interphase cell look like? Can one observe the many events of the subphases with the light
microscope?
7. Describe the events of prophase.
8. Why do the sister chromatids of a chromosome become visible in prophase?
9. Describe the events of metaphase.
10. What are astral rays and what type of cells have them?
11. Describe the events of anaphase.
12. What is karyokinesis and how is it related to anaphase?
13. Describe the events of telophase.
14. How is telophase different in plants and animals?
15. What is cytokinesis, and how is it related to daughter cells and telophase?
16. What is the relationship between daughter cells and G1.
17. Genetically, are the daughter cells similar, identical, or different? Explain your answer.
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18. When you take a cutting from a plant, stick it into the ground and it grows into a new plant, it does so by
mitotic division. What type of reproduction is this, sexual or asexual, and are the plants genetically
identical to one another? Explain your answer.
Lesson 17, Meiosis
1. What are gonads and gametes?
2. What is an homologous pair of chromosomes?
3. Describe the role of meiosis and mitosis in sexual reproduction.
4. What are the steps of meiosis 1 and meiosis 2?
5. Describe the events of interphase 1.
6. Describe the events of prophase 1. What is a tetrad?
7. Describe crossing over, and why is it important?
8. How is metaphase 1 different from mitotic metaphase?
9. Describe independent assortment, and what .
10. Describe the events of anaphase 1 and telophase 1.
11. How is interphase 2 different from interphase 1?
12. How is prophase 2 different from prophase 1?
13. How is metaphase 2 different from metaphase 1?
14. Describe the events of anaphase 2 and telophase 2.
15. How many gametes are produced from each cell that undergoes meiosis?
16. What is the ploidy of the gametes produced by meiosis?
17. Why are the gametes produced by meiosis genetically different from one another, and the cell from which
they developed?
18. Describe some of the problems that can occur in meiosis (translocation, inversion, non-disjunction,
trisomies).
19. What is the benefit of meiosis and sexual reproduction in nature.
20. Why is genetic diversity important?
Lesson 18, Mendel and Single Trait Inheritance
1. Define the following terms: offspring, gene, locus, homologous pair, diploid, haploid, paternal, maternal,
gene pair, alleles, genotype, phenotype.
2. How are dominant alleles different from recessive alleles?
3. What is the relationship between the following: gene pair, alleles, pp, Pp, PP, homozygous dominant,
homozygous recessive, heterozygous.
4. What is the genotype of an individual who shows a recessive phenotype?
5. Who was Gregor Mendel and why is he held in such high esteem by biologists?
6. What was Mendel’s “tool” and what traits did he study?
7. What was the blending theory of inheritance of the 19th century?
8. Describe how Mendel tested the blending theory.
9. What is Mendel’s Law of Segregation?
10. What is meant by P, F1, and F2 generations?
11. Describe the predicted genotypic and phenotypic outcomes in the offspring of the following parental
crosses (I = pea pods inflated, i= pea pods constricted): II x II, ii x ii, II x ii, Ii x Ii, Ii x II, Ii x ii
Lesson 19, Solving Problems and Multiple Trait Inheritance
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1. How is complete dominance different from incomplete dominance?
2. What would be the expected genotypic outcome of a heterozygous hornless bull mating with a horned cow?
Hornless is dominant to horned.
3. A hornless bull has a horned sire(father) and a hornless dam(mother). What are the genotypes of all
individuals mentioned above?
4. In animals, albinism (no pigment) is recessive to normal coloration (brown or black pigment). Three
brown guinea pigs and two white guinea pigs were born to parents, one of which was brown and the other
an albino.What was the genotype of each parent? What are the genotypes of the offspring?
5. In guinea pigs, rough coat (long fur) is dominat to smooth coat (short fur). An animal with a smooth coat is
mated to an animal that is homozygous for rough coat. What are possible phenotypes and genotypes of
offspring?
6. A woman who is Rh- had parents with Rh+ blood (both parents). She married a man with Rh+ blood,
whose father was Rh+ and whose mother was Rh-. If the man and the woman have a child, what are the
possible genotypes of the child? What are the probabilities that the child will have a particular Rh
genotype?
7. Determine all genotypes in the pedigree for eye color below. Brown (and hazel) eyes are dominant to nonbrown (green or blue). Shaded individuals have brown or hazel eyes.
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8.
Determine whether the shaded trait is encoded by a dominant or recessive allele. This is a pedigree for fur
color in guinea pigs. Shaded individuals have brown fur, and the unshaded individuals have black fur.
Determine the genotypes of all individuals.
9. Determine the genotypes of the individuals in the pedigree below. This pedigree is for Landsteiner blood
types. A and B alleles are (co)dominant, and i alleles are recessive.
A
B
B
O
O
B
Lesson 20, Multiple Trait Inheritance and Epistasis
1. What is Mendel’s Law of Independent Assortment, and how is it related to meiosis?
2. If you cross dihybrids and the traits assort independently, what phenotypic ratio would you expect to see in
the offspring?
3. If traits assort independently, are they on the same or different chromosomes? Explain.
4. If traits assort independently, why can one use random laws of probability, like the multiplicative property
of probability, to predict outcomes?
5. If traits are on the same chromosome, can one multiply probabilities to predict outcomes? Explain your
answer.
6. What would be the predicted genotypic and phenotypic outcomes in the following parental cross of traits
that assort independently (R=round seed, r=wrinkled seed; Y=yellow seed, y=green seed; T=tall plant,
t=short plant: RrYyTt x RrYYTt.
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7. What would be the probability of getting a RrYy offspring from the following cross: RrYy x RRYY
(assume independent assortment)? What would be the probability of getting the following offsring from
the same parental cross: rryy, RRYY?
8. Why do linked traits not assort independently, and if dihybrids for linked genes were crossed would the
offspring show a 9:3:3:1 phenotypic ratio? Explain.
9. Describe epistasis.
10. Can a black horse mated to a chesnut horse produce a bay foal? Can two bay horses produce a chesnut
horse? Can two black horses produce a chesnut? Explain your answers.
11. In the following cross of two horses (involving the horse Agouti and Extension genes), AaEe x AaEe, what
color are the parents, and what is the probability of having a bay foal?
Lesson 21, X-Linked Traits Imprinting and More
1.
2.
3.
4.
5.
6.
How are sex chromosomes different from autosomes, and how many pairs of each do human cells possess?
What sex chromosome combinations produce males and females?
Who pioneered X linked traits and what was his tool?
Why do males show a greater incidence of X linked recessive phenotypes?
Who determines the sex of a child, the father or the mother? Explain.
In cats coat color is X linked. XB=black fur, XY=yellow fur. Describe the phenotypes of the following cats
and the role of Lyonization: XBY, XYY, XB XB, XY XY, XB XY
7. Colorblindness is an x-linked trait. Normal vision is dominant to colorblindness. A colorblind man
marries and has a daughter with a woman who has normal vision, whose father was colorblind. What is the
probability of their daughter being colorblind? If they had a son, could he be colorblind?
8. Determine all genotypes in the pedigree, below. Shaded individuals have hemophilia, an x-linked recessive
bleeding disorder. H= normal blood, h = hemophilia.
9. Define the following: pleiotropy, polygenic inheritance, conditional gene expresssion, sex limited traits,
and sex influenced traits.
10. Describe how a maternally imprinted gene gets passed from generation to generation.
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11. Lets say that an allele for hair color is paternally imprinted. Dark hair (H)is dominant over light hair(h). An
individual is heterozygous for dark hair (Hh), but the dominant allele was inherited from the father. What
color hair will the individual have? Explain your answer.
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