EU Test Material

Cells

WHAT IS A CELL?

Unicellular organisms are capable of (i) independent existence and (ii) performing the
essential functions of life. Anything less than a complete structure of a cell does not ensure
independent living. Hence, cell is the fundamental structural and functional unit of all living
organisms. Anton Von Leeuwenhoek first saw and described a live cell. Robert Brown later
discovered the nucleus. The invention of the microscope and its improvement leading to the
electron microscope revealed all the structural details of the cell.

CELL THEORY

In 1838, Matthias Schleiden, a German botanist, examined a large number of plants and
observed that all plants are composed of different kinds of cells which form the tissues of the
plant. At about the same time, Theodore Schwann (1839), a British Zoologist, studied
different types of animal cells and reported that cells had a thin outer layer which is today
known as the ‘plasma membrane’. He also concluded, based on his studies on plant tissues,
that the presence of cell wall is a unique character of the plant cells. On the basis of this,
Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells
and products of cells. Schleiden and Schwann together formulated the cell theory. This theory
however, did not explain as to how new cells were formed. Rudolf Virchow (1855) first
explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e
cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final
shape. Cell theory as understood today is: (i) all living organisms are composed of cells and
products of cells. (ii) all cells arise from pre-existing cells.

Comparing Prokaryotic and Eukaryotic Cells

Cells fall into one of two broad categories: prokaryotic and eukaryotic. The predominantly
single-celled organisms of the domains Bacteria and Archaea are classified as prokaryotes
(pro– = before; –karyon– = nucleus). Animal cells, plant cells, fungi, and protists are
eukaryotes (eu– = true).

Components of Prokaryotic Cells

All cells share four common components: 1) a plasma membrane, an outer covering that
separates the cell’s interior from its surrounding environment; 2) cytoplasm, consisting of a
jelly-like region within the cell in which other cellular components are found; 3) DNA, the
genetic material of the cell; and 4) ribosomes, particles that synthesize proteins. However,
prokaryotes differ from eukaryotic cells in several ways.

A prokaryotic cell is a simple, single-celled (unicellular) organism that lacks a nucleus, or

any other membrane-bound organelle. We will shortly come to see that this is significantly
different in eukaryotes. Prokaryotic DNA is found in the central part of the cell: a darkened
region called the nucleoid.
 EU Test Material

This figure shows the generalized structure of a prokaryotic cell.

Unlike Archaea and eukaryotes, bacteria have a cell wall made of peptidoglycan, comprised
of sugars and amino acids, and many have a polysaccharide capsule (Figure). The cell wall
acts as an extra layer of protection, helps the cell maintain its shape, and prevents
dehydration. The capsule enables the cell to attach to surfaces in its environment. Some
prokaryotes have flagella, pili, or fimbriae. Flagella are used for locomotion, while most pili
are used to exchange genetic material during a type of reproduction called conjugation.

Eukaryotic Cells

In nature, the relationship between form and function is apparent at all levels, including the
level of the cell, and this will become clear as we explore eukaryotic cells. The principle
“form follows function” is found in many contexts. For example, birds and fish have
streamlined bodies that allow them to move quickly through the medium in which they live,
be it air or water. It means that, in general, one can deduce the function of a structure by
looking at its form, because the two are matched.

A eukaryotic cell is a cell that has a membrane-bound nucleus and other membrane-
bound compartments or sacs, called organelles, which have specialized functions. The
word eukaryotic means “true kernel” or “true nucleus,” alluding to the presence of the
membrane-bound nucleus in these cells. The word “organelle” means “little organ,” and, as
already mentioned, organelles have specialized cellular functions, just as the organs of your
body have specialized functions.

Cell Size

At 0.1–5.0 µm in diameter, prokaryotic cells are significantly smaller than eukaryotic cells,
which have diameters ranging from 10–100 µm (Figure). The small size of prokaryotes
allows ions and organic molecules that enter them to quickly spread to other parts of the cell.
Similarly, any wastes produced within a prokaryotic cell can quickly move out. However,
larger eukaryotic cells have evolved different structural adaptations to enhance cellular
 EU Test Material

transport. Indeed, the large size of these cells would not be possible without these
adaptations. In general, cell size is limited because volume increases much more quickly than
does cell surface area. As a cell becomes larger, it becomes more and more difficult for the
cell to acquire sufficient materials to support the processes inside the cell, because the
relative size of the surface area across which materials must be transported declines.

This figure shows the relative sizes of different kinds of cells and cellular components. An
adult human is shown for comparison.

Section Summary

Prokaryotes are predominantly single-celled organisms of the domains Bacteria and Archaea.
All prokaryotes have plasma membranes, cytoplasm, ribosomes, a cell wall, DNA, and lack
membrane-bound organelles. Many also have polysaccharide capsules. Prokaryotic cells
range in diameter from 0.1–5.0 µm.

Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes,
but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its
DNA is surrounded by a membrane), and has other membrane-bound organelles that allow
for compartmentalization of functions. Eukaryotic cells tend to be 10 to 100 times the size of
prokaryotic cells.

Eukaryotic cells have a more complex structure than do prokaryotic cells. Organelles allow
for various functions to occur in the cell at the same time. Before discussing the functions of
organelles within a eukaryotic cell, let us first examine two important components of the cell:
the plasma membrane and the cytoplasm.
 EU Test Material

Figure (a) This figure shows a typical animal cell

Figure (b) This figures shows a typical plant cell.

 EU Test Material

What structures does a plant cell have that an animal cell does not have? What structures does
an animal cell have that a plant cell does not have? Plant cells have plasmodesmata, a cell
wall, a large central vacuole, chloroplasts, and plastids. Animal cells have lysosomes and
centrosomes.

The Plasma Membrane

Like prokaryotes, eukaryotic cells have a plasma membrane (Figure 1) made up of

a phospholipid bilayer with embedded proteins that separates the internal contents of the
cell from its surrounding environment. A phospholipid is a lipid molecule composed of two
fatty acid chains, a glycerol backbone, and a phosphate group. The plasma membrane
regulates the passage of some substances, such as organic molecules, ions, and water,
preventing the passage of some to maintain internal conditions, while actively bringing in or
removing others. Other compounds move passively across the membrane.

Figure 1 The plasma membrane is a phospholipid bilayer with embedded proteins. There are
other components, such as cholesterol and carbohydrates, which can be found in the
membrane in addition to phospholipids and protein.

The plasma membranes of cells that specialize in absorption are folded into fingerlike
projections called microvilli (singular = microvillus). This folding increases the surface area
of the plasma membrane. Such cells are typically found lining the small intestine, the organ
that absorbs nutrients from digested food. This is an excellent example of form matching the
function of a structure.

People with celiac disease have an immune response to gluten, which is a protein found in
wheat, barley, and rye. The immune response damages microvilli, and thus, afflicted
individuals cannot absorb nutrients. This leads to malnutrition, cramping, and diarrhea.
Patients suffering from celiac disease must follow a gluten-free diet.

The Cytoplasm

The cytoplasm comprises the contents of a cell between the plasma membrane and the
nuclear envelope (a structure to be discussed shortly). It is made up of organelles suspended
in the gel-like cytosol, the cytoskeleton, and various chemicals. Even though the cytoplasm
 EU Test Material

consists of 70 to 80 percent water, it has a semi-solid consistency, which comes from the
proteins within it. However, proteins are not the only organic molecules found in the
cytoplasm. Glucose and other simple sugars, polysaccharides, amino acids, nucleic acids,
fatty acids, and derivatives of glycerol are found there too. Ions of sodium, potassium,
calcium, and many other elements are also dissolved in the cytoplasm. Many metabolic
reactions, including protein synthesis, take place in the cytoplasm.

The Cytoskeleton

If you were to remove all the organelles from a cell, would the plasma membrane and the
cytoplasm be the only components left? No. Within the cytoplasm, there would still be ions
and organic molecules, plus a network of protein fibers that helps to maintain the shape of
the cell, secures certain organelles in specific positions, allows cytoplasm and vesicles to
move within the cell, and enables unicellular organisms to move independently. Collectively,
this network of protein fibers is known as the cytoskeleton. There are three types of fibers
within the cytoskeleton: microfilaments, also known as actin filaments, intermediate
filaments, and microtubules (Figure 2).

Figure 2 Microfilaments, intermediate filaments, and microtubules compose a cell’s

cytoskeleton.

Microfilaments are the thinnest of the cytoskeletal fibers and function in moving cellular
components, for example, during cell division. They also maintain the structure of microvilli,
the extensive folding of the plasma membrane found in cells dedicated to absorption. These
components are also common in muscle cells and are responsible for muscle cell contraction.
Intermediate filaments are of intermediate diameter and have structural functions, such as
maintaining the shape of the cell and anchoring organelles. Keratin, the compound that
strengthens hair and nails, forms one type of intermediate filament. Microtubules are the
thickest of the cytoskeletal fibers. These are hollow tubes that can dissolve and reform
quickly. Microtubules guide organelle movement and are the structures that pull
chromosomes to their poles during cell division. They are also the structural components of
 EU Test Material

flagella and cilia. In cilia and flagella, the microtubules are organized as a circle of nine
double microtubules on the outside and two microtubules in the center.

The centrosome is a region near the nucleus of animal cells that functions as a microtubule-
organizing center. It contains a pair of centrioles, two structures that lie perpendicular to each
other. Each centriole is a cylinder of nine triplets of microtubules.

The centrosome replicates itself before a cell divides, and the centrioles play a role in pulling
the duplicated chromosomes to opposite ends of the dividing cell. However, the exact
function of the centrioles in cell division is not clear, since cells that have the centrioles
removed can still divide, and plant cells, which lack centrioles, are capable of cell division.

Flagella and Cilia

Flagella (singular = flagellum) are long, hair-like structures that extend from the plasma
membrane and are used to move an entire cell, (for example, sperm, Euglena). When present,
the cell has just one flagellum or a few flagella. When cilia (singular = cilium) are present,
however, they are many in number and extend along the entire surface of the plasma
membrane. They are short, hair-like structures that are used to move entire cells (such as
paramecium) or move substances along the outer surface of the cell (for example, the cilia of
cells lining the fallopian tubes that move the ovum toward the uterus, or cilia lining the cells
of the respiratory tract that move particulate matter toward the throat that mucus has trapped).

The Endomembrane System

The endomembrane system (endo = within) is a group of membranes and organelles in

eukaryotic cells that work together to modify, package, and transport lipids and
proteins. It includes the nuclear envelope, lysosomes, vesicles, endoplasmic reticulum and
the Golgi apparatus, which we will cover shortly. Although not technically within the cell, the
plasma membrane is included in the endomembrane system because, as you will see, it
interacts with the other endomembranous organelles.

The Nucleus

Typically, the nucleus is the most prominent organelle in a cell. The nucleus (plural =
nuclei) houses the cell’s DNA in the form of chromatin and directs the synthesis of
ribosomes and proteins. Let us look at it in more detail (Figure 3).
 EU Test Material

Figure 3 The outermost boundary of the nucleus is the nuclear envelope. Notice that the
nuclear envelope consists of two phospholipid bilayers (membranes)—an outer membrane
and an inner membrane—in contrast to the plasma membrane, which consists of only one
phospholipid bilayer.

The nuclear envelope is a double-membrane structure that constitutes the outermost

portion of the nucleus (Figure 3). Both the inner and outer membranes of the nuclear
envelope are phospholipid bilayers.

The nuclear envelope is punctuated with pores that control the passage of ions, molecules,
and RNA between the nucleoplasm and the cytoplasm.

To understand chromatin, it is helpful to first consider chromosomes. Chromosomes are

structures within the nucleus that are made up of DNA, the hereditary material, and proteins.
This combination of DNA and proteins is called chromatin. In eukaryotes, chromosomes are
linear structures. Every species has a specific number of chromosomes in the nucleus of its
body cells. For example, in humans, the chromosome number is 46, whereas in fruit flies, the
chromosome number is eight.

Chromosomes are only visible and distinguishable from one another when the cell is getting
ready to divide. When the cell is in the growth and maintenance phases of its life cycle, the
chromosomes resemble an unwound, jumbled bunch of threads.
 EU Test Material

Figure 4 This image shows various levels of the organization of chromatin (DNA and

protein).

Figure 5 This image shows paired chromosomes. (credit: modification of work by NIH;
scale-bar data from Matt Russell)

We already know that the nucleus directs the synthesis of ribosomes, but how does it do this?
Some chromosomes have sections of DNA that encode ribosomal RNA. A darkly stained
area within the nucleus, called the nucleolus (plural = nucleoli), aggregates the ribosomal
RNA with associated proteins to assemble the ribosomal subunits that are then transported
through the nuclear pores into the cytoplasm.

The Endoplasmic Reticulum

The endoplasmic reticulum (ER) is a series of interconnected membranous tubules that

collectively modify proteins and synthesize lipids. However, these two functions are
performed in separate areas of the endoplasmic reticulum: the rough endoplasmic reticulum
and the smooth endoplasmic reticulum, respectively.

The hollow portion of the ER tubules is called the lumen or cisternal space. The membrane of
the ER, which is a phospholipid bilayer embedded with proteins, is continuous with the
nuclear envelope.

The rough endoplasmic reticulum (RER) is so named because the ribosomes attached to its
cytoplasmic surface give it a studded appearance when viewed through an electron
microscope.

The ribosomes synthesize proteins while attached to the ER, resulting in the transfer of their
newly synthesized proteins into the lumen of the RER where they undergo modifications
such as folding or addition of sugars. The RER also makes phospholipids for cell membranes.

If the phospholipids or modified proteins are not destined to stay in the RER, they will be
packaged within vesicles and transported from the RER by budding from the membrane.
 EU Test Material

Since the RER is engaged in modifying proteins that will be secreted from the cell, it is
abundant in cells that secrete proteins, such as the liver.

The smooth endoplasmic reticulum (SER) is continuous with the RER but has few or no
ribosomes on its cytoplasmic surface. The SER’s functions include synthesis of
carbohydrates, lipids (including phospholipids), and steroid hormones; detoxification of
medications and poisons; alcohol metabolism; and storage of calcium ions.

The Golgi Apparatus

We have already mentioned that vesicles can bud from the ER, but where do the vesicles go?
Before reaching their final destination, the lipids or proteins within the transport vesicles need
to be sorted, packaged, and tagged so that they wind up in the right place. The sorting,
tagging, packaging, and distribution of lipids and proteins take place in the Golgi
apparatus (also called the Golgi body), a series of flattened membranous sacs.

Figure 6 The Golgi apparatus in this transmission electron micrograph of a white blood cell
is visible as a stack of semicircular flattened rings in the lower portion of this image. Several
vesicles can be seen near the Golgi apparatus. (credit: modification of work by Louisa
Howard; scale-bar data from Matt Russell)

The Golgi apparatus has a receiving face near the endoplasmic reticulum and a releasing face
on the side away from the ER, toward the cell membrane. The transport vesicles that form
from the ER travel to the receiving face, fuse with it, and empty their contents into the lumen
of the Golgi apparatus. As the proteins and lipids travel through the Golgi, they undergo
further modifications. The most frequent modification is the addition of short chains of sugar
molecules. The newly modified proteins and lipids are then tagged with small molecular
groups to enable them to be routed to their proper destinations.

Finally, the modified and tagged proteins are packaged into vesicles that bud from the
opposite face of the Golgi. While some of these vesicles, transport vesicles, deposit their
contents into other parts of the cell where they will be used, others, secretory vesicles, fuse
with the plasma membrane and release their contents outside the cell.

The amount of Golgi in different cell types again illustrates that form follows function within
cells. Cells that engage in a great deal of secretory activity (such as cells of the salivary
glands that secrete digestive enzymes or cells of the immune system that secrete antibodies)
have an abundant number of Golgi.
 EU Test Material

In plant cells, the Golgi has an additional role of synthesizing polysaccharides, some of which
are incorporated into the cell wall and some of which are used in other parts of the cell.

Lysosomes

In animal cells, the lysosomes are the cell’s “garbage disposal.” Digestive enzymes within
the lysosomes aid the breakdown of proteins, polysaccharides, lipids, nucleic acids, and even
worn-out organelles. In single-celled eukaryotes, lysosomes are important for digestion of the
food they ingest and the recycling of organelles. These enzymes are active at a much lower
pH (more acidic) than those located in the cytoplasm. Many reactions that take place in the
cytoplasm could not occur at a low pH, thus the advantage of compartmentalizing the
eukaryotic cell into organelles is apparent.

Lysosomes also use their hydrolytic enzymes to destroy disease-causing organisms that might
enter the cell. A good example of this occurs in a group of white blood cells called
macrophages, which are part of your body’s immune system. In a process known as
phagocytosis, a section of the plasma membrane of the macrophage invaginates (folds in) and
engulfs a pathogen. The invaginated section, with the pathogen inside, then pinches itself off
from the plasma membrane and becomes a vesicle. The vesicle fuses with a lysosome. The
lysosome’s hydrolytic enzymes then destroy the pathogen (Figure 7).

Figure 7 A macrophage has phagocytized a potentially pathogenic bacterium into a vesicle,

which then fuses with a lysosome within the cell so that the pathogen can be destroyed. Other
organelles are present in the cell, but for simplicity, are not shown.

Vesicles and Vacuoles

Vesicles and vacuoles are membrane-bound sacs that function in storage and transport.
Vacuoles are somewhat larger than vesicles, and the membrane of a vacuole does not fuse
with the membranes of other cellular components. Vesicles can fuse with other membranes
within the cell system. Additionally, enzymes within plant vacuoles can break down
macromolecules.
 EU Test Material

Figure 8 The endomembrane system works to modify, package, and transport lipids and
proteins.

Why does the cis face of the Golgi not face the plasma membrane?

<!– Because that face receives chemicals from the ER, which is toward the center of the cell.
–>

Ribosomes

Ribosomes are the cellular structures responsible for protein synthesis. When viewed
through an electron microscope, free ribosomes appear as either clusters or single tiny dots
floating freely in the cytoplasm. Ribosomes may be attached to either the cytoplasmic side of
the plasma membrane or the cytoplasmic side of the endoplasmic reticulum. Electron
microscopy has shown that ribosomes consist of large and small subunits. Ribosomes are
enzyme complexes that are responsible for protein synthesis.

Because protein synthesis is essential for all cells, ribosomes are found in practically every
cell, although they are smaller in prokaryotic cells. They are particularly abundant in
immature red blood cells for the synthesis of hemoglobin, which functions in the transport of
oxygen throughout the body.
 EU Test Material

Mitochondria

Mitochondria (singular = mitochondrion) are often called the “powerhouses” or “energy

factories” of a cell because they are responsible for making adenosine triphosphate (ATP),
the cell’s main energy-carrying molecule. The formation of ATP from the breakdown of
glucose is known as cellular respiration. Mitochondria are oval-shaped, double-membrane
organelles (Figure 9) that have their own ribosomes and DNA. Each membrane is a
phospholipid bilayer embedded with proteins. The inner layer has folds called cristae, which
increase the surface area of the inner membrane. The area surrounded by the folds is called
the mitochondrial matrix. The cristae and the matrix have different roles in cellular
respiration.

In keeping with our theme of form following function, it is important to point out that muscle
cells have a very high concentration of mitochondria because muscle cells need a lot of
energy to contract.

Figure 9 This transmission electron micrograph shows a mitochondrion as viewed with an

electron microscope. Notice the inner and outer membranes, the cristae, and the
mitochondrial matrix.

Peroxisomes

Peroxisomes are small, round organelles enclosed by single membranes. They carry out
oxidation reactions that break down fatty acids and amino acids. They also detoxify many
poisons that may enter the body. Alcohol is detoxified by peroxisomes in liver cells. A
byproduct of these oxidation reactions is hydrogen peroxide, H2O2, which is contained within
the peroxisomes to prevent the chemical from causing damage to cellular components outside
of the organelle. Hydrogen peroxide is safely broken down by peroxisomal enzymes into
water and oxygen.

Animal Cells versus Plant Cells

Despite their fundamental similarities, there are some striking differences between animal
and plant cells (see Table 1). Animal cells have centrioles, centrosomes (discussed under the
cytoskeleton), and lysosomes, whereas plant cells do not. Plant cells have a cell wall,
 EU Test Material

chloroplasts, plasmodesmata, and plastids used for storage, and a large central vacuole,
whereas animal cells do not.

The Cell Wall

In Figure (b), the diagram of a plant cell, you see a structure external to the plasma
membrane called the cell wall. The cell wall is a rigid covering that protects the cell, provides
structural support, and gives shape to the cell. Fungal and protist cells also have cell walls.

While the chief component of prokaryotic cell walls is peptidoglycan, the major organic
molecule in the plant cell wall is cellulose, a polysaccharide made up of long, straight chains
of glucose units. When nutritional information refers to dietary fiber, it is referring to the
cellulose content of food.

Chloroplasts

Like mitochondria, chloroplasts also have their own DNA and ribosomes. Chloroplasts
function in photosynthesis and can be found in eukaryotic cells such as plants and algae. In
photosynthesis, carbon dioxide, water, and light energy are used to make glucose and oxygen.
This is the major difference between plants and animals: Plants (autotrophs) are able to make
their own food, like glucose, whereas animals (heterotrophs) must rely on other organisms for
their organic compounds or food source.

Like mitochondria, chloroplasts have outer and inner membranes, but within the space
enclosed by a chloroplast’s inner membrane is a set of interconnected and stacked, fluid-filled
membrane sacs called thylakoids (Figure 10). Each stack of thylakoids is called a granum
(plural = grana). The fluid enclosed by the inner membrane and surrounding the grana is
called the stroma.

Figure 3.10 This simplified diagram of a chloroplast shows the outer membrane, inner
membrane, thylakoids, grana, and stroma.

The chloroplasts contain a green pigment called chlorophyll, which captures the energy of
sunlight for photosynthesis. Like plant cells, photosynthetic protists also have chloroplasts.
 EU Test Material

Some bacteria also perform photosynthesis, but they do not have chloroplasts. Their
photosynthetic pigments are located in the thylakoid membrane within the cell itself.

Evolution in Action

Endosymbiosis: We have mentioned that both mitochondria and chloroplasts contain DNA
and ribosomes. Have you wondered why? Strong evidence points to endosymbiosis as the
explanation.

Symbiosis is a relationship in which organisms from two separate species live in close
association and typically exhibit specific adaptations to each other. Endosymbiosis (endo-=
within) is a relationship in which one organism lives inside the other. Endosymbiotic
relationships abound in nature. Microbes that produce vitamin K live inside the human gut.
This relationship is beneficial for us because we are unable to synthesize vitamin K. It is also
beneficial for the microbes because they are protected from other organisms and are provided
a stable habitat and abundant food by living within the large intestine.

Scientists have long noticed that bacteria, mitochondria, and chloroplasts are similar in size.
We also know that mitochondria and chloroplasts have DNA and ribosomes, just as bacteria
do and they resemble the types found in bacteria. Scientists believe that host cells and
bacteria formed a mutually beneficial endosymbiotic relationship when the host cells ingested
aerobic bacteria and cyanobacteria but did not destroy them. Through evolution, these
ingested bacteria became more specialized in their functions, with the aerobic bacteria
becoming mitochondria and the photosynthetic bacteria becoming chloroplasts.

The Central Vacuole

Previously, we mentioned vacuoles as essential components of plant cells. If you look

at Figure (b), you will see that plant cells each have a large, central vacuole that occupies
most of the cell. The central vacuole plays a key role in regulating the cell’s concentration of
water in changing environmental conditions. In plant cells, the liquid inside the central
vacuole provides turgor pressure, which is the outward pressure caused by the fluid inside the
cell. Have you ever noticed that if you forget to water a plant for a few days, it wilts? That is
because as the water concentration in the soil becomes lower than the water concentration in
the plant, water moves out of the central vacuoles and cytoplasm and into the soil. As the
central vacuole shrinks, it leaves the cell wall unsupported. This loss of support to the cell
walls of a plant results in the wilted appearance. Additionally, this fluid has a very bitter
taste, which discourages consumption by insects and animals. The central vacuole also
functions to store proteins in developing seed cells.
 EU Test Material

Extracellular Matrix of Animal Cells

Most animal cells release materials into the extracellular space. The primary components of
these materials are glycoproteins and the protein collagen. Collectively, these materials are
called the extracellular matrix (Figure 11). Not only does the extracellular matrix hold the
cells together to form a tissue, but it also allows the cells within the tissue to communicate
with each other.

Figure 11 The extracellular matrix consists of a network of substances secreted by cells.

Blood clotting provides an example of the role of the extracellular matrix in cell
communication. When the cells lining a blood vessel are damaged, they display a protein
receptor called tissue factor. When tissue factor binds with another factor in the extracellular
matrix, it causes platelets to adhere to the wall of the damaged blood vessel, stimulates
adjacent smooth muscle cells in the blood vessel to contract (thus constricting the blood
vessel), and initiates a series of steps that stimulate the platelets to produce clotting factors.

Intercellular Junctions

Cells can also communicate with each other by direct contact, referred to as intercellular
junctions. There are some differences in the ways that plant and animal cells do this.
Plasmodesmata (singular = plasmodesma) are junctions between plant cells, whereas animal
cell contacts include tight and gap junctions, and desmosomes.

In general, long stretches of the plasma membranes of neighboring plant cells cannot touch
one another because they are separated by the cell walls surrounding each cell.
Plasmodesmata are numerous channels that pass between the cell walls of adjacent plant
 EU Test Material

cells, connecting their cytoplasm and enabling signal molecules and nutrients to be
transported from cell to cell (Figure 12 a).

Figure 12 There are four kinds of connections between cells. (a) A plasmodesma is a channel
between the cell walls of two adjacent plant cells. (b) Tight junctions join adjacent animal
cells. (c) Desmosomes join two animal cells together. (d) Gap junctions act as channels
between animal cells.

A tight junction is a watertight seal between two adjacent animal cells (Figure 12 b). Proteins
hold the cells tightly against each other. This tight adhesion prevents materials from leaking
between the cells. Tight junctions are typically found in the epithelial tissue that lines internal
organs and cavities, and composes most of the skin. For example, the tight junctions of the
epithelial cells lining the urinary bladder prevent urine from leaking into the extracellular
space.

Also found only in animal cells are desmosomes, which act like spot welds between adjacent
epithelial cells (Figure 12 c). They keep cells together in a sheet-like formation in organs and
tissues that stretch, like the skin, heart, and muscles.

Gap junctions in animal cells are like plasmodesmata in plant cells in that they are channels
between adjacent cells that allow for the transport of ions, nutrients, and other substances that
enable cells to communicate (Figure 12 d). Structurally, however, gap junctions and
plasmodesmata differ.

Table 1 Components of Prokaryotic and Eukaryotic Cells and Their Functions

 EU Test Material

Present
Present in
Cell Present in in
Function Plant
Component Prokaryotes? Animal
Cells?
Cells?

Separates cell from external

environment; controls passage
Plasma
of organic molecules, ions, Yes Yes Yes
membrane
water, oxygen, and wastes into
and out of the cell

Provides structure to cell; site of

many metabolic reactions;
Cytoplasm Yes Yes Yes
medium in which organelles are
found

Nucleoid Location of DNA Yes No No

Cell organelle that houses DNA

Nucleus and directs synthesis of No Yes Yes
ribosomes and proteins

Ribosomes Protein synthesis Yes Yes Yes

ATP production/cellular
Mitochondria No Yes Yes
respiration

Oxidizes and breaks down fatty

Peroxisomes acids and amino acids, and No Yes Yes
detoxifies poisons

Vesicles and Storage and transport; digestive

No Yes Yes
vacuoles function in plant cells

Unspecified role in cell division

in animal cells; organizing
Centrosome No Yes No
center of microtubules in
animal cells

Lysosomes Digestion of macromolecules; No Yes No

recycling of worn-out
 EU Test Material

Table 1 Components of Prokaryotic and Eukaryotic Cells and Their Functions

Present
Present in
Cell Present in in
Function Plant
Component Prokaryotes? Animal
Cells?
Cells?

organelles

Yes, primarily
Yes,
Protection, structural support peptidoglycan in
Cell wall No primarily
and maintenance of cell shape bacteria but not
cellulose
Archaea

Chloroplasts Photosynthesis No No Yes

Endoplasmic Modifies proteins and

No Yes Yes
reticulum synthesizes lipids

Modifies, sorts, tags, packages,

Golgi
and distributes lipids and No Yes Yes
apparatus
proteins

Maintains cell’s shape, secures

organelles in specific positions,
allows cytoplasm and vesicles
Cytoskeleton Yes Yes Yes
to move within the cell, and
enables unicellular organisms to
move independently

No, except
for some
Flagella Cellular locomotion Some Some
plant
sperm.

Cellular locomotion, movement

of particles along extracellular
Cilia No Some No
surface of plasma membrane,
and filtration
 EU Test Material

Section Summary

Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes,
but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its
DNA is surrounded by a membrane), and has other membrane-bound organelles that allow
for compartmentalization of functions. The plasma membrane is a phospholipid bilayer
embedded with proteins. The nucleolus within the nucleus is the site for ribosome assembly.
Ribosomes are found in the cytoplasm or are attached to the cytoplasmic side of the plasma
membrane or endoplasmic reticulum. They perform protein synthesis. Mitochondria perform
cellular respiration and produce ATP. Peroxisomes break down fatty acids, amino acids, and
some toxins. Vesicles and vacuoles are storage and transport compartments. In plant cells,
vacuoles also help break down macromolecules.

Animal cells also have a centrosome and lysosomes. The centrosome has two bodies, the
centrioles, with an unknown role in cell division. Lysosomes are the digestive organelles of
animal cells.

Plant cells have a cell wall, chloroplasts, and a central vacuole. The plant cell wall, whose
primary component is cellulose, protects the cell, provides structural support, and gives shape
to the cell. Photosynthesis takes place in chloroplasts. The central vacuole expands, enlarging
the cell without the need to produce more cytoplasm.

The endomembrane system includes the nuclear envelope, the endoplasmic reticulum, Golgi
apparatus, lysosomes, vesicles, as well as the plasma membrane. These cellular components
work together to modify, package, tag, and transport membrane lipids and proteins.

The cytoskeleton has three different types of protein elements. Microfilaments provide
rigidity and shape to the cell, and facilitate cellular movements. Intermediate filaments bear
tension and anchor the nucleus and other organelles in place. Microtubules help the cell resist
compression, serve as tracks for motor proteins that move vesicles through the cell, and pull
replicated chromosomes to opposite ends of a dividing cell. They are also the structural
elements of centrioles, flagella, and cilia.

Animal cells communicate through their extracellular matrices and are connected to each
other by tight junctions, desmosomes, and gap junctions. Plant cells are connected and
communicate with each other by plasmodesmata.

CELL DIVISION
Cell division happens when a parent cell divides into two or more cells called daughter cells.
Cell division usually occurs as part of a larger cell cycle. All cells reproduce by splitting into
two, where each parental cell gives rise to two daughter cells.
 EU Test Material

These newly formed daughter cells could themselves divide and grow, giving rise to a new
cell population that is formed by the division and growth of a single parental cell and its
descendant.
In other words, such cycles of growth and division allow a single cell to form a structure
consisting of millions of cells.
Explore the cell division notes to learn about the types and phases of cell division.

Types of Cell Division

There are two distinct types of cell division out of which the first one is vegetative division,
wherein each daughter cell duplicates the parent cell called mitosis. The second one is
meiosis, which divides into four haploid daughter cells.

Mitosis: The process cells use to make exact replicas of themselves. Mitosis is observed in
almost all the body’s cells, including eyes, skin, hair, and muscle cells.
 EU Test Material

Meiosis: In this type of cell division, sperm or egg cells are produced instead of identical
daughter cells as in mitosis.
Binary Fission: Single-celled organisms like bacteria replicate themselves for reproduction.

Phases of the Cell Cycle

There are two primary phases in the cell cycle:

1. Interphase: This phase was thought to represent the resting stage between subsequent
cell divisions, but new research has shown that it is a very active phase.
2. M Phase (Mitosis phase): This is where the actual cell division occurs. There are two
key steps in this phase, namely cytokinesis and karyokinesis.

The interphase further comprises three phases:

1. G0 Phase (Resting Phase): The cell neither divides nor prepares itself for the
division.
2. G1 Phase (Gap 1): The cell is metabolically active and grows continuously during
this phase.
3. S phase (Synthesis): The DNA replication or synthesis occurs during this stage.
4. G2 phase (Gap 2): Protein synthesis happens in this phase.
5. Quiescent Stage (G0): The cells that do not undergo further division exits the G1
phase and enters an inactive stage. This stage is known as the quiescent stage (G0) of
the cell cycle.

There are four stages in the M Phase, namely:

1. Prophase
2. Metaphase
3. Anaphase
4. Telophase
 EU Test Material

Scientists use the term bioenergetics to describe the concept of energy flow (Figure 1)
through living systems, such as cells. Cellular processes such as the building and breaking
down of complex molecules occur through stepwise chemical reactions. Some of these
chemical reactions are spontaneous and release energy, whereas others require energy to
proceed. Just as living things must continually consume food to replenish their energy
supplies, cells must continually produce more energy to replenish that used by the many
energy-requiring chemical reactions that constantly take place. Together, all of the chemical
reactions that take place inside cells, including those that consume or generate energy, are
referred to as the cell’s metabolism.

Figure 1 Ultimately, most life forms get their energy from the sun. Plants use photosynthesis
to capture sunlight, and herbivores eat the plants to obtain energy. Carnivores eat the
herbivores, and eventual decomposition of plant and animal material contributes to the
nutrient pool.

Metabolic Pathways

Consider the metabolism of sugar. This is a classic example of one of the many cellular
processes that use and produce energy. Living things consume sugars as a major energy
source, because sugar molecules have a great deal of energy stored within their bonds. For
the most part, photosynthesizing organisms like plants produce these sugars. During
photosynthesis, plants use energy (originally from sunlight) to convert carbon dioxide gas
(CO2) into sugar molecules (like glucose: C6H12O6). They consume carbon dioxide and
produce oxygen as a waste product. This reaction is summarized as:
 EU Test Material

6CO2 + 6H2O + energy ——-> C6H12O6+ 6O2

Because this process involves synthesizing an energy-storing molecule, it requires energy

input to proceed. During the light reactions of photosynthesis, energy is provided by a
molecule called adenosine triphosphate (ATP), which is the primary energy currency of all
cells. Just as the dollar is used as currency to buy goods, cells use molecules of ATP as
energy currency to perform immediate work. In contrast, energy-storage molecules such as
glucose are consumed only to be broken down to use their energy. The reaction that harvests
the energy of a sugar molecule in cells requiring oxygen to survive can be summarized by the
reverse reaction to photosynthesis. In this reaction, oxygen is consumed and carbon dioxide is
released as a waste product. The reaction is summarized as:

C6H12O6 + 6O2 ——> 6CO2 + 6H2O + energy

Both of these reactions involve many steps.

The processes of making and breaking down sugar molecules illustrate two examples of
metabolic pathways. A metabolic pathway is a series of chemical reactions that takes a
starting molecule and modifies it, step-by-step, through a series of metabolic intermediates,
eventually yielding a final product. In the example of sugar metabolism, the first metabolic
pathway synthesized sugar from smaller molecules, and the other pathway broke sugar down
into smaller molecules. These two opposite processes—the first requiring energy and the
second producing energy—are referred to as anabolic pathways (building polymers) and
catabolic pathways (breaking down polymers into their monomers), respectively.
Consequently, metabolism is composed of synthesis (anabolism) and degradation
(catabolism) (Figure 2).

It is important to know that the chemical reactions of metabolic pathways do not take place
on their own. Each reaction step is facilitated, or catalyzed, by a protein called an
enzyme. Enzymes are important for catalyzing all types of biological reactions—those
that require energy as well as those that release energy.

Figure 2 Catabolic pathways are those that generate energy by breaking down larger
molecules. Anabolic pathways are those that require energy to synthesize larger molecules.
Both types of pathways are required for maintaining the cell’s energy balance.
 EU Test Material

Energy

Thermodynamics refers to the study of energy and energy transfer involving physical matter.
The matter relevant to a particular case of energy transfer is called a system, and everything
outside of that matter is called the surroundings. For instance, when heating a pot of water on
the stove, the system includes the stove, the pot, and the water. Energy is transferred within
the system (between the stove, pot, and water). There are two types of systems: open and
closed. In an open system, energy can be exchanged with its surroundings. The stovetop
system is open because heat can be lost to the air. A closed system cannot exchange energy
with its surroundings.

Biological organisms are open systems. Energy is exchanged between them and their
surroundings as they use energy from the sun to perform photosynthesis or consume energy-
storing molecules and release energy to the environment by doing work and releasing heat.
Like all things in the physical world, energy is subject to physical laws. The laws of
thermodynamics govern the transfer of energy in and among all systems in the universe.

In general, energy is defined as the ability to do work, or to create some kind of change.
Energy exists in different forms. For example, electrical energy, light energy, and heat energy
are all different types of energy. To appreciate the way energy flows into and out of
biological systems, it is important to understand two of the physical laws that govern energy.

Thermodynamics

The first law of thermodynamics states that the total amount of energy in the universe is
constant and conserved. In other words, there has always been, and always will be, exactly
the same amount of energy in the universe. Energy exists in many different forms.
According to the first law of thermodynamics, energy may be transferred from place to place
or transformed into different forms, but it cannot be created or destroyed. The transfers
and transformations of energy take place around us all the time. Light bulbs transform
electrical energy into light and heat energy. Gas stoves transform chemical energy from
natural gas into heat energy. Plants perform one of the most biologically useful energy
transformations on earth: that of converting the energy of sunlight to chemical energy stored
within organic molecules (Figure 1). Some examples of energy transformations are shown
in Figure 3.

The challenge for all living organisms is to obtain energy from their surroundings in forms
that they can transfer or transform into usable energy to do work. Living cells have evolved
to meet this challenge. Chemical energy stored within organic molecules such as sugars and
fats is transferred and transformed through a series of cellular chemical reactions into energy
within molecules of ATP. Energy in ATP molecules is easily accessible to do work.
Examples of the types of work that cells need to do include building complex molecules,
transporting materials, powering the motion of cilia or flagella, and contracting muscle fibers
to create movement.
 EU Test Material

Figure 3 Shown are some examples of energy transferred and transformed from one system to
another and from one form to another. The food we consume provides our cells with the
energy required to carry out bodily functions, just as light energy provides plants with the
means to create the chemical energy they need. (credit “ice cream”: modification of work by
D. Sharon Pruitt; credit “kids”: modification of work by Max from Providence; credit “leaf”:
modification of work by Cory Zanker)

A living cell’s primary tasks of obtaining, transforming, and using energy to do work may
seem simple. However, the second law of thermodynamics explains why these tasks are
harder than they appear. All energy transfers and transformations are never completely
efficient. In every energy transfer, some amount of energy is lost in a form that is unusable.
In most cases, this form is heat energy. Thermodynamically, heat energy is defined as the
energy transferred from one system to another that is not work. For example, when a light
bulb is turned on, some of the energy being converted from electrical energy into light energy
is lost as heat energy. Likewise, some energy is lost as heat energy during cellular metabolic
reactions.

An important concept in physical systems is that of order and disorder. The more energy that
is lost by a system to its surroundings, the less ordered and more random the system is.
 EU Test Material

Scientists refer to the measure of randomness or disorder within a system as entropy.

High entropy means high disorder and low energy. Molecules and chemical reactions have
varying entropy as well. For example, entropy increases as molecules at a high concentration
in one place diffuse and spread out. The second law of thermodynamics says that energy will
always be lost as heat in energy transfers or transformations.

Living things are highly ordered, requiring constant energy input to be maintained in a state
of low entropy.

Potential and Kinetic Energy

When an object is in motion, there is energy associated with that object. Think of a wrecking
ball. Even a slow-moving wrecking ball can do a great deal of damage to other objects.
Energy associated with objects in motion is called kinetic energy (Figure 4). A speeding
bullet, a walking person, and the rapid movement of molecules in the air (which produces
heat) all have kinetic energy.

Now what if that same motionless wrecking ball is lifted two stories above ground with a
crane? If the suspended wrecking ball is unmoving, is there energy associated with it? The
answer is yes. The energy that was required to lift the wrecking ball did not disappear, but is
now stored in the wrecking ball by virtue of its position and the force of gravity acting on it.
This type of energy is called potential energy (Figure 4). If the ball were to fall, the potential
energy would be transformed into kinetic energy until all of the potential energy was
exhausted when the ball rested on the ground. Wrecking balls also swing like a pendulum;
through the swing, there is a constant change of potential energy (highest at the top of the
swing) to kinetic energy (highest at the bottom of the swing). Other examples of potential
energy include the energy of water held behind a dam or a person about to skydive out of an
airplane.

Figure 4 Still water has potential energy; moving water, such as in a waterfall or a rapidly
flowing river, has kinetic energy. (credit “dam”: modification of work by “Pascal”/Flickr;
credit “waterfall”: modification of work by Frank Gualtieri)

Potential energy is not only associated with the location of matter, but also with the structure
of matter. Even a spring on the ground has potential energy if it is compressed; so does a
 EU Test Material

rubber band that is pulled taut. On a molecular level, the bonds that hold the atoms of
molecules together exist in a particular structure that has potential energy. Remember that
anabolic cellular pathways require energy to synthesize complex molecules from simpler
ones and catabolic pathways release energy when complex molecules are broken down. The
fact that energy can be released by the breakdown of certain chemical bonds implies that
those bonds have potential energy. In fact, there is potential energy stored within the bonds of
all the food molecules we eat, which is eventually harnessed for use. This is because these
bonds can release energy when broken. The type of potential energy that exists within
chemical bonds, and is released when those bonds are broken, is called chemical energy.
Chemical energy is responsible for providing living cells with energy from food. The release
of energy occurs when the molecular bonds within food molecules are broken.

Free and Activation Energy

After learning that chemical reactions release energy when energy-storing bonds are broken,
an important next question is the following: How is the energy associated with these chemical
reactions quantified and expressed? How can the energy released from one reaction be
compared to that of another reaction? A measurement of free energy is used to quantify these
energy transfers. Recall that according to the second law of thermodynamics, all energy
transfers involve the loss of some amount of energy in an unusable form such as heat. Free
energy specifically refers to the energy associated with a chemical reaction that is available
after the losses are accounted for. In other words, free energy is usable energy, or energy that
is available to do work.

If energy is released during a chemical reaction, then the change in free energy, signified as
∆G (delta G) will be a negative number. A negative change in free energy also means that the
products of the reaction have less free energy than the reactants, because they release some
free energy during the reaction. Reactions that have a negative change in free energy and
consequently release free energy are called exergonic reactions. Think: exergonic means
energy is exiting the system. These reactions are also referred to as spontaneous reactions,
and their products have less stored energy than the reactants. An important distinction must
be drawn between the term spontaneous and the idea of a chemical reaction occurring
immediately. Contrary to the everyday use of the term, a spontaneous reaction is not one that
suddenly or quickly occurs. The rusting of iron is an example of a spontaneous reaction that
occurs slowly, little by little, over time.

If a chemical reaction absorbs energy rather than releases energy on balance, then the ∆G for
that reaction will be a positive value. In this case, the products have more free energy than the
reactants. Thus, the products of these reactions can be thought of as energy-storing
molecules. These chemical reactions are called endergonic reactions and they are non-
spontaneous. An endergonic reaction will not take place on its own without the addition of
free energy.
 EU Test Material

Figure 5 Shown are some examples of endergonic processes (ones that require energy) and
exergonic processes (ones that release energy). (credit a: modification of work by Natalie
Maynor; credit b: modification of work by USDA; credit c: modification of work by Cory
Zanker; credit d: modification of work by Harry Malsch)

Look at each of the processes shown and decide if it is endergonic or exergonic.

There is another important concept that must be considered regarding endergonic and
exergonic reactions. Exergonic reactions require a small amount of energy input to get going,
before they can proceed with their energy-releasing steps. These reactions have a net release
of energy, but still require some energy input in the beginning. This small amount of energy
input necessary for all chemical reactions to occur is called the activation energy.

Enzymes

A substance that helps a chemical reaction to occur is called a catalyst, and the molecules that
catalyze biochemical reactions are called enzymes. Most enzymes are proteins and perform
the critical task of lowering the activation energies of chemical reactions inside the cell.
Most of the reactions critical to a living cell happen too slowly at normal temperatures to be
of any use to the cell. Without enzymes to speed up these reactions, life could not persist.
Enzymes do this by binding to the reactant molecules and holding them in such a way as to
make the chemical bond-breaking and -forming processes take place more easily. It is
 EU Test Material

important to remember that enzymes do not change whether a reaction is exergonic

(spontaneous) or endergonic. This is because they do not change the free energy of the
reactants or products. They only reduce the activation energy required for the reaction to go
forward (Figure 6). In addition, an enzyme itself is unchanged by the reaction it catalyzes.
Once one reaction has been catalyzed, the enzyme is able to participate in other reactions.

Figure 6 Enzymes lower the activation energy of the reaction but do not change the free
energy of the reaction.

The chemical reactants to which an enzyme binds are called the enzyme’s substrates. There
may be one or more substrates, depending on the particular chemical reaction. In some
reactions, a single reactant substrate is broken down into multiple products. In others, two
substrates may come together to create one larger molecule. Two reactants might also enter a
reaction and both become modified, but they leave the reaction as two products. The location
within the enzyme where the substrate binds is called the enzyme’s active site. The active
site is where the “action” happens. Since enzymes are proteins, there is a unique combination
of amino acid side chains within the active site. Each side chain is characterized by different
properties. They can be large or small, weakly acidic or basic, hydrophilic or hydrophobic,
positively or negatively charged, or neutral. The unique combination of side chains creates a
very specific chemical environment within the active site. This specific environment is suited
to bind to one specific chemical substrate (or substrates).

Active sites are subject to influences of the local environment. Increasing the environmental
temperature generally increases reaction rates, enzyme-catalyzed or otherwise. However,
temperatures outside of an optimal range reduce the rate at which an enzyme catalyzes a
reaction. Hot temperatures will eventually cause enzymes to denature, an irreversible change
in the three-dimensional shape and therefore the function of the enzyme. Enzymes are also
suited to function best within a certain pH and salt concentration range, and, as with
temperature, extreme pH, and salt concentrations can cause enzymes to denature.
 EU Test Material

For many years, scientists thought that enzyme-substrate binding took place in a simple “lock
and key” fashion. This model asserted that the enzyme and substrate fit together perfectly in
one instantaneous step. However, current research supports a model called induced fit (Figure
7). The induced-fit model expands on the lock-and-key model by describing a more dynamic
binding between enzyme and substrate. As the enzyme and substrate come together, their
interaction causes a mild shift in the enzyme’s structure that forms an ideal binding
arrangement between enzyme and substrate.

When an enzyme binds its substrate, an enzyme-substrate complex is formed. This complex
lowers the activation energy of the reaction and promotes its rapid progression in one of
multiple possible ways. On a basic level, enzymes promote chemical reactions that involve
more than one substrate by bringing the substrates together in an optimal orientation for
reaction. Another way in which enzymes promote the reaction of their substrates is by
creating an optimal environment within the active site for the reaction to occur. The chemical
properties that emerge from the particular arrangement of amino acid R groups within an
active site create the perfect environment for an enzyme’s specific substrates to react.

The enzyme-substrate complex can also lower activation energy by compromising the bond
structure so that it is easier to break. Finally, enzymes can also lower activation energies by
taking part in the chemical reaction itself. In these cases, it is important to remember that the
enzyme will always return to its original state by the completion of the reaction. One of the
hallmark properties of enzymes is that they remain ultimately unchanged by the reactions
they catalyze. After an enzyme has catalyzed a reaction, it releases its product(s) and can
catalyze a new reaction.

Figure 7 The induced-fit model is an adjustment to the lock-and-key model and explains how
enzymes and substrates undergo dynamic modifications during the transition state to increase
the affinity of the substrate for the active site.

It would seem ideal to have a scenario in which all of an organism’s enzymes existed in
abundant supply and functioned optimally under all cellular conditions, in all cells, at all
times. However, a variety of mechanisms ensures that this does not happen. Cellular needs
and conditions constantly vary from cell to cell, and change within individual cells over time.
The required enzymes of stomach cells differ from those of fat storage cells, skin cells, blood
cells, and nerve cells. Furthermore, a digestive organ cell works much harder to process and
break down nutrients during the time that closely follows a meal compared with many hours
 EU Test Material

after a meal. As these cellular demands and conditions vary, so must the amounts and
functionality of different enzymes.

Since the rates of biochemical reactions are controlled by activation energy, and enzymes
lower and determine activation energies for chemical reactions, the relative amounts and
functioning of the variety of enzymes within a cell ultimately determine which reactions will
proceed and at what rates. This determination is tightly controlled in cells. In certain cellular
environments, enzyme activity is partly controlled by environmental factors like pH,
temperature, salt concentration, and, in some cases, cofactors or coenzymes.

Enzymes can also be regulated in ways that either promote or reduce enzyme activity.
There are many kinds of molecules that inhibit or promote enzyme function, and various
mechanisms by which they do so. In some cases of enzyme inhibition, an inhibitor molecule
is similar enough to a substrate that it can bind to the active site and simply block the
substrate from binding. When this happens, the enzyme is inhibited through competitive
inhibition, because an inhibitor molecule competes with the substrate for binding to the
active site.

On the other hand, in noncompetitive inhibition, an inhibitor molecule binds to the enzyme
in a location other than the active site, called an allosteric site, but still manages to block
substrate binding to the active site. Some inhibitor molecules bind to enzymes in a location
where their binding induces a conformational change that reduces the affinity of the enzyme
for its substrate. This type of inhibition is called allosteric inhibition (Figure 8). Most
allosterically regulated enzymes are made up of more than one polypeptide, meaning that
they have more than one protein subunit. When an allosteric inhibitor binds to a region on an
enzyme, all active sites on the protein subunits are changed slightly such that they bind their
substrates with less efficiency. There are allosteric activators as well as inhibitors. Allosteric
activators bind to locations on an enzyme away from the active site, inducing a
conformational change that increases the affinity of the enzyme’s active site(s) for its
substrate(s) (Figure 8).
 EU Test Material

Figure 8 Allosteric inhibition works by indirectly inducing a conformational change to the

active site such that the substrate no longer fits. In contrast, in allosteric activation, the
activator molecule modifies the shape of the active site to allow a better fit of the substrate.

Through the Indigenous Lens

Plants cannot run or hide from their predators and have evolved many strategies to deter
those who would eat them. Think of thorns, irritants and secondary metabolites: these are
compounds that do not directly help the plant grow, but are made specifically to keep
predators away. Secondary metabolites are the most common way plants deter
predators. Some examples of secondary metabolites are atropine, nicotine, THC and
caffeine. Humans have found these secondary metabolite compounds a rich source of
materials for medicines. It is estimated that 90% of the drugs in the modern pharmacy have
their “roots” in these secondary metabolites.

First peoples herbal treatments revealed these secondary metabolites to the world. For
example, Indigenous peoples have long used the bark of willow shrubs and alder trees for a
tea, tonic or poultice to reduce inflammation.
 EU Test Material

Figure 9 Pacific willow bark contains the compound salicin.

Both willow and alder bark contain the compound salicin. Most of us have this compound in
our medicine cupboard in the form of salicylic acid or aspirin. Aspirin has been proved to
reduce pain and inflammation, and once in our cells salicin converts to salicylic acid.

So how does it work? Salicin or aspirin acts as an enzyme inhibitor. In the inflammatory
response two enzymes, COX1 and COX2 are key to this process. Salicin or aspirin
specifically modifies an amino acid (serine) in the active site of these two related enzymes.
This modification of the active sites does not allow the normal substrate to bind and so the
inflammatory process is disrupted. As you have read in this chapter, this makes it competitive
enzyme inhibitor.

Pharmaceutical Drug Developer

Figure 10 Have you ever wondered how pharmaceutical drugs are developed? (credit:
Deborah Austin)
 EU Test Material

Enzymes are key components of metabolic pathways. Understanding how enzymes work and
how they can be regulated are key principles behind the development of many of the
pharmaceutical drugs on the market today. Biologists working in this field collaborate with
other scientists to design drugs (Figure 10).

Consider statins for example—statins is the name given to one class of drugs that can reduce
cholesterol levels. These compounds are inhibitors of the enzyme HMG-CoA reductase,
which is the enzyme that synthesizes cholesterol from lipids in the body. By inhibiting this
enzyme, the level of cholesterol synthesized in the body can be reduced. Similarly,
acetaminophen, popularly marketed under the brand name Tylenol, is an inhibitor of the
enzyme cyclooxygenase. While it is used to provide relief from fever and inflammation
(pain), its mechanism of action is still not completely understood.

How are drugs discovered? One of the biggest challenges in drug discovery is identifying a
drug target. A drug target is a molecule that is literally the target of the drug. In the case of
statins, HMG-CoA reductase is the drug target. Drug targets are identified through
painstaking research in the laboratory. Identifying the target alone is not enough; scientists
also need to know how the target acts inside the cell and which reactions go awry in the case
of disease. Once the target and the pathway are identified, then the actual process of drug
design begins. In this stage, chemists and biologists work together to design and synthesize
molecules that can block or activate a particular reaction. However, this is only the
beginning: If and when a drug prototype is successful in performing its function, then it is
subjected to many tests from in vitro experiments to clinical trials before it can get approval
from the U.S. Food and Drug Administration to be on the market.

Many enzymes do not work optimally, or even at all, unless bound to other specific non-
protein helper molecules. They may bond either temporarily through ionic or hydrogen
bonds, or permanently through stronger covalent bonds. Binding to these molecules promotes
optimal shape and function of their respective enzymes. Two examples of these types of
helper molecules are cofactors and coenzymes. Cofactors are inorganic ions such as ions of
iron and magnesium. Coenzymes are organic helper molecules, those with a basic atomic
structure made up of carbon and hydrogen. Like enzymes, these molecules participate in
reactions without being changed themselves and are ultimately recycled and reused. Vitamins
are the source of coenzymes. Some vitamins are the precursors of coenzymes and others act
directly as coenzymes. Vitamin C is a direct coenzyme for multiple enzymes that take part in
building the important connective tissue, collagen. Therefore, enzyme function is, in part,
regulated by the abundance of various cofactors and coenzymes, which may be supplied by
an organism’s diet or, in some cases, produced by the organism.

Figure 11 Vitamins are important coenzymes or precursors of coenzymes, and are required
for enzymes to function properly. Multivitamin capsules usually contain mixtures of all the
vitamins at different percentages.
 EU Test Material

Feedback Inhibition in Metabolic Pathways

Molecules can regulate enzyme function in many ways. The major question remains,
however: What are these molecules and where do they come from? Some are cofactors and
coenzymes, as you have learned. What other molecules in the cell provide enzymatic
regulation such as allosteric modulation, and competitive and non-competitive inhibition?
Perhaps the most relevant sources of regulatory molecules, with respect to enzymatic cellular
metabolism, are the products of the cellular metabolic reactions themselves. In a most
efficient and elegant way, cells have evolved to use the products of their own reactions for
feedback inhibition of enzyme activity. Feedback inhibition involves the use of a reaction
product to regulate its own further production (Figure 11). The cell responds to an abundance
of the products by slowing down production during anabolic or catabolic reactions. Such
reaction products may inhibit the enzymes that catalyzed their production through the
mechanisms described above.

Figure 12 Metabolic pathways are a series of reactions catalyzed by multiple enzymes.

Feedback inhibition, where the end product of the pathway inhibits an upstream process, is an
important regulatory mechanism in cells.

The production of both amino acids and nucleotides is controlled through feedback
inhibition. Additionally, ATP is an allosteric regulator of some of the enzymes involved in
the catabolic breakdown of sugar, the process that creates ATP. In this way, when ATP is in
abundant supply, the cell can prevent the production of ATP. On the other hand, ADP serves
as a positive allosteric regulator (an allosteric activator) for some of the same enzymes that
are inhibited by ATP. Thus, when relative levels of ADP are high compared to ATP, the cell
is triggered to produce more ATP through sugar catabolism.

Section Summary

Cells perform the functions of life through various chemical reactions. A cell’s metabolism
refers to the combination of chemical reactions that take place within it. Catabolic reactions
break down complex chemicals into simpler ones and are associated with energy release.
Anabolic processes build complex molecules out of simpler ones and require energy.

In studying energy, the term system refers to the matter and environment involved in energy
transfers. Entropy is a measure of the disorder of a system. The physical laws that describe
the transfer of energy are the laws of thermodynamics. The first law states that the total
amount of energy in the universe is constant. The second law of thermodynamics states that
every energy transfer involves some loss of energy in an unusable form, such as heat energy.
 EU Test Material

Energy comes in different forms: kinetic, potential, and free. The change in free energy of a
reaction can be negative (releases energy, exergonic) or positive (consumes energy,
endergonic). All reactions require an initial input of energy to proceed, called the activation
energy.

Enzymes are chemical catalysts that speed up chemical reactions by lowering their activation
energy. Enzymes have an active site with a unique chemical environment that fits particular
chemical reactants for that enzyme, called substrates. Enzymes and substrates are thought to
bind according to an induced-fit model. Enzyme action is regulated to conserve resources and
respond optimally to the environment.

Glycolysis

Energy production within a cell involves many coordinated chemical pathways. Most of these
pathways are combinations of oxidation and reduction reactions. Oxidation and reduction
occur in tandem. An oxidation reaction strips an electron from an atom in a compound, and
the addition of this electron to another compound is a reduction reaction. Because oxidation
and reduction usually occur together, these pairs of reactions are called oxidation-reduction
reactions, or redox reactions.

Electrons and Energy

The removal of an electron from a molecule, oxidizing it, results in a decrease in potential
energy in the oxidized compound. The electron (sometimes as part of a hydrogen atom) does
not remain unbonded, however, in the cytoplasm of a cell. Rather, the electron is shifted to a
second compound, reducing the second compound. The shift of an electron from one
compound to another removes some potential energy from the first compound
(the oxidized compound) and increases the potential energy of the second compound
(the reduced compound). The transfer of electrons between molecules is important because
most of the energy stored in atoms and used to fuel cell functions is in the form of high-
energy electrons. The transfer of energy in the form of electrons allows the cell to transfer
and use energy in an incremental fashion—in small packages rather than in a single,
destructive burst. This chapter focuses on the extraction of energy from food. You will see
that as you track the path of the transfers, you are tracking the path of electrons moving
through metabolic pathways.

Electron Carriers

In living systems, a small class of compounds functions as electron shuttles: they bind and
carry high-energy electrons between compounds in pathways. The principal electron carriers
we will consider are derived from the B vitamin group and are derivatives of nucleotides.
These compounds can be easily reduced (that is, they accept electrons) or oxidized (they lose
electrons). Nicotinamide adenine dinucleotide (NAD) (Figure 1) is derived from vitamin B3,
niacin. NAD+ is the oxidized form of the molecule; NADH is the reduced form of the
molecule after it has accepted two electrons and a proton (which together are the equivalent
of a hydrogen atom with an extra electron).
 EU Test Material

NAD+ can accept electrons from an organic molecule according to the general equation:

RH (Reducing Agent) + NAD + (Oxidizing Agent) —-> NADH (Reduced) + R (Oxidized)

When electrons are added to a compound, they are reduced. A compound that reduces
another is called a reducing agent. In the above equation, RH is a reducing agent, and
NAD+ is reduced to NADH. When electrons are removed from compound, it is oxidized.
A compound that oxidizes another is called an oxidizing agent. In the above equation,
NAD+ is an oxidizing agent, and RH is oxidized to R.

Similarly, flavin adenine dinucleotide (FAD+) is derived from vitamin B2, also called
riboflavin. Its reduced form is FADH2. A second variation of NAD, NADP, contains an extra
phosphate group. Both NAD+ and FAD+ are extensively used in energy extraction from
sugars, and NADP plays an important role in anabolic reactions and photosynthesis.

Figure 1 The oxidized form of the electron carrier (NAD+) is shown on the left and the
reduced form (NADH) is shown on the right. The nitrogenous base in NADH has one more
hydrogen ion and two more electrons than in NAD+.

ATP in Living Systems

A living cell cannot store significant amounts of free energy. Excess free energy would result
in an increase of heat in the cell, which would result in excessive thermal motion that could
damage and then destroy the cell. Rather, a cell must be able to handle that energy in a way
that enables the cell to store the energy safely and release it for use only as needed. Living
cells accomplish this by using the compound adenosine triphosphate (ATP). ATP is often
called the “energy currency” of the cell, and, like currency, this versatile compound can be
used to fill any energy need of the cell. How? It functions similarly to a rechargeable battery.

When ATP is broken down, usually by the removal of its terminal phosphate group, energy is
released. The cell uses the energy to do work, usually by the released phosphate binding to
another molecule, activating it. For example, in the mechanical work of muscle contraction,
ATP supplies the energy to move the contractile muscle proteins. Recall the active transport
work of the sodium-potassium pump in cell membranes. ATP alters the structure of the
integral protein that functions as the pump, changing its affinity for sodium and potassium. In
this way, the cell performs work, pumping ions against their electrochemical gradients.

ATP Structure and Function

 EU Test Material

At the heart of ATP is a molecule of adenosine monophosphate (AMP), which is composed

of an adenine molecule bonded to a ribose molecule and a single phosphate group (Figure 2).
Ribose is a five-carbon sugar found in RNA, and AMP is one of the nucleotides in RNA. The
addition of a second phosphate group to this core molecule results in the formation of
adenosine diphosphate (ADP); the addition of a third phosphate group forms
adenosine triphosphate (ATP).

Figure 2 ATP (adenosine triphosphate) has three phosphate groups that can be removed by
hydrolysis to form ADP (adenosine diphosphate) or AMP (adenosine monophosphate).The
negative charges on the phosphate group naturally repel each other, requiring energy to bond
them together and releasing energy when these bonds are broken.

The addition of a phosphate group to a molecule requires energy. Phosphate groups are
negatively charged and thus repel one another when they are arranged in series, as they are in
ADP and ATP. This repulsion makes the ADP and ATP molecules inherently unstable. The
release of one or two phosphate groups from ATP, a process called dephosphorylation,
releases energy.

Even exergonic, energy-releasing reactions require a small amount of activation energy to

proceed. However, consider endergonic reactions, which require much more energy input
because their products have more free energy than their reactants. Within the cell, where does
energy to power such reactions come from? The answer lies with an energy-supplying
molecule called adenosine triphosphate, or ATP. ATP is a small, relatively simple molecule,
but within its bonds contains the potential for a quick burst of energy that can be
harnessed to perform cellular work. This molecule can be thought of as the primary energy
currency of cells in the same way that money is the currency that people exchange for things
they need. ATP is used to power the majority of energy-requiring cellular reactions.

ATP in Living Systems

A living cell cannot store significant amounts of free energy. Excess free energy would result
in an increase of heat in the cell, which would denature enzymes and other proteins, and thus
destroy the cell. Rather, a cell must be able to store energy safely and release it for use only
as needed. Living cells accomplish this using ATP, which can be used to fill any energy need
of the cell. How? It functions as a rechargeable battery.

When ATP is broken down, usually by the removal of its terminal phosphate group, energy is
released. This energy is used to do work by the cell, usually by the binding of the released
 EU Test Material

phosphate to another molecule, thus activating it. For example, in the mechanical work of
muscle contraction, ATP supplies energy to move the contractile muscle proteins.

ATP Structure and Function

At the heart of ATP is a molecule of adenosine monophosphate (AMP), which is composed

of an adenine molecule bonded to both a ribose molecule and a single phosphate group
(Figure 3). Ribose is a five-carbon sugar found in RNA and AMP is one of the nucleotides in
RNA. The addition of a second phosphate group to this core molecule results in
adenosine diphosphate (ADP); the addition of a third phosphate group forms
adenosine triphosphate (ATP).

Figure 3 The structure of ATP shows the basic components of a two-ring adenine, five-
carbon ribose, and three phosphate groups.

The addition of a phosphate group to a molecule requires a high amount of energy and results
in a high-energy bond. Phosphate groups are negatively charged and thus repel one another
when they are arranged in series, as they are in ADP and ATP. This repulsion makes the ADP
and ATP molecules inherently unstable. The release of one or two phosphate groups from
ATP, a process called hydrolysis, releases energy.

Glycolysis

You have read that nearly all of the energy used by living things comes to them in the
bonds of the sugar, glucose. Glycolysis is the first step in the breakdown of glucose to
extract energy for cell metabolism. Many living organisms carry out glycolysis as part of
their metabolism. Glycolysis takes place in the cytoplasm of most prokaryotic and all
eukaryotic cells.

Glycolysis begins with the six-carbon, ring-shaped structure of a single glucose molecule and
ends with two molecules of a three-carbon sugar called pyruvate. Glycolysis consists of two
distinct phases. In the first part of the glycolysis pathway, energy is used to make adjustments
so that the six-carbon sugar molecule can be split evenly into two three-carbon pyruvate
molecules. In the second part of glycolysis, ATP and nicotinamide-adenine dinucleotide
(NADH) are produced (Figure 4).
 EU Test Material

If the cell cannot catabolize the pyruvate molecules further, it will harvest only two ATP
molecules from one molecule of glucose. For example, mature mammalian red blood cells
are only capable of glycolysis, which is their sole source of ATP. If glycolysis is interrupted,
these cells would eventually die.

Figure 4 In glycolysis, a glucose molecule is converted into two pyruvate molecules.

Section Summary

ATP functions as the energy currency for cells. It allows cells to store energy briefly and
transport it within itself to support endergonic chemical reactions. The structure of ATP is
that of an RNA nucleotide with three phosphate groups attached. As ATP is used for energy,
a phosphate group is detached, and ADP is produced. Energy derived from glucose
catabolism is used to recharge ADP into ATP.

Glycolysis is the first pathway used in the breakdown of glucose to extract energy. Because it
is used by nearly all organisms on earth, it must have evolved early in the history of life.
Glycolysis consists of two parts: The first part prepares the six-carbon ring of glucose for
separation into two three-carbon sugars. Energy from ATP is invested into the molecule
during this step to energize the separation. The second half of glycolysis extracts ATP and
high-energy electrons from hydrogen atoms and attaches them to NAD+. Two ATP molecules
are invested in the first half and four ATP molecules are formed during the second half. This
produces a net gain of two ATP molecules per molecule of glucose for the cell.
 EU Test Material

The Citric Acid Cycle

In eukaryotic cells, the pyruvate molecules produced at the end of glycolysis are transported
into mitochondria, which are sites of cellular respiration. If oxygen is available, aerobic
respiration will go forward. In mitochondria, pyruvate will be transformed into a two-carbon
acetyl group (by removing a molecule of carbon dioxide) that will be picked up by a carrier
compound called coenzyme A (CoA), which is made from vitamin B5. The resulting
compound is called acetyl CoA. (Figure 1). Acetyl CoA can be used in a variety of ways by
the cell, but its major function is to deliver the acetyl group derived from pyruvate to the next
pathway in glucose catabolism.

Figure 1 Pyruvate is converted into acetyl-CoA before entering the citric acid cycle.

Like the conversion of pyruvate to acetyl CoA, the citric acid cycle in eukaryotic cells takes
place in the matrix of the mitochondria. Unlike glycolysis, the citric acid cycle is a closed
loop: The last part of the pathway regenerates the compound used in the first step. The eight
steps of the cycle are a series of chemical reactions that produces two carbon dioxide
molecules, one ATP molecule (or an equivalent), and reduced forms (NADH and FADH2) of
NAD+ and FAD+, important coenzymes in the cell. Part of this is considered an aerobic
pathway (oxygen-requiring) because the NADH and FADH2 produced must transfer their
electrons to the next pathway in the system, which will use oxygen. If oxygen is not present,
this transfer does not occur.

Two carbon atoms come into the citric acid cycle from each acetyl group. Two carbon
dioxide molecules are released on each turn of the cycle; however, these do not contain the
same carbon atoms contributed by the acetyl group on that turn of the pathway. The two
acetyl-carbon atoms will eventually be released on later turns of the cycle; in this way, all six
carbon atoms from the original glucose molecule will be eventually released as carbon
dioxide. It takes two turns of the cycle to process the equivalent of one glucose molecule.
Each turn of the cycle forms three high-energy NADH molecules and one high-energy
FADH2 molecule. These high-energy carriers will connect with the last portion of aerobic
respiration to produce ATP molecules. One ATP (or an equivalent) is also made in each
cycle. Several of the intermediate compounds in the citric acid cycle can be used in
synthesizing non-essential amino acids; therefore, the cycle is both anabolic and catabolic.
 EU Test Material

Figure 2 In eukaryotes, oxidative phosphorylation takes place in mitochondria. In

prokaryotes, this process takes place in the plasma membrane. (Credit: modification of work
by Mariana Ruiz Villareal)

Oxidative Phosphorylation

You have just read about two pathways in glucose catabolism—glycolysis and the citric acid
cycle—that generate ATP. Most of the ATP generated during the aerobic catabolism of
glucose, however, is not generated directly from these pathways. Rather, it derives from a
process that begins with passing electrons through a series of chemical reactions to a final
electron acceptor, oxygen. These reactions take place in specialized protein complexes
located in the inner membrane of the mitochondria of eukaryotic organisms and on the
inner part of the cell membrane of prokaryotic organisms. The energy of the electrons is
harvested and used to generate a electrochemical gradient across the inner mitochondrial
membrane. The potential energy of this gradient is used to generate ATP. The entirety of
this process is called oxidative phosphorylation.

The electron transport chain (Figure 3 a) is the last component of aerobic respiration and is
the only part of metabolism that uses atmospheric oxygen. Oxygen continuously diffuses into
plants for this purpose. In animals, oxygen enters the body through the respiratory system.
Electron transport is a series of chemical reactions that resembles a bucket brigade in that
electrons are passed rapidly from one component to the next, to the endpoint of the chain
where oxygen is the final electron acceptor and water is produced. There are four complexes
composed of proteins, labeled I through IV in Figure 3 c, and the aggregation of these four
complexes, together with associated mobile, accessory electron carriers, is called the electron
transport chain. The electron transport chain is present in multiple copies in the inner
mitochondrial membrane of eukaryotes and in the plasma membrane of prokaryotes. In each
transfer of an electron through the electron transport chain, the electron loses energy, but with
some transfers, the energy is stored as potential energy by using it to pump hydrogen ions
across the inner mitochondrial membrane into the intermembrane space, creating an
electrochemical gradient.
 EU Test Material

Figure 3 (a) The electron transport chain is a set of molecules that supports a series of
oxidation-reduction reactions. (b) ATP synthase is a complex, molecular machine that uses an
H+ gradient to regenerate ATP from ADP. (c) Chemiosmosis relies on the potential energy
provided by the H+ gradient across the membrane.

Cyanide inhibits cytochrome c oxidase, a component of the electron transport chain. If

cyanide poisoning occurs, would you expect the pH of the intermembrane space to increase
or decrease? What affect would cyanide have on ATP synthesis?

Electrons from NADH and FADH2 are passed to protein complexes in the electron transport
chain. As they are passed from one complex to another (there are a total of four), the
electrons lose energy, and some of that energy is used to pump hydrogen ions from the
 EU Test Material

mitochondrial matrix into the intermembrane space. In the fourth protein complex, the
electrons are accepted by oxygen, the terminal acceptor. The oxygen with its extra electrons
then combines with two hydrogen ions, further enhancing the electrochemical gradient, to
form water. If there were no oxygen present in the mitochondrion, the electrons could not be
removed from the system, and the entire electron transport chain would back up and stop.
The mitochondria would be unable to generate new ATP in this way, and the cell would
ultimately die from lack of energy. This is the reason we must breathe to draw in new
oxygen.

In the electron transport chain, the free energy from the series of reactions just described is
used to pump hydrogen ions across the membrane. The uneven distribution of H+ ions across
the membrane establishes an electrochemical gradient, owing to the H+ ions’ positive charge
and their higher concentration on one side of the membrane.

Hydrogen ions diffuse through the inner membrane through an integral membrane protein
called ATP synthase (Figure 3b). This complex protein acts as a tiny generator, turned by the
force of the hydrogen ions diffusing through it, down their electrochemical gradient from the
intermembrane space, where there are many mutually repelling hydrogen ions to the matrix,
where there are few. The turning of the parts of this molecular machine regenerate ATP from
ADP. This flow of hydrogen ions across the membrane through ATP synthase is called
chemiosmosis.

Chemiosmosis (Figure 3c) is used to generate 90 percent of the ATP made during aerobic
glucose catabolism. The result of the reactions is the production of ATP from the energy of
the electrons removed from hydrogen atoms. These atoms were originally part of a glucose
molecule. At the end of the electron transport system, the electrons are used to reduce an
oxygen molecule to oxygen ions. The extra electrons on the oxygen ions attract hydrogen
ions (protons) from the surrounding medium, and water is formed. The electron transport
chain and the production of ATP through chemiosmosis are collectively called oxidative
phosphorylation.

ATP Yield

The number of ATP molecules generated from the catabolism of glucose varies. For example,
the number of hydrogen ions that the electron transport chain complexes can pump through
the membrane varies between species. Another source of variance stems from the shuttle of
electrons across the mitochondrial membrane. The NADH generated from glycolysis cannot
easily enter mitochondria. Thus, electrons are picked up on the inside of the mitochondria by
either NAD+ or FAD+. Fewer ATP molecules are generated when FAD+ acts as a carrier.
NAD+ is used as the electron transporter in the liver and FAD+ in the brain, so ATP yield
depends on the tissue being considered.

Another factor that affects the yield of ATP molecules generated from glucose is that
intermediate compounds in these pathways are used for other purposes. Glucose catabolism
connects with the pathways that build or break down all other biochemical compounds in
cells, and the result is somewhat messier than the ideal situations described thus far. For
example, sugars other than glucose are fed into the glycolytic pathway for energy extraction.
Other molecules that would otherwise be used to harvest energy in glycolysis or the citric
acid cycle may be removed to form nucleic acids, amino acids, lipids, or other compounds.
 EU Test Material

Overall, in living systems, these pathways of glucose catabolism extract about 34 percent of
the energy contained in glucose.

Mitochondrial Disease Physician

What happens when the critical reactions of cellular respiration do not proceed correctly?
Mitochondrial diseases are genetic disorders of metabolism. Mitochondrial disorders can
arise from mutations in nuclear or mitochondrial DNA, and they result in the production of
less energy than is normal in body cells. Symptoms of mitochondrial diseases can include
muscle weakness, lack of coordination, stroke-like episodes, and loss of vision and hearing.
Most affected people are diagnosed in childhood, although there are some adult-onset
diseases. Identifying and treating mitochondrial disorders is a specialized medical field. The
educational preparation for this profession requires a college education, followed by medical
school with a specialization in medical genetics. Medical geneticists can be board certified by
the American Board of Medical Genetics and go on to become associated with professional
organizations devoted to the study of mitochondrial disease, such as the Mitochondrial
Medicine Society and the Society for Inherited Metabolic Disease.

Section Summary

The citric acid cycle is a series of chemical reactions that removes high-energy electrons and
uses them in the electron transport chain to generate ATP. One molecule of ATP (or an
equivalent) is produced per each turn of the cycle.

The electron transport chain is the portion of aerobic respiration that uses free oxygen as the
final electron acceptor for electrons removed from the intermediate compounds in glucose
catabolism. The electrons are passed through a series of chemical reactions, with a small
amount of free energy used at three points to transport hydrogen ions across the membrane.
This contributes to the gradient used in chemiosmosis. As the electrons are passed from
NADH or FADH2 down the electron transport chain, they lose energy. The products of the
electron transport chain are water and ATP. A number of intermediate compounds can be
diverted into the anabolism of other biochemical molecules, such as nucleic acids, non-
essential amino acids, sugars, and lipids. These same molecules, except nucleic acids, can
serve as energy sources for the glucose pathway.

Fermentation

In aerobic respiration, the final electron acceptor is an oxygen molecule, O2. If aerobic
respiration occurs, then ATP will be produced using the energy of the high-energy electrons
carried by NADH or FADH2 to the electron transport chain. If aerobic respiration does not
occur, NADH must be reoxidized to NAD+ for reuse as an electron carrier for glycolysis to
continue. How is this done? Some living systems use an organic molecule as the final
electron acceptor. Processes that use an organic molecule to regenerate NAD+ from NADH
are collectively referred to as fermentation. In contrast, some living systems use an inorganic
molecule as a final electron acceptor; both methods are a type of anaerobic cellular
 EU Test Material

respiration. Anaerobic respiration enables organisms to convert energy for their use in the
absence of oxygen.

Lactic Acid Fermentation

The fermentation method used by animals and some bacteria like those in yogurt is lactic acid
fermentation (Figure 1). This occurs routinely in mammalian red blood cells and in skeletal
muscle that has insufficient oxygen supply to allow aerobic respiration to continue (that is, in
muscles used to the point of fatigue). In muscles, lactic acid produced by fermentation must
be removed by the blood circulation and brought to the liver for further metabolism. The
chemical reaction of lactic acid fermentation is the following:
Pyruvic acid +NADH↔lactic acid+NAD+Pyruvic acid +NADH↔lactic acid+NAD+

The enzyme that catalyzes this reaction is lactate dehydrogenase. The reaction can proceed in
either direction, but the left-to-right reaction is inhibited by acidic conditions. This lactic acid
build-up causes muscle stiffness and fatigue. Once the lactic acid has been removed from the
muscle and is circulated to the liver, it can be converted back to pyruvic acid and further
catabolized for energy.

Figure 1

Tremetol, a metabolic poison found in white snake root plant, prevents the metabolism of
lactate. When cows eat this plant, Tremetol is concentrated in the milk. Humans who
consume the milk become ill. Symptoms of this disease, which include vomiting, abdominal
pain, and tremors, become worse after exercise.

Why do you think this is the case?

 EU Test Material

<!– The illness is caused by lactic acid build-up. Lactic acid levels rise after exercise, making
the symptoms worse. Milk sickness is rare today, but was common in the Midwestern United
States in the early 1800s. –>

Alcohol Fermentation

Another familiar fermentation process is alcohol fermentation (Figure 2), which produces
ethanol, an alcohol. The alcohol fermentation reaction is the following:

Figure 2 The reaction resulting in alcohol fermentation is shown.

In the first reaction, a carboxyl group is removed from pyruvic acid, releasing carbon dioxide
as a gas. The loss of carbon dioxide reduces the molecule by one carbon atom, making
acetaldehyde. The second reaction removes an electron from NADH, forming NAD+ and
producing ethanol from the acetaldehyde, which accepts the electron. The fermentation of
pyruvic acid by yeast produces the ethanol found in alcoholic beverages (Figure 3). If the
carbon dioxide produced by the reaction is not vented from the fermentation chamber, for
example in beer and sparkling wines, it remains dissolved in the medium until the pressure is
released. Ethanol above 12 percent is toxic to yeast, so natural levels of alcohol in wine occur
at a maximum of 12 percent.

Figure 3 Fermentation of grape juice to make wine produces CO2 as a byproduct.

Fermentation tanks have valves so that pressure inside the tanks can be released.
 EU Test Material

Anaerobic Cellular Respiration

Certain prokaryotes, including some species of bacteria and Archaea, use anaerobic
respiration. For example, the group of Archaea called methanogens reduces carbon dioxide to
methane to oxidize NADH. These microorganisms are found in soil and in the digestive tracts
of ruminants, such as cows and sheep. Similarly, sulfate-reducing bacteria and Archaea, most
of which are anaerobic (Figure 4), reduce sulfate to hydrogen sulfide to regenerate
NAD+ from NADH.

Figure 4 The green color seen in these coastal waters is from an eruption of hydrogen sulfide.
Anaerobic, sulfate-reducing bacteria release hydrogen sulfide gas as they decompose algae in
the water. (credit: NASA image courtesy Jeff Schmaltz, MODIS Land Rapid Response Team
at NASA GSFC)

Other fermentation methods occur in bacteria. Many prokaryotes are facultatively anaerobic.
This means that they can switch between aerobic respiration and fermentation, depending on
the availability of oxygen. Certain prokaryotes, like Clostridia bacteria, are obligate
anaerobes. Obligate anaerobes live and grow in the absence of molecular oxygen. Oxygen is
a poison to these microorganisms and kills them upon exposure. It should be noted that all
forms of fermentation, except lactic acid fermentation, produce gas. The production of
particular types of gas is used as an indicator of the fermentation of specific carbohydrates,
which plays a role in the laboratory identification of the bacteria. The various methods of
fermentation are used by different organisms to ensure an adequate supply of NAD+ for the
sixth step in glycolysis. Without these pathways, that step would not occur, and no ATP
would be harvested from the breakdown of glucose.
 EU Test Material

Section Summary

If NADH cannot be metabolized through aerobic respiration, another electron acceptor is

used. Most organisms will use some form of fermentation to accomplish the regeneration of
NAD+, ensuring the continuation of glycolysis. The regeneration of NAD+ in fermentation is
not accompanied by ATP production; therefore, the potential for NADH to produce ATP
using an electron transport chain is not utilized

Nucleic acids are the organic materials present in all organisms in the form of DNA or RNA.
These nucleic acids are formed by the combination of nitrogenous bases, sugar molecules and
phosphate groups that are linked by different bonds in a series of sequences. The DNA
structure defines the basic genetic makeup of our body. In fact, it defines the genetic makeup
of nearly all life on earth.
Table of Contents

 What is DNA?
 Discovery
 Diagram
 DNA Structure
 Chargaff’s Rule
 DNA Replication
 Function of DNA
 Why DNA is called a Polynucleotide Molecule?
Read on to explore DNA meaning, structure, function, DNA discovery and diagram in
complete detail.

What is DNA?
“DNA is a group of molecules that is responsible for carrying and transmitting the
hereditary materials or the genetic instructions from parents to offsprings.”
This is also true for viruses, as most of these entities have either RNA or DNA as their
genetic material. For instance, some viruses may have RNA as their genetic material, while
others have DNA as the genetic material. The Human Immunodeficiency Virus (HIV)
contains RNA, which is then converted into DNA after attaching itself to the host cell.
Apart from being responsible for the inheritance of genetic information in all living beings,
DNA also plays a crucial role in the production of proteins. Nuclear DNA is the DNA
contained within the nucleus of every cell in a eukaryotic organism. It codes for the majority
of the organism’s genomes while the mitochondrial DNA and plastid DNA handles the rest.
The DNA present in the mitochondria of the cell is termed mitochondrial DNA. It is inherited
from the mother to the child. In humans, there are approximately 16,000 base pairs of
mitochondrial DNA. Similarly, plastids have their own DNA, and they play an essential role
in photosynthesis.
Also Read: Difference between gene and DNA

Full-Form of DNA
 EU Test Material

DNA is known as Deoxyribonucleic Acid. It is an organic compound that has a unique

molecular structure. It is found in all prokaryotic cells and eukaryotic cells.

DNA Types
There are three different DNA types:

 A-DNA: It is a right-handed double helix similar to the B-DNA form. Dehydrated

DNA takes an A form that protects the DNA during extreme conditions such as
desiccation. Protein binding also removes the solvent from DNA, and the DNA takes
an A form.
 B-DNA: This is the most common DNA conformation and is a right-handed helix.
The majority of DNA has a B type conformation under normal physiological
conditions.
 Z-DNA: Z-DNA is a left-handed DNA where the double helix winds to the left in a
zig-zag pattern. It was discovered by Andres Wang and Alexander Rich. It is found
ahead of the start site of a gene and hence, is believed to play some role in gene
regulation.

Who Discovered DNA?

DNA was first recognized and identified by the Swiss biologist Johannes Friedrich
Miescher in 1869 during his research on white blood cells.
The double helix structure of a DNA molecule was later discovered through the experimental
data by James Watson and Francis Crick. Finally, it was proved that DNA is responsible for
storing genetic information in living organisms.
Also Read: Difference between deoxyribose and ribose

DNA Diagram
The following diagram explains the DNA structure representing the different parts of the
DNA. DNA comprises a sugar-phosphate backbone and the nucleotide bases (guanine,
cytosine, adenine and thymine).
 EU Test Material

DNA Diagram representing the DNA Structure

Read more: Properties of DNA

DNA Structure
The DNA structure can be thought of as a twisted ladder. This structure is described as a
double-helix, as illustrated in the figure above. It is a nucleic acid, and all nucleic acids are
made up of nucleotides. The DNA molecule is composed of units called nucleotides, and
each nucleotide is composed of three different components such as sugar, phosphate groups
and nitrogen bases.
The basic building blocks of DNA are nucleotides, which are composed of a sugar group, a
phosphate group, and a nitrogen base. The sugar and phosphate groups link the nucleotides
together to form each strand of DNA. Adenine (A), Thymine (T), Guanine (G) and Cytosine
(C) are four types of nitrogen bases.
These 4 Nitrogenous bases pair together in the following
way: A with T, and C with G. These base pairs are essential for the DNA’s double helix
structure, which resembles a twisted ladder.
The order of the nitrogenous bases determines the genetic code or the DNA’s instructions.

Components of DNA Structure

Among the three components of DNA structure, sugar is the one which forms the backbone
of the DNA molecule. It is also called deoxyribose. The nitrogenous bases of the opposite
strands form hydrogen bonds, forming a ladder-like structure.
 EU Test Material

DNA Structure Backbone

The DNA molecule consists of 4 nitrogen bases, namely adenine (A), thymine (T), cytosine
(C) and Guanine (G), which ultimately form the structure of a nucleotide. The A and G are
purines, and the C and T are pyrimidines.
The two strands of DNA run in opposite directions. These strands are held together by the
hydrogen bond that is present between the two complementary bases. The strands are
helically twisted, where each strand forms a right-handed coil, and ten nucleotides make up a
single turn.
The pitch of each helix is 3.4 nm. Hence, the distance between two consecutive base pairs
(i.e., hydrogen-bonded bases of the opposite strands) is 0.34 nm.

The DNA coils up, forming chromosomes, and each chromosome has a single molecule of
DNA in it. Overall, human beings have around twenty-three pairs of chromosomes in the
nucleus of cells. DNA also plays an essential role in the process of cell division.
Also Read: DNA Packaging

Chargaff’s Rule
 EU Test Material

Erwin Chargaff, a biochemist, discovered that the number of nitrogenous bases in the
DNA was present in equal quantities. The amount of A is equal to T, whereas the amount of
C is equal to G.
A=T; C=G
In other words, the DNA of any cell from any organism should have a 1:1 ratio of purine and
pyrimidine bases.

DNA Replication
DNA replication is an important process that occurs during cell division. It is also known
as semi-conservative replication, during which DNA makes a copy of itself.

DNA replication takes place in three stages:

Step 1: Initiation
The replication of DNA begins at a point known as the origin of replication. The two DNA
strands are separated by the DNA helicase. This forms the replication fork.

Step 2: Elongation
DNA polymerase III reads the nucleotides on the template strand and makes a new strand by
adding complementary nucleotides one after the other. For eg., if it reads an Adenine on the
template strand, it will add a Thymine on the complementary strand.
While adding nucleotides to the lagging strand, gaps are formed between the strands. These
gaps are known as Okazaki fragments. These gaps or nicks are sealed by ligase.

Step 3: Termination
The termination sequence present opposite to the origin of replication terminates the
replication process. The TUS protein (terminus utilization substance) binds to terminator
sequence and halts DNA polymerase movement. It induces termination.
 EU Test Material

Also Read: DNA Replication

DNA Function
DNA is the genetic material which carries all the hereditary information. Genes are the small
segments of DNA, consisting mostly of 250 – 2 million base pairs. A gene code for a
polypeptide molecule, where three nitrogenous bases sequence stands for one amino acid.
Polypeptide chains are further folded in secondary, tertiary and quaternary structures to form
different proteins. As every organism contains many genes in its DNA, different types of
proteins can be formed. Proteins are the main functional and structural molecules in most
organisms. Apart from storing genetic information, DNA is involved in:

 Replication process: Transferring the genetic information from one cell to its
daughters and from one generation to the next and equal distribution of DNA during
the cell division
 Mutations: The changes which occur in the DNA sequences
 Transcription
 Cellular Metabolism
 DNA Fingerprinting
 Gene Therapy
Also Read: r-factor

Why DNA is called a Polynucleotide Molecule?

The DNA is called a polynucleotide because the DNA molecule is composed of nucleotides
– deoxyadenylate (A) deoxyguanylate (G) deoxycytidylate (C) and deoxythymidylate
(T), which are combined to create long chains called a polynucleotide. As per the DNA
structure, the DNA consists of two chains of polynucleotides.
Also Read: Genetic Material
For more detailed information on DNA meaning, diagram, its types, DNA structure and
function, or any other related topics, explore at BYJU’S Biology.
Explore more

 Difference between Replication and Transcription

 DNA Cloning
 DNA Fingerprinting
 DNA As Genetic Material
 DNA Structure and Polynucleotide
 How is DNA inherited from each parent?
 Do you get more DNA from your mother or father?

Frequently Asked Questions

What is the structure of DNA?

 EU Test Material

DNA is a double helical structure composed of nucleotides. The two helices are joined
together by hydrogen bonds. The DNA also bears a sugar-phosphate backbone.

What are the three different types of DNA?

The three different types of DNA include:

 A-DNA
 B-DNA
 Z-DNA

How is Z-DNA different from other forms of DNA?

Z-DNA is a left-handed double helix. The helix winds to the left in a zig-zag manner. On the
contrary, A and B-DNA are right-handed DNA.

What are the functions of DNA?

The functions of DNA include:

 Replication
 Gene expression
 Mutation
 Transcription

What type of DNA is found in humans?

B-DNA is found in humans. It is a right-handed double-helical structure.
 EU Test Material

Introduction to Photosynthesis

Figure 1 This sage thrasher’s diet, like that of almost all organisms, depends on
photosynthesis. (credit: modification of work by Dave Menke, U.S. Fish and Wildlife
Service)

No matter how complex or advanced a machine, such as the latest cellular phone, the device
cannot function without energy. Living things, similar to machines, have many complex
components; they too cannot do anything without energy, which is why humans and all other
organisms must “eat” in some form or another. That may be common knowledge, but how
many people realize that every bite of every meal ingested depends on the process of
photosynthesis?

All living organisms on earth consist of one or more cells. Each cell runs on the chemical
energy found mainly in carbohydrate molecules (food), and the majority of these molecules
are produced by one process: photosynthesis. Through photosynthesis, certain organisms
convert solar energy (sunlight) into chemical energy, which is then used to build
carbohydrate molecules. The energy used to hold these molecules together is released when
an organism breaks down food. Cells then use this energy to perform work, such as cellular
respiration.

The energy that is harnessed from photosynthesis enters the ecosystems of our planet
continuously and is transferred from one organism to another. Therefore, directly or
indirectly, the process of photosynthesis provides most of the energy required by living things
on earth.

Photosynthesis also results in the release of oxygen into the atmosphere. In short, to eat and
breathe, humans depend almost entirely on the organisms that carry out photosynthesis.

Solar Dependence and Food Production

Some organisms can carry out photosynthesis, whereas others cannot. An autotroph is an
organism that can produce its own food. The Greek roots of the word autotroph mean “self”
(auto) “feeder” (troph). Plants are the best-known autotrophs, but others exist, including
certain types of bacteria and algae (Figure 2). Oceanic algae contribute enormous quantities
of food and oxygen to global food chains. Plants are also photoautotrophs, a type of autotroph
 EU Test Material

that uses sunlight and carbon from carbon dioxide to synthesize chemical energy in the form
of carbohydrates. All organisms carrying out photosynthesis require sunlight.

Figure 2 (a) Plants, (b) algae, and (c) certain bacteria, called cyanobacteria, are
photoautotrophs that can carry out photosynthesis. Algae can grow over enormous areas in
water, at times completely covering the surface. (credit a: Steve Hillebrand, U.S. Fish and
Wildlife Service; credit b: “eutrophication&hypoxia”/Flickr; credit c: NASA; scale-bar data
from Matt Russell)

Heterotrophs are organisms incapable of photosynthesis that must therefore obtain energy
and carbon from food by consuming other organisms. The Greek roots of the
word heterotroph mean “other” (hetero) “feeder” (troph), meaning that their food comes
from other organisms. Even if the food organism is another animal, this food traces its origins
back to autotrophs and the process of photosynthesis. Humans are heterotrophs, as are all
animals. Heterotrophs depend on autotrophs, either directly or indirectly. Deer and wolves
are heterotrophs. A deer obtains energy by eating plants. A wolf eating a deer obtains energy
that originally came from the plants eaten by that deer. The energy in the plant came from
photosynthesis, and therefore it is the only autotroph in this example (Figure 3). Using this
reasoning, all food eaten by humans also links back to autotrophs that carry out
photosynthesis.

Figure 3 The energy stored in carbohydrate molecules from photosynthesis passes through
the food chain. The predator that eats these deer is getting energy that originated in the
photosynthetic vegetation that the deer consumed. (credit: Steve VanRiper, U.S. Fish and
Wildlife Service)
 EU Test Material

Biology in Action

Photosynthesis at the Grocery Store

Figure 4 Photosynthesis is the origin of the products that comprise the main elements of the
human diet. (credit: Associação Brasileira de Supermercados)

Major grocery stores in the United States are organized into departments, such as dairy,
meats, produce, bread, cereals, and so forth. Each aisle contains hundreds, if not thousands,
of different products for customers to buy and consume (Figure 4).

Although there is a large variety, each item links back to photosynthesis. Meats and dairy
products link to photosynthesis because the animals were fed plant-based foods. The breads,
cereals, and pastas come largely from grains, which are the seeds of photosynthetic plants.
What about desserts and drinks? All of these products contain sugar—the basic carbohydrate
molecule produced directly from photosynthesis. The photosynthesis connection applies to
every meal and every food a person consumes.

Main Structures and Summary of Photosynthesis

Photosynthesis requires sunlight, carbon dioxide, and water as starting reactants (Figure 5).
After the process is complete, photosynthesis releases oxygen and produces carbohydrate
molecules, most commonly glucose. These sugar molecules contain the energy that living
things need to survive.
 EU Test Material

Figure 5 Photosynthesis uses solar energy, carbon dioxide, and water to release oxygen to
produce energy-storing sugar molecules. Photosynthesis is the origin of the products that
comprise the main elements of the human diet. (credit: Associação Brasileira de
Supermercados)

The complex reactions of photosynthesis can be summarized by the chemical equation shown
in Figure 6.

Figure 6 The process of photosynthesis can be represented by an equation, wherein carbon

dioxide and water produce sugar and oxygen using energy from sunlight.
 EU Test Material

Although the equation looks simple, the many steps that take place during photosynthesis are
actually quite complex, as in the way that the reaction summarizing cellular respiration
represented many individual reactions. Before learning the details of how photoautotrophs
turn sunlight into food, it is important to become familiar with the physical structures
involved.

In plants, photosynthesis takes place primarily in leaves, which consist of many layers of
cells and have differentiated top and bottom sides. The process of photosynthesis occurs not
on the surface layers of the leaf, but rather in a middle layer called the mesophyll (Figure 7).
The gas exchange of carbon dioxide and oxygen occurs through small, regulated openings
called stomata.

In all autotrophic eukaryotes, photosynthesis takes place inside an organelle called a

chloroplast. In plants, chloroplast-containing cells exist in the mesophyll. Chloroplasts have a
double (inner and outer) membrane. Within the chloroplast is a third membrane that forms
stacked, disc-shaped structures called thylakoids. Embedded in the thylakoid membrane are
molecules of chlorophyll, a pigment (a molecule that absorbs light) through which the entire
process of photosynthesis begins. Chlorophyll is responsible for the green color of plants.
The thylakoid membrane encloses an internal space called the thylakoid space. Other types of
pigments are also involved in photosynthesis, but chlorophyll is by far the most important. As
shown in Figure 7, a stack of thylakoids is called a granum, and the space surrounding the
granum is called stroma (not to be confused with stomata, the openings on the leaves).
 EU Test Material

Figure 7 Not all cells of a leaf carry out photosynthesis. Cells within the middle layer of a
leaf have chloroplasts, which contain the photosynthetic apparatus. (credit “leaf”:
modification of work by Cory Zanker)

On a hot, dry day, plants close their stomata to conserve water. What impact will this have on
photosynthesis?

The Two Parts of Photosynthesis

Photosynthesis takes place in two stages: the light-dependent reactions and the Calvin cycle.
In the light-dependent reactions, which take place at the thylakoid membrane, chlorophyll
absorbs energy from sunlight and then converts it into chemical energy with the use of water.
The light-dependent reactions release oxygen from the hydrolysis of water as a byproduct. In
the Calvin cycle, which takes place in the stroma, the chemical energy derived from the light-
dependent reactions drives both the capture of carbon in carbon dioxide molecules and the
subsequent assembly of sugar molecules. The two reactions use carrier molecules to transport
the energy from one to the other. The carriers that move energy from the light-dependent
reactions to the Calvin cycle reactions can be thought of as “full” because they bring energy.
After the energy is released, the “empty” energy carriers return to the light-dependent
reactions to obtain more energy.

Section Summary

The process of photosynthesis transformed life on earth. By harnessing energy from the sun,
photosynthesis allowed living things to access enormous amounts of energy. Because of
photosynthesis, living things gained access to sufficient energy, allowing them to evolve new
structures and achieve the biodiversity that is evident today.

Only certain organisms, called autotrophs, can perform photosynthesis; they require the
presence of chlorophyll, a specialized pigment that can absorb light and convert light energy
into chemical energy. Photosynthesis uses carbon dioxide and water to assemble
carbohydrate molecules (usually glucose) and releases oxygen into the air. Eukaryotic
autotrophs, such as plants and algae, have organelles called chloroplasts in which
photosynthesis takes place.

How can light energy be used to make food? When a person turns on a lamp, electrical
energy becomes light energy. Like all other forms of kinetic energy, light can travel, change
form, and be harnessed to do work. In the case of photosynthesis, light energy is converted
into chemical energy, which photoautotrophs use to build basic carbohydrate molecules
(Figure 1). However, autotrophs only use a few specific wavelengths of sunlight.
 EU Test Material

Figure 1 Photoautotrophs can capture visible light energy in specific wavelengths from the
sun, converting it into the chemical energy used to build food molecules. (credit: Gerry
Atwell)

What Is Light Energy?

The sun emits an enormous amount of electromagnetic radiation (solar energy in a spectrum
from very short gamma rays to very long radio waves). Humans can see only a tiny fraction
of this energy, which we refer to as “visible light.” The manner in which solar energy travels
is described as waves. Scientists can determine the amount of energy of a wave by measuring
its wavelength (shorter wavelengths are more powerful than longer wavelengths)—the
distance between consecutive crest points of a wave. Therefore, a single wave is measured
from two consecutive points, such as from crest to crest or from trough to trough (Figure 2).

Figure 2 The wavelength of a single wave is the distance between two consecutive points of
similar position (two crests or two troughs) along the wave.

Visible light constitutes only one of many types of electromagnetic radiation emitted from the
sun and other stars. Scientists differentiate the various types of radiant energy from the sun
within the electromagnetic spectrum. The electromagnetic spectrum is the range of all
possible frequencies of radiation (Figure 3). The difference between wavelengths relates to
the amount of energy carried by them.
 EU Test Material

Figure 3 The sun emits energy in the form of electromagnetic radiation. This radiation exists
at different wavelengths, each of which has its own characteristic energy. All electromagnetic
radiation, including visible light, is characterized by its wavelength.

Each type of electromagnetic radiation travels at a particular wavelength. The longer the
wavelength, the less energy it carries. Short, tight waves carry the most energy. This may
seem illogical, but think of it in terms of a piece of moving heavy rope. It takes little effort by
a person to move a rope in long, wide waves. To make a rope move in short, tight waves, a
person would need to apply significantly more energy.

The electromagnetic spectrum (Figure 3) shows several types of electromagnetic radiation

originating from the sun, including X-rays and ultraviolet (UV) rays. The higher-energy
waves can penetrate tissues and damage cells and DNA, which explains why both X-rays and
UV rays can be harmful to living organisms.

Absorption of Light

Light energy initiates the process of photosynthesis when pigments absorb specific
wavelengths of visible light. Organic pigments, whether in the human retina or the
chloroplast thylakoid, have a narrow range of energy levels that they can absorb. Energy
levels lower than those represented by red light are insufficient to raise an orbital electron to
an excited (quantum) state. Energy levels higher than those in blue light will physically tear
the molecules apart, in a process called bleaching. Our retinal pigments can only “see”
(absorb) wavelengths between 700 nm and 400 nm of light, a spectrum that is therefore called
visible light. For the same reasons, plants, pigment molecules absorb only light in the
wavelength range of 700 nm to 400 nm; plant physiologists refer to this range for plants as
photosynthetically active radiation.

The visible light seen by humans as white light actually exists in a rainbow of colors. Certain
objects, such as a prism or a drop of water, disperse white light to reveal the colors to the
human eye. The visible light portion of the electromagnetic spectrum shows the rainbow of
colors, with violet and blue having shorter wavelengths, and therefore higher energy. At the
other end of the spectrum toward red, the wavelengths are longer and have lower energy
(Figure 4).
 EU Test Material

Figure 4 Light energy can excite electrons. When a photon of light energy interacts with an
electron, the electron may absorb the energy and jump from its lowest energy ground state to
an excited state. Credit: Rao, A. and Ryan, K. Department of Biology, Texas A&M
University.

Figure 5 The colors of visible light do not carry the same amount of energy. Violet has the
shortest wavelength and therefore carries the most energy, whereas red has the longest
wavelength and carries the least amount of energy. (credit: modification of work by NASA)

Understanding Pigments

Different kinds of pigments exist, and each absorbs only specific wavelengths (colors) of
visible light. Pigments reflect or transmit the wavelengths they cannot absorb, making them
appear a mixture of the reflected or transmitted light colors.

Chlorophylls and carotenoids are the two major classes of photosynthetic pigments found in
plants and algae; each class has multiple types of pigment molecules. There are five major
chlorophylls: a, b, c and d and a related molecule found in prokaryotes
 EU Test Material

called bacteriochlorophyll. Chlorophyll a and chlorophyll b are found in higher plant

chloroplasts and will be the focus of the following discussion.

With dozens of different forms, carotenoids are a much larger group of pigments. The
carotenoids found in fruit—such as the red of tomato (lycopene), the yellow of corn seeds
(zeaxanthin), or the orange of an orange peel (β-carotene)—are used as advertisements to
attract seed dispersers. In photosynthesis, carotenoids function as photosynthetic pigments
that are very efficient molecules for the disposal of excess energy. When a leaf is exposed to
full sun, the light-dependent reactions are required to process an enormous amount of energy;
if that energy is not handled properly, it can do significant damage. Therefore, many
carotenoids reside in the thylakoid membrane, absorb excess energy, and safely dissipate that
energy as heat.

Each type of pigment can be identified by the specific pattern of wavelengths it absorbs from
visible light: This is termed the absorption spectrum. The graph in Figure 6 shows the
absorption spectra for chlorophyll a, chlorophyll b, and a type of carotenoid pigment called
β-carotene (which absorbs blue and green light). Notice how each pigment has a distinct set
of peaks and troughs, revealing a highly specific pattern of absorption. Chlorophyll a absorbs
wavelengths from either end of the visible spectrum (blue and red), but not green. Because
green is reflected or transmitted, chlorophyll appears green. Carotenoids absorb in the short-
wavelength blue region, and reflect the longer yellow, red, and orange wavelengths.

Figure 6 (a) Chlorophyll a, (b) chlorophyll b, and (c) β-carotene are hydrophobic organic
pigments found in the thylakoid membrane. Chlorophyll a and b, which are identical except
for the part indicated in the red box, are responsible for the green color of leaves. β-carotene
is responsible for the orange color in carrots. Each pigment has (d) a unique absorbance
spectrum. Credit: Rao, A., Ryan, K., Tag, A., Fletcher, S. and Hawkins, A. Department of
Biology, Texas A&M University.
 EU Test Material

Many photosynthetic organisms have a mixture of pigments, and by using these pigments, the
organism can absorb energy from a wider range of wavelengths. Not all photosynthetic
organisms have full access to sunlight. Some organisms grow underwater where light
intensity and quality decrease and change with depth. Other organisms grow in competition
for light. Plants on the rainforest floor must be able to absorb any bit of light that comes
through, because the taller trees absorb most of the sunlight and scatter the remaining solar
radiation (Figure 7).

Figure 7 Plants that commonly grow in the shade have adapted to low levels of light by
changing the relative concentrations of their chlorophyll pigments. (credit: Jason Hollinger)

When studying a photosynthetic organism, scientists can determine the types of pigments
present by generating absorption spectra. An instrument called a spectrophotometer can
differentiate which wavelengths of light a substance can absorb. Spectrophotometers measure
transmitted light and compute from it the absorption. By extracting pigments from leaves and
placing these samples into a spectrophotometer, scientists can identify which wavelengths of
light an organism can absorb. Additional methods for the identification of plant pigments
include various types of chromatography that separate the pigments by their relative affinities
to solid and mobile phases.

How Light-Dependent Reactions Work

The overall function of light-dependent reactions is to convert solar energy into chemical
energy in the form of NADPH and ATP. This chemical energy supports the light-independent
reactions and fuels the assembly of sugar molecules. The light-dependent reactions are
depicted in (Figure 8). Protein complexes and pigment molecules work together to produce
NADPH and ATP. The numbering of the photosystems is derived from the order in which
they were discovered, not in the order of the transfer of electrons.
 EU Test Material

Figure 8 A photosystem consists of 1) a light-harvesting complex and 2) a reaction center.

Pigments in the light-harvesting complex pass light energy to two special
chlorophyll a molecules in the reaction center. The light excites an electron from the
chlorophyll a pair, which passes to the primary electron acceptor. The excited electron must
then be replaced. Credit: Rao, A., Ryan, K., Fletcher, S. and Hawkins, A. Department of
Biology, Texas A&M University.

The actual step that converts light energy into chemical energy takes place in a multiprotein
complex called a photosystem, two types of which are found embedded in the thylakoid
membrane: photosystem II (PSII) and photosystem I (PSI) (Figure 9). The two complexes
differ on the basis of what they oxidize (that is, the source of the low-energy electron supply)
and what they reduce (the place to which they deliver their energized electrons).

Both photosystems have the same basic structure; a number of antenna proteins to which the
chlorophyll molecules are bound surround the reaction center where the photochemistry
takes place. Each photosystem is serviced by the light-harvesting complex, which passes
energy from sunlight to the reaction center; it consists of multiple antenna proteins that
contain a mixture of 300 to 400 chlorophyll a and b molecules as well as other pigments like
carotenoids. The absorption of a single photon or distinct quantity or “packet” of light by any
of the chlorophylls pushes that molecule into an excited state. In short, the light energy has
now been captured by biological molecules but is not stored in any useful form yet. The
energy is transferred from chlorophyll to chlorophyll until eventually (after about a millionth
of a second), it is delivered to the reaction center. Up to this point, only energy has been
transferred between molecules, not electrons.

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure 9 In the photosystem II (PSII) reaction center, energy from sunlight is used to extract
electrons from water. The electrons travel through the chloroplast electron transport chain to
photosystem I (PSI), which reduces NADP+ to NADPH. The electron transport chain moves
protons across the thylakoid membrane into the lumen. At the same time, splitting of water
adds protons to the lumen, and reduction of NADPH removes protons from the stroma. The
net result is a low pH in the thylakoid lumen, and a high pH in the stroma. ATP synthase uses
this electrochemical gradient to make ATP. Credit: Rao, A., Ryan, K., Fletcher, S.
Department of Biology, Texas A&M University.

What is the initial source of electrons for the chloroplast electron transport chain?

a. water
b. oxygen
c. carbon dioxide
d. NADPH

The reaction center contains a pair of chlorophyll a molecules with a special property. Those
two chlorophylls can undergo oxidation upon excitation; they can actually give up an electron
in a process called a photoact. It is at this step in the reaction center during photosynthesis
that light energy is converted into an excited electron. All of the subsequent steps involve
getting that electron onto the energy carrier NADPH for delivery to the Calvin cycle where
the electron is deposited onto carbon for long-term storage in the form of a carbohydrate.
PSII and PSI are two major components of the photosynthetic electron transport chain,
which also includes the cytochrome complex. The cytochrome complex, an enzyme
composed of two protein complexes, transfers the electrons from the carrier molecule
plastoquinone (Pq) to the protein plastocyanin (Pc), thus enabling both the transfer of protons
across the thylakoid membrane and the transfer of electrons from PSII to PSI.

The reaction center of PSII (called P680) delivers its high-energy electrons, one at the time,
to the primary electron acceptor, and through the electron transport chain (Pq to
 EU Test Material

cytochrome complex to plastocyanine) to PSI. P680’s missing electron is replaced by

extracting a low-energy electron from water; thus, water is “split” during this stage of
photosynthesis, and PSII is re-reduced after every photoact. Splitting one H2O molecule
releases two electrons, two hydrogen atoms, and one atom of oxygen. However, splitting two
molecules is required to form one molecule of diatomic O2 gas. About 10 percent of the
oxygen is used by mitochondria in the leaf to support oxidative phosphorylation. The
remainder escapes to the atmosphere where it is used by aerobic organisms to support
respiration.

As electrons move through the proteins that reside between PSII and PSI, they lose energy.
This energy is used to move hydrogen atoms from the stromal side of the membrane to the
thylakoid lumen. Those hydrogen atoms, plus the ones produced by splitting water,
accumulate in the thylakoid lumen and will be used synthesize ATP in a later step. Because
the electrons have lost energy prior to their arrival at PSI, they must be re-energized by PSI,
hence, another photon is absorbed by the PSI antenna. That energy is relayed to the PSI
reaction center (called P700). P700 is oxidized and sends a high-energy electron to NADP+ to
form NADPH. Thus, PSII captures the energy to create proton gradients to make ATP, and
PSI captures the energy to reduce NADP+ into NADPH. The two photosystems work in
concert, in part, to guarantee that the production of NADPH will roughly equal the
production of ATP. Other mechanisms exist to fine-tune that ratio to exactly match the
chloroplast’s constantly changing energy needs.

Generating an Energy Carrier: ATP

As in the intermembrane space of the mitochondria during cellular respiration, the buildup of
hydrogen ions inside the thylakoid lumen creates a concentration gradient. The passive
diffusion of hydrogen ions from high concentration (in the thylakoid lumen) to low
concentration (in the stroma) is harnessed to create ATP, just as in the electron transport
chain of cellular respiration. The ions build up energy because of diffusion and because they
all have the same electrical charge, repelling each other.

To release this energy, hydrogen ions will rush through any opening, similar to water jetting
through a hole in a dam. In the thylakoid, that opening is a passage through a specialized
protein channel called the ATP synthase. The energy released by the hydrogen ion stream
allows ATP synthase to attach a third phosphate group to ADP, which forms a molecule of
ATP (Figure 9). The flow of hydrogen ions through ATP synthase is called chemiosmosis
because the ions move from an area of high to an area of low concentration through a semi-
permeable structure of the thylakoid.

Chapter Outline

1 The Process of Meiosis

2 Sexual Reproduction
 EU Test Material

The ability to reproduce is a basic characteristic of all organisms: Hippopotamuses give birth
to hippopotamus calves; Joshua trees produce seeds from which Joshua tree seedlings
emerge; and adult flamingos lay eggs that hatch into flamingo chicks. However, unlike the
organisms shown above, offspring may or may not resemble their parents. For example, in
the case of most insects such as butterflies (with a complete metamorphosis), the larval forms
rarely resemble the adult forms.

Although many unicellular organisms and a few multicellular organisms can produce
genetically identical clones of themselves through asexual reproduction, many single-celled
organisms and most multicellular organisms reproduce regularly using another method—
sexual reproduction. This highly evolved method involves the production by parents of two
haploid cells and the fusion of two haploid cells to form a single, genetically recombined
diploid cell—a genetically unique organism. In almost all sexually reproducing species, these
two haploid cells differ in size, with the smaller cell called “male” and the larger one called
“female." These haploid cells are produced by a type of cell division called meiosis. Sexual
reproduction, involving both meiosis and fertilization, introduces variation into offspring that
may account for the evolutionary success of sexual reproduction. The vast majority of
eukaryotic organisms, both multicellular and unicellular, can or must employ some form of
meiosis and fertilization to reproduce.

In most plants and animals the zygote formed by fertilization, through thousands of rounds of
mitotic cell division, will develop into an adult organism.

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe the behavior of chromosomes during meiosis, and the differences between
the first and second meiotic divisions
 Describe the cellular events that take place during meiosis
 Explain the differences between meiosis and mitosis
 Explain the mechanisms within the meiotic process that produce genetic variation
among the haploid gametes

Sexual reproduction requires the union of two specialized cells, called gametes, each of
which contains one set of chromosomes. When gametes unite, they form a zygote, or
fertilized egg that contains two sets of chromosomes. (Note: Cells that contain one set of
chromosomes are called haploid; cells containing two sets of chromosomes are
called diploid.) If the reproductive cycle is to continue for any sexually reproducing species,
then the diploid cell must somehow reduce its number of chromosome sets to produce
haploid gametes; otherwise, the number of chromosome sets will double with every future
round of fertilization. Therefore, sexual reproduction requires a nuclear division that reduces
the number of chromosome sets by half.

Most animals and plants and many unicellular organisms are diploid and therefore have two
sets of chromosomes. In each somatic cell of the organism (all cells of a multicellular
organism except the gametes or reproductive cells), the nucleus contains two copies of each
chromosome, called homologous chromosomes. Homologous chromosomes are matched
pairs containing the same genes in identical locations along their lengths. Diploid organisms
inherit one copy of each homologous chromosome from each genetic contributor.
 EU Test Material

Meiosis is the nuclear division that forms haploid cells from diploid cells, and it employs
many of the same cellular mechanisms as mitosis. However, as you have
learned, mitosis produces daughter cells whose nuclei are genetically identical to the original
parent nucleus. In mitosis, both the parent and the daughter nuclei are at the same “ploidy
level”—diploid in the case of most multicellular animals. Plants use mitosis to grow as
sporophytes, and to grow and produce eggs and sperm as gametophytes; so they use mitosis
for both haploid and diploid cells (as well as for all other ploidies). In meiosis, the starting
nucleus is always diploid and the daughter nuclei that result are haploid. To achieve this
reduction in chromosome number, meiosis consists of one round of chromosome replication
followed by two rounds of nuclear division. Because many events that occur during each of
the division stages are analogous to the events of mitosis, the same stage names are assigned.
However, because there are two rounds of division, the major process and the stages are
designated with a “I” or a “II.” Thus, meiosis I is the first round of meiotic division and
consists of prophase I, prometaphase I, and so on. Likewise, Meiosis II (during which the
second round of meiotic division takes place) includes prophase II, prometaphase II, and so
on.

Figure 2 Overview of Meiosis. The production of gametes is a crucial process for sexually
reproducing organisms. Meiosis is the mechanism used to reduce diploid cells to haploid
gametes while introducing genetic diversity. Prior to meiosis, chromosomes are replicated in
S-phase to ensure proper number of chromosomes in the resulting gametes. During meiosis,
two successive rounds of division reduces the number of chromosomes (ploidy) of the cell by
 EU Test Material

half. going from diploid cells to haploid gametes. Credit: Rao, A., Tag, A, Fletcher, S., and
Ryan, K. Department of Biology, Texas A&M University.

Meiosis I

Meiosis is preceded by an interphase consisting of G1, S, and G2 phases, which are nearly
identical to the phases preceding mitosis. The G1 phase (the “first gap phase”) is focused on
cell growth. During the S phase—the second phase of interphase—the cell copies
or replicates the DNA of the chromosomes. Finally, in the G2 phase (the “second gap phase”)
the cell undergoes the final preparations for meiosis.

During DNA duplication in the S phase, each chromosome is replicated to produce two
identical copies—sister chromatids that are held together at the centromere
by cohesin proteins, which hold the chromatids together until anaphase II. (Note: these
chromosome copies are called sister chromatids regardless of whether they are in a female
gamete or a male gamete.)

Prophase I

Early in prophase I, before the chromosomes can be seen clearly with a microscope, the
homologous chromosomes are attached at their tips to the nuclear envelope by proteins. As
the nuclear envelope begins to break down, the proteins associated with homologous
chromosomes bring the pair closer together. Recall that in mitosis, homologous chromosomes
do not pair together. The synaptonemal complex, a lattice of proteins between the
homologous chromosomes, first forms at specific locations and then spreads outward to cover
the entire length of the chromosomes. The tight pairing of the homologous chromosomes is
called synapsis. In synapsis, the genes on the chromatids of the homologous chromosomes
are aligned precisely with each other. The synaptonemal complex supports the exchange of
chromosomal segments between homologous nonsister chromatids—a process
called crossing over. Crossing over can be observed visually after the exchange
as chiasmata (singular = chiasma) (Figure 3).

In humans, even though the X and Y sex chromosomes are not completely homologous (that
is, most of their genes differ), there is a small region of homology that allows the X and Y
chromosomes to pair up during prophase I. A partial synaptonemal complex develops only
between the regions of homology.
 EU Test Material

Figure 3 Early in prophase I, homologous chromosomes come together to form a synapse.

The chromosomes are bound tightly together and in perfect alignment by a protein lattice
called a synaptonemal complex and by cohesin proteins at the centromere.

Located at intervals along the synaptonemal complex are large protein assemblies
called recombination nodules. These assemblies mark the points of later chiasmata and
mediate the multistep process of crossover—or genetic recombination—between the
nonsister chromatids. Near the recombination nodule, the double-stranded DNA of each
chromatid is cleaved, the cut ends are modified, and a new connection is made between the
nonsister chromatids. As prophase I progresses, the synaptonemal complex begins to break
down and the chromosomes begin to condense. When the synaptonemal complex is gone, the
homologous chromosomes remain attached to each other at the centromere and at chiasmata.
The chiasmata remain until anaphase I. The number of chiasmata varies according to the
species and the length of the chromosome. There must be at least one chiasma per
chromosome for proper separation of homologous chromosomes during meiosis I, but there
may be as many as 25. Following crossover, the synaptonemal complex breaks down and the
cohesin connection between homologous pairs is removed. At the end of prophase I, the pairs
are held together only at the chiasmata (Figure 4). These pairs are called tetrads because a
total of four sister chromatids of each pair of homologous chromosomes are now visible.

The crossover events are the first source of genetic variation in the nuclei produced by
meiosis. A single crossover event between homologous nonsister chromatids leads to a
reciprocal exchange of equivalent DNA between an egg-derived chromosome and a sperm-
derived chromosome. When a recombinant sister chromatid is moved into a gamete cell it
will carry a combination of maternal and paternal genes that did not exist before the
crossover. Crossover events can occur almost anywhere along the length of the synapsed
chromosomes. Different cells undergoing meiosis will therefore produce different
recombinant chromatids, with varying combinations of maternal and parental genes. Multiple
crossovers in an arm of the chromosome have the same effect, exchanging segments of DNA
to produce genetically recombined chromosomes.
 EU Test Material

Figure 4 Crossover occurs between nonsister chromatids of homologous chromosomes. The

result is an exchange of genetic material between homologous chromosomes.

Prometaphase I

The key event in prometaphase I is the attachment of the spindle fiber microtubules to the
kinetochore proteins at the centromeres. Kinetochore proteins are multiprotein complexes
that bind the centromeres of a chromosome to the microtubules of the mitotic spindle.
Microtubules grow from microtubule-organizing centers (MTOCs). In animal cells, MTOCs
are centrosomes located at opposite poles of the cell. The microtubules from each pole move
toward the middle of the cell and attach to one of the kinetochores of the two fused
homologous chromosomes. Each member of the homologous pair attaches to a microtubule
extending from opposite poles of the cell so that in the next phase, the microtubules can pull
the homologous pair apart. A spindle fiber that has attached to a kinetochore is called
a kinetochore microtubule. At the end of prometaphase I, each tetrad is attached to
microtubules from both poles, with one homologous chromosome facing each pole. The
homologous chromosomes are still held together at the chiasmata. In addition, the nuclear
membrane has broken down entirely.

Metaphase I

During metaphase I, the homologous chromosomes are arranged at the metaphase plate—
roughly in the midline of the cell, with the kinetochores facing opposite poles. Each
homologous pair is oriented randomly at the equator. For example, if the two homologous
members of chromosome 1 are labeled a and b, then the chromosomes could line up a-b or b-
 EU Test Material

a. This is important in determining the genes carried by a gamete, as each will only receive
one of the two homologous chromosomes. (Recall that homologous chromosomes are not
identical. They contain different versions of the same genes, and after recombination during
crossing over, each gamete will have a unique genetic makeup that has never existed before.)

The randomness in the alignment of recombined chromosomes at the metaphase plate,

coupled with the crossing over events between nonsister chromatids, are responsible for
much of the genetic variation in the offspring. To clarify this further, remember that the
homologous chromosomes of a sexually reproducing organism are originally inherited as two
separate sets, one from each parent. Using humans as an example, one set of 23 chromosomes
is present in the egg cell, often called maternal chromosomes because the genetic contributor
is often the mother. The other set of 23 chromosomes is contained in the sperm, and the
genetic contributor is called a father who provides the paternal chromosomes. Every cell of
the multicellular offspring has copies of the original two sets of homologous chromosomes.
When the offspring human creates their own gametes through meiosis, the two sets of
chromosomes will be rearranged. In prophase I of meiosis, the homologous chromosomes
form the tetrads. In metaphase I, these pairs line up at the midway point between the two
poles of the cell to form the metaphase plate. Because there is an equal chance that a
microtubule fiber will encounter a maternally or paternally inherited chromosome, the
arrangement of the tetrads at the metaphase plate is random. Thus, any maternally inherited
chromosome may face either pole. Likewise, any paternally inherited chromosome may also
face either pole. The orientation of each tetrad is independent of the orientation of the other
22 tetrads.

This event—the random (or independent) assortment of homologous chromosomes at the

metaphase plate—is the second mechanism that introduces variation into the gametes or
spores. In each cell that undergoes meiosis, the arrangement of the tetrads is different. The
number of variations is dependent on the number of chromosomes making up a set. There are
two possibilities for orientation at the metaphase plate; the possible number of alignments
therefore equals 2n in a diploid cell, where n is the number of chromosomes per haploid set.
Humans have 23 chromosome pairs, which results in over eight million (223) possible
genetically-distinct gametes just from the random alignment of chromosomes at the
metaphase plate. This number does not include the variability that was previously produced
by crossing over between the nonsister chromatids. Given these two mechanisms, it is highly
unlikely that any two haploid cells resulting from meiosis will have the same genetic
composition (Figure 5).

To summarize, meiosis I creates genetically diverse gametes in two ways. First, during
prophase I, crossover events between the nonsister chromatids of each homologous pair of
chromosomes generate recombinant chromatids with new combinations of maternal and
paternal genes. Second, the random assortment of tetrads on the metaphase plate produces
unique combinations of maternal and paternal chromosomes that will make their way into the
gametes.
 EU Test Material

Figure .5 Random, independent assortment during metaphase I is demonstrated by

considering a cell with a set of two chromosomes (n = 2). There are two possible homologous
chromosome arrangements at the equatorial plane in metaphase I, that are then separated
during anaphase I. The total possible number of different gametes is 2n, where n equals the
number of chromosomes in a set. In this example, there are four possible genetic
combinations for the gametes. With n = 23 in human cells, there are over eight million
possible combinations of paternal and maternal chromosomes. Credit: Rao, A.and Fletcher, S.
Department of Biology, Texas A&M University.

Anaphase I

In anaphase I, the microtubules pull the linked chromosomes apart. The sister chromatids
remain tightly bound together at the centromere. The chiasmata are broken in anaphase I as
the microtubules attached to the fused kinetochores pull the homologous chromosomes apart
(Figure 6).

Telophase I and Cytokinesis

In telophase, the separated chromosomes arrive at opposite poles. The remainder of the
typical telophase events may or may not occur, depending on the species. In some organisms,
the chromosomes “decondense” and nuclear envelopes form around the separated sets of
chromatids produced during telophase I. In other organisms, cytokinesis—the physical
separation of the cytoplasmic components into two daughter cells—occurs without
reformation of the nuclei. In nearly all species of animals and some fungi, cytokinesis
separates the cell contents via a cleavage furrow (constriction of the actin ring that leads to
cytoplasmic division). In plants, a cell plate is formed during cell cytokinesis by Golgi
vesicles fusing at the metaphase plate. This cell plate will ultimately lead to the formation of
cell walls that separate the two daughter cells.
 EU Test Material

Two haploid cells are the result of the first meiotic division of a diploid cell. The cells are
haploid because at each pole, there is just one of each pair of the homologous chromosomes.
Therefore, only one full set of the chromosomes is present. This is why the cells are
considered haploid—there is only one chromosome set, even though each chromosome still
consists of two sister chromatids. Recall that sister chromatids are merely duplicates of one of
the two homologous chromosomes (except for changes that occurred during crossing over).
In meiosis II, these two sister chromatids will separate, creating four haploid daughter cells.

Meiosis II

In some species, cells enter a brief interphase, or interkinesis, before entering meiosis II.
Interkinesis lacks an S phase, so chromosomes are not duplicated. The two cells produced in
meiosis I go through the events of meiosis II in synchrony. During meiosis II, the sister
chromatids within the two daughter cells separate, forming four new haploid gametes. The
mechanics of meiosis II are similar to mitosis, except that each dividing cell has only one set
of homologous chromosomes, each with two chromatids. Therefore, each cell has half the
number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of
chromosomal content, cells at the start of meiosis II are similar to haploid cells in G2,
preparing to undergo mitosis.

Prophase II

If the chromosomes decondensed in telophase I, they condense again. If nuclear envelopes

were formed, they fragment into vesicles. The MTOCs that were duplicated during
interkinesis move away from each other toward opposite poles, and new spindles are formed.

Prometaphase II

The nuclear envelopes are completely broken down, and the spindle is fully formed. Each
sister chromatid forms an individual kinetochore that attaches to microtubules from opposite
poles.

Metaphase II

The sister chromatids are maximally condensed and aligned at the equator of the cell.

Anaphase II

The sister chromatids are pulled apart by the kinetochore microtubules and move toward
opposite poles. Nonkinetochore microtubules elongate the cell.
 EU Test Material

Figure 6 The process of chromosome alignment differs between meiosis I and meiosis II. In
prophase I, pairs of homologous chromosomes form chiasmata which allow for crossing over
events (genetic diversity). These pairs of homologous chromosomes arrange at the metaphase
plate in metaphase I. In anaphase I, homologous chromosomes separate. Telophase I and
cytokinesis result in haploid cells with 2 sister chromatids of each chromosome. In prophase
II, spindle microtubules form and elongate and any nuclear envelope disappears. Sister
chromatids arrange at the midpoint of the cells in metaphase II. In anaphase II, the sister
chromatids separate. Telophase II and cytokinesis result in haploid cells with a single copy of
each chromosome. Credit: Rao, A., Ryan, K., Fletcher, S. and Tag, A. Department of
Biology, Texas A&M University.

Telophase II and Cytokinesis

The chromosomes arrive at opposite poles and begin to decondense. Nuclear envelopes form
around the chromosomes. If the parent cell was diploid, as is most commonly the case, then
cytokinesis now separates the two cells into four unique haploid cells. The cells produced are
genetically unique because of the random assortment of paternal and maternal homologs and
because of the recombination of maternal and paternal segments of chromosomes (with their
sets of genes) that occurs during crossover. The entire process of meiosis is outlined in Figure
7.
 EU Test Material

Figure 7 An animal cell with a diploid number of four (2n = 4) proceeds through the stages
of meiosis to form four haploid daughter cells.

Comparing Meiosis and Mitosis

Mitosis and meiosis are both forms of division of the nucleus in eukaryotic cells. They share
some similarities, but also exhibit a number of distinct processes that lead to very different
outcomes (Figure 8). Mitosis is a single nuclear division that results in two nuclei that are
usually partitioned into two new cells. The nuclei resulting from a mitotic division are
genetically identical to the original nucleus. They have the same number of sets of
chromosomes: one set in the case of haploid cells and two sets in the case of diploid cells. In
contrast, meiosis consists of two nuclear divisions resulting in four nuclei that are usually
partitioned into four new, genetically distinct cells. The four nuclei produced during meiosis
 EU Test Material

are not genetically identical, and they contain one chromosome set only. This is half the
number of chromosome sets of the original cell, which is diploid.

The main differences between mitosis and meiosis occur in meiosis I, which is a very
different nuclear division than mitosis. In meiosis I, the homologous chromosome pairs
physically meet and are bound together with the synaptonemal complex. Following this, the
chromosomes develop chiasmata and undergo crossover between nonsister chromatids. In the
end, the chromosomes line up along the metaphase plate as tetrads—with kinetochore fibers
from opposite spindle poles attached to each kinetochore of a homolog to form a tetrad. All of
these events occur only in meiosis I.

When the chiasmata resolve and the tetrad is broken up with the homologous chromosomes
moving to one pole or another, the ploidy level—the number of sets of chromosomes in each
future nucleus—has been reduced from two to one. For this reason, meiosis I is referred to as
a reductional division. There is no such reduction in ploidy level during mitosis.

Meiosis II is analogous to a mitotic division. In this case, the duplicated chromosomes (only
one set of them) line up on the metaphase plate with divided kinetochores attached to
kinetochore fibers from opposite poles. During anaphase II, as in mitotic anaphase, the
kinetochores divide and one sister chromatid—now referred to as a chromosome—is pulled
to one pole while the other sister chromatid is pulled to the other pole. If it were not for the
fact that there had been crossover, the two products of each individual meiosis II division
would be identical (as in mitosis). Instead, they are different because there has always been at
least one crossover per chromosome. Meiosis II is not a reduction division because although
there are fewer copies of the genome in the resulting cells, there is still one set of
chromosomes, as there was at the end of meiosis I.
 EU Test Material

Figure 8 Meiosis and mitosis are both preceded by one cycle of DNA replication; however,
meiosis includes two nuclear divisions. The four daughter cells resulting from meiosis are
haploid and genetically distinct. The daughter cells resulting from mitosis are diploid and
identical to the parent cell.

EVOLUTION CONNECTION

Evolution Connection

The Mystery of the Evolution of MeiosisSome characteristics of organisms are so

widespread and fundamental that it is sometimes difficult to remember that they evolved like
other simple traits. Meiosis is such an extraordinarily complex series of cellular events that
biologists have had trouble testing hypotheses concerning how it may have evolved.
Although meiosis is inextricably entwined with sexual reproduction and its advantages and
disadvantages, it is important to separate the questions of the evolution of meiosis and the
evolution of sex, because early meiosis may have been advantageous for different reasons
than it is now. Thinking outside the box and imagining what the early benefits from meiosis
might have been is one approach to uncovering how it may have evolved.
 EU Test Material

Meiosis and mitosis share obvious cellular processes, and it makes sense that meiosis evolved
from mitosis. The difficulty lies in the clear differences between meiosis I and mitosis. Adam
Wilkins and Robin Holliday1 summarized the unique events that needed to occur for the
evolution of meiosis from mitosis. These steps are homologous chromosome pairing and
synapsis, crossover exchanges, sister chromatids remaining attached during anaphase, and
suppression of DNA replication in interphase. They argue that the first step is the hardest and
most important and that understanding how it evolved would make the evolutionary process
clearer. They suggest genetic experiments that might shed light on the evolution of synapsis.

There are other approaches to understanding the evolution of meiosis in progress. Different
forms of meiosis exist in single-celled protists. Some appear to be simpler or more
“primitive” forms of meiosis. Comparing the meiotic divisions of different protists may shed
light on the evolution of meiosis. Marilee Ramesh and colleagues2 compared the genes
involved in meiosis in protists to understand when and where meiosis might have evolved.
Although research is still ongoing, recent scholarship into meiosis in protists suggests that
some aspects of meiosis may have evolved later than others. This kind of genetic comparison
can tell us what aspects of meiosis are the oldest and what cellular processes they may have
borrowed from in earlier cells.

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain that meiosis and sexual reproduction are highly evolved traits
 Identify variation among offspring as a potential evolutionary advantage of sexual
reproduction
 Describe the three different life-cycle types among sexually reproducing multicellular
organisms.

Sexual reproduction was likely an early evolutionary innovation after the appearance of
eukaryotic cells. It appears to have been very successful because most eukaryotes are able to
reproduce sexually and, in many animals, it is the only mode of reproduction. And yet,
scientists also recognize some real disadvantages to sexual reproduction. On the surface,
creating offspring that are genetic clones of the parent appears to be a better system. If the
parent organism is successfully occupying a habitat, offspring with the same traits should be
similarly successful. There is also the obvious benefit to an organism that can produce
offspring whenever circumstances are favorable by asexual budding, fragmentation, or by
producing eggs asexually. These methods of reproduction do not require a partner with which
to reproduce. Indeed, some organisms that lead a solitary lifestyle have retained the ability to
reproduce asexually. In addition, in asexual populations, every individual is capable of
reproduction. In sexual populations, the males are not producing the offspring themselves, so
hypothetically an asexual population could grow twice as fast.

However, multicellular organisms that exclusively depend on asexual reproduction are

exceedingly rare. Why are meiosis and sexual reproductive strategies so common? These are
important (and as yet unanswered) questions in biology, even though they have been the
focus of much research beginning in the latter half of the 20th century. There are several
possible explanations, one of which is that the variation that sexual reproduction creates
 EU Test Material

among offspring is very important to the survival and reproduction of the population. Thus,
on average, a sexually reproducing population will leave more descendants than an otherwise
similar asexually reproducing population. The only source of variation in asexual organisms
is mutation. Mutations that take place during the formation of germ cell lines are also a
source of variation in sexually reproducing organisms. However, in contrast to mutation
during asexual reproduction, the mutations during sexual reproduction can be continually
reshuffled from one generation to the next when different parents combine their unique
genomes and the genes are mixed into different combinations by crossovers during prophase
I and random assortment at metaphase I.

EVOLUTION CONNECTION

Evolution Connection

The Red Queen HypothesisGenetic variation is the outcome of sexual reproduction, but
why are ongoing variations necessary, even under seemingly stable environmental
conditions? Enter the Red Queen hypothesis, first proposed by Leigh Van Valen in
1973.3 The concept was named in reference to the Red Queen's race in Lewis Carroll's
book, Through the Looking-Glass.

All species coevolve (evolve together) with other organisms. For example, predators evolve
with their prey, and parasites evolve with their hosts. Each tiny advantage gained by
favorable variation gives a species a reproductive edge over close competitors, predators,
parasites, or even prey. However, survival of any given genotype or phenotype in a
population is dependent on the reproductive fitness of other genotypes or phenotypes within a
given species. The only method that will allow a coevolving species to maintain its own share
of the resources is to also continually improve its fitness (the capacity of the members to
produce more reproductively viable offspring relative to others within a species). As one
species gains an advantage, this increases selection on the other species; they must also
develop an advantage or they will be outcompeted. No single species progresses too far ahead
because genetic variation among the progeny of sexual reproduction provides all species with
a mechanism to improve rapidly. Species that cannot keep up become extinct. The Red
Queen’s catchphrase was, “It takes all the running you can do to stay in the same place.” This
is an apt description of coevolution between competing species.

Life Cycles of Sexually Reproducing Organisms

Fertilization and meiosis alternate in sexual life cycles. What happens between these two
events depends on the organism’s “reproductive strategy.” The process of meiosis reduces the
chromosome number by half. Fertilization, the joining of two haploid gametes, restores the
diploid condition. Some organisms have a multicellular diploid stage that is most obvious and
only produce haploid reproductive cells. Animals, including humans, have this type of life
cycle. Other organisms, such as fungi, have a multicellular haploid stage that is most obvious.
Plants and some algae have alternation of generations, in which they have multicellular
diploid and haploid life stages that are apparent to different degrees depending on the group.

Nearly all animals employ a diploid-dominant life-cycle strategy in which the only haploid
cells produced by the organism are the gametes. Early in the development of the embryo,
specialized diploid cells, called germ cells, are produced within the gonads (such as the testes
and ovaries). Germ cells are capable of mitosis to perpetuate the germ cell line and meiosis to
 EU Test Material

produce haploid gametes. Once the haploid gametes are formed, they lose the ability to divide
again. There is no multicellular haploid life stage. Fertilization occurs with the fusion of two
gametes, usually from different individuals, restoring the diploid state (Figure 9).

Figure .9 In animals, sexually reproducing adults form haploid gametes, called egg and
sperm, from diploid germ cells. Fusion of the two gametes gives rise to a fertilized egg cell,
or zygote. The zygote will undergo multiple rounds of mitosis to produce a multicellular
offspring. The germ cells are generated early in the development of the zygote.

Most fungi and algae employ a life-cycle type in which the “body” of the organism—the
ecologically important part of the life cycle—is haploid. The haploid cells that make up the
tissues of the dominant multicellular stage are formed by mitosis. During sexual
reproduction, specialized haploid cells from two individuals—designated the (+) and (−)
mating types—join to form a diploid zygote. The zygote immediately undergoes meiosis to
form four haploid cells called spores. Although these spores are haploid like the “parents,”
they contain a new genetic combination from two parents. The spores can remain dormant for
various time periods. Eventually, when conditions are favorable, the spores form
multicellular haploid structures through many rounds of mitosis (Figure 10).
 EU Test Material

Visual Connection

Figure .10 Fungi, such as black bread mold (Rhizopus nigricans), have a haploid
multicellular stage that produces specialized haploid cells by mitosis that fuse to form a
diploid zygote. The zygote undergoes meiosis to produce haploid spores. Each spore gives
rise to a multicellular haploid organism by mitosis. Above, different mating hyphae types
(denoted as + and –) join to form a zygospore through nuclear fusion. (credit “zygomycota”
micrograph: modification of work by “Fanaberka”/Wikimedia Commons)

If a mutation occurs so that a fungus is no longer able to produce a minus mating type, will it
still be able to reproduce?

The third life-cycle type, employed by some algae and all plants, is a blend of the haploid-
dominant and diploid-dominant extremes. Species with alternation of generations have both
haploid and diploid multicellular organisms as part of their life cycle. The haploid
multicellular plants are called gametophytes, because they produce gametes from specialized
cells. Meiosis is not directly involved in the production of gametes in this case, because the
organism that produces the gametes is already haploid. Fertilization between the gametes
forms a diploid zygote. The zygote will undergo many rounds of mitosis and give rise to a
diploid multicellular plant called a sporophyte. Specialized cells of the sporophyte will
undergo meiosis and produce haploid spores. The spores will subsequently develop into the
gametophytes (Figure 11).
 EU Test Material

Figure .11 Plants have a life cycle that alternates between a multicellular haploid organism
and a multicellular diploid organism. In some plants, such as ferns, both the haploid and
diploid plant stages are free-living. The diploid plant is called a sporophyte because it
produces haploid spores by meiosis. The spores develop into multicellular, haploid plants that
are called gametophytes because they produce gametes. The gametes of two individuals will
fuse to form a diploid zygote that becomes the sporophyte. (credit “fern”: modification of
work by Cory Zanker; credit “sporangia”: modification of work by "Obsidian
Soul"/Wikimedia Commons; credit “gametophyte and sporophyte”: modification of work by
“Vlmastra”/Wikimedia Commons)

Although all plants utilize some version of the alternation of generations, the relative size of
the sporophyte and the gametophyte and the relationship between them vary greatly. In plants
such as moss, the gametophyte organism is the free-living plant and the sporophyte is
physically dependent on the gametophyte. In other plants, such as ferns, both the
gametophyte and sporophyte plants are free-living; however, the sporophyte is much larger.
In seed plants, such as magnolia trees and daisies, the gametophyte is composed of only a few
cells and, in the case of the female gametophyte, is completely retained within the
sporophyte.

Sexual reproduction takes many forms in multicellular organisms. The fact that nearly every
multicellular organism on Earth employs sexual reproduction is strong evidence for the
benefits of producing offspring with unique gene combinations, though there are other
possible benefits as well.
 EU Test Material

Chapter Outline

.1 Mendel’s Experiments and the Laws of Probability

.2 Characteristics and Traits

.3 Laws of Inheritance

Genetics is the study of heredity. Johann Gregor Mendel set the framework for genetics long
before chromosomes or genes had been identified, at a time when meiosis was not well
understood. Mendel selected a simple biological system and conducted methodical,
quantitative analyses using large sample sizes. Because of Mendel’s work, the fundamental
principles of heredity were revealed. We now know that genes, carried on chromosomes, are
the basic functional units of heredity with the capability to be replicated, expressed, or
mutated. Today, the postulates put forth by Mendel form the basis of classical, or Mendelian,
genetics. Not all genes are transmitted from parents to offspring according to Mendelian
genetics, but Mendel’s experiments serve as an excellent starting point for thinking about
inheritance.

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe the scientific reasons for the success of Mendel’s experimental work
 Describe the expected outcomes of monohybrid crosses involving dominant and
recessive alleles
 Apply the sum and product rules to calculate probabilities

Figure .2 Johann Gregor Mendel is considered the father of genetics.

Johann Gregor Mendel (1822–1884) (Figure .2) was a lifelong learner, teacher, scientist, and
man of faith. As a young adult, he joined the Augustinian Abbey of St. Thomas in Brno in
 EU Test Material

what is now the Czech Republic. Supported by the monastery, he taught physics, botany, and
natural science courses at the secondary and university levels. In 1856, he began a decade-
long research pursuit involving inheritance patterns in honeybees and plants, ultimately
settling on pea plants as his primary model system (a system with convenient characteristics
used to study a specific biological phenomenon to be applied to other systems). In 1865,
Mendel presented the results of his experiments with nearly 30,000 pea plants to the local
Natural History Society. He demonstrated that traits are transmitted from parents to offspring
independently of other traits and in dominant and recessive patterns. In 1866, he published
his work, Experiments in Plant Hybridization,1 in the proceedings of the Natural History
Society of Brünn.

Mendel’s work went virtually unnoticed by the scientific community, which believed,
incorrectly, that the process of inheritance involved a blending of parental traits that produced
an intermediate physical appearance in offspring. The blending theory of
inheritance asserted that the original parental traits were lost or absorbed by the blending in
the offspring, but we now know that this is not the case. This hypothetical process appeared
to be correct because of what we know now as continuous variation. Continuous
variation results from the action of many genes to determine a characteristic like human
height. Offspring appear to be a “blend” of their parents’ traits.

Instead of continuous characteristics, Mendel worked with traits that were inherited in
distinct classes (specifically, violet versus white flowers); this is referred to as discontinuous
variation. Mendel’s choice of these kinds of traits allowed him to see experimentally that the
traits were not blended in the offspring, nor were they absorbed, but rather that they kept their
distinctness and could be passed on. In 1868, Mendel became abbot of the monastery and
exchanged his scientific pursuits for his pastoral duties. He was not recognized for his
extraordinary scientific contributions during his lifetime. In fact, it was not until 1900 that his
work was rediscovered, reproduced, and revitalized by scientists on the brink of discovering
the chromosomal basis of heredity.

Mendel’s Model System

Mendel’s seminal work was accomplished using the garden pea, Pisum sativum, to study
inheritance. This species naturally self-fertilizes, such that pollen encounters ova within
individual flowers. The flower petals remain sealed tightly until after pollination, preventing
pollination from other plants. The result is highly inbred, or “true-breeding,” pea plants.
These are plants that always produce offspring that look like the parent. By experimenting
with true-breeding pea plants, Mendel avoided the appearance of unexpected traits in
offspring that might occur if the plants were not true breeding. The garden pea also grows to
maturity within one season, meaning that several generations could be evaluated over a
relatively short time. Finally, large quantities of garden peas could be cultivated
simultaneously, allowing Mendel to conclude that his results did not come about simply by
chance.
 EU Test Material

Mendelian Crosses

Mendel performed hybridizations, which involve mating two true-breeding individuals that
have different traits. In the pea, which is naturally self-pollinating, this is done by manually
transferring pollen from the anther of a mature pea plant of one variety to the stigma of a
separate mature pea plant of the second variety. In plants, pollen carries the male gametes
(sperm) to the stigma, a sticky organ that traps pollen and allows the sperm to move down the
pistil to the female gametes (ova) below. To prevent the pea plant that was receiving pollen
from self-fertilizing and confounding his results, Mendel painstakingly removed all of the
anthers from the plant’s flowers before they had a chance to mature.

Plants used in first-generation crosses were called P0, or parental generation one (Figure 3).
After each cross, Mendel collected the seeds belonging to the P0 plants and grew them the
following season. These offspring were called the F1, or the first filial (filial = offspring,
daughter or son) generation. Once Mendel examined the characteristics in the F1 generation
of plants, he allowed them to self-fertilize naturally. He then collected and grew the seeds
from the F1 plants to produce the F2, or second filial, generation. Mendel’s experiments
extended beyond the F2 generation to the F3 and F4 generations, and so on, but it was the ratio
of characteristics in the P0−F1−F2 generations that were the most intriguing and became the
basis for Mendel’s postulates.
 EU Test Material

Figure .3 In one of his experiments on inheritance patterns, Mendel crossed plants that were
true-breeding for violet flower color with plants true-breeding for white flower color (the P
generation). The resulting hybrids in the F1 generation all had violet flowers. In the
F2 generation, approximately three quarters of the plants had violet flowers, and one quarter
had white flowers.

Garden Pea Characteristics Revealed the Basics of Heredity

In his 1865 publication, Mendel reported the results of his crosses involving seven different
characteristics, each with two contrasting traits. A trait is defined as a variation in the
physical appearance of a heritable characteristic. The characteristics included plant height,
seed texture, seed color, flower color, pea pod size, pea pod color, and flower position. For
the characteristic of flower color, for example, the two contrasting traits were white versus
 EU Test Material

violet. To fully examine each characteristic, Mendel generated large numbers of F1 and
F2 plants, reporting results from 19,959 F2 plants alone. His findings were consistent.

What results did Mendel find in his crosses for flower color? First, Mendel confirmed that he
had plants that bred true for white or violet flower color. Regardless of how many generations
Mendel examined, all self-crossed offspring of parents with white flowers had white flowers,
and all self-crossed offspring of parents with violet flowers had violet flowers. In addition,
Mendel confirmed that, other than flower color, the pea plants were physically identical.

Once these validations were complete, Mendel applied the pollen from a plant with violet
flowers to the stigma of a plant with white flowers. After gathering and sowing the seeds that
resulted from this cross, Mendel found that 100 percent of the F1 hybrid generation had violet
flowers. Conventional wisdom at that time (the blending theory) would have predicted the
hybrid flowers to be pale violet or for hybrid plants to have equal numbers of white and violet
flowers. In other words, the contrasting parental traits were expected to blend in the
offspring. Instead, Mendel’s results demonstrated that the white flower trait in the
F1 generation had completely disappeared.

Importantly, Mendel did not stop his experimentation there. He allowed the F1 plants to self-
fertilize and found that, of F2-generation plants, 705 had violet flowers and 224 had white
flowers. This was a ratio of 3.15 violet flowers per one white flower, or approximately 3:1.
When Mendel transferred pollen from a plant with violet flowers to the stigma of a plant with
white flowers and vice versa, he obtained about the same ratio regardless of which parent,
male or female, contributed which trait. This is called a reciprocal cross—a paired cross in
which the respective traits of the male and female in one cross become the respective traits of
the female and male in the other cross. For the other six characteristics Mendel examined, the
F1 and F2 generations behaved in the same way as they had for flower color. One of the two
traits would disappear completely from the F1 generation only to reappear in the
F2 generation at a ratio of approximately 3:1 (Table 1).
 EU Test Material

The Results of Mendel’s Garden Pea Hybridizations

Table 12.1

Upon compiling his results for many thousands of plants, Mendel concluded that the
characteristics could be divided into expressed and latent traits. He called these, respectively,
dominant and recessive traits. Dominant traits are those that are inherited unchanged in a
hybridization. Recessive traits become latent, or disappear, in the offspring of a
hybridization. The recessive trait does, however, reappear in the progeny of the hybrid
offspring. An example of a dominant trait is the violet-flower trait. For this same
characteristic (flower color), white-colored flowers are a recessive trait. The fact that the
recessive trait reappeared in the F2 generation meant that the traits remained separate (not
blended) in the plants of the F1 generation. Mendel also proposed that plants possessed two
copies of the trait for the flower-color characteristic, and that each parent transmitted one of
its two copies to its offspring, where they came together. Moreover, the physical observation
of a dominant trait could mean that the genetic composition of the organism included two
dominant versions of the characteristic or that it included one dominant and one recessive
version. Conversely, the observation of a recessive trait meant that the organism lacked any
dominant versions of this characteristic.

So why did Mendel repeatedly obtain 3:1 ratios in his crosses? To understand how Mendel
deduced the basic mechanisms of inheritance that lead to such ratios, we must first review the
laws of probability.

Probability Basics
 EU Test Material

Probabilities are mathematical measures of likelihood. The empirical probability of an event

is calculated by dividing the number of times the event occurs by the total number of
opportunities for the event to occur. It is also possible to calculate theoretical probabilities by
dividing the number of times that an event is expected to occur by the number of times that it
could occur. Empirical probabilities come from observations, like those of Mendel.
Theoretical probabilities, on the other hand, come from knowing how the events are produced
and assuming that the probabilities of individual outcomes are equal. A probability of one for
some event indicates that it is guaranteed to occur, whereas a probability of zero indicates
that it is guaranteed not to occur. An example of a genetic event is a round seed produced by
a pea plant.

In one experiment, Mendel demonstrated that the probability of the event “round seed”
occurring was one in the F1 offspring of true-breeding parents, one of which has round seeds
and one of which has wrinkled seeds. When the F1 plants were subsequently self-crossed, the
probability of any given F2 offspring having round seeds was now three out of four. In other
words, in a large population of F2 offspring chosen at random, 75 percent were expected to
have round seeds, whereas 25 percent were expected to have wrinkled seeds. Using large
numbers of crosses, Mendel was able to calculate probabilities and use these to predict the
outcomes of other crosses.

The Product Rule and Sum Rule

Mendel demonstrated that pea plants transmit characteristics as discrete units from parent to
offspring. As will be discussed, Mendel also determined that different characteristics, like
seed color and seed texture, were transmitted independently of one another and could be
considered in separate probability analyses. For instance, performing a cross between a plant
with green, wrinkled seeds and a plant with yellow, round seeds still produced offspring that
had a 3:1 ratio of yellow:green seeds (ignoring seed texture) and a 3:1 ratio of wrinkled:round
seeds (ignoring seed color). The characteristics of color and texture did not influence each
other.

The product rule of probability can be applied to this phenomenon of the independent
transmission of characteristics. The product rule states that the probability of two independent
events occurring together can be calculated by multiplying the individual probabilities of
each event occurring alone. To demonstrate the product rule, imagine that you are rolling a
six-sided die (D) and flipping a penny (P) at the same time. The die may roll any number
from 1–6 (D#), whereas the penny may turn up heads (PH) or tails (PT). The outcome of
rolling the die has no effect on the outcome of flipping the penny and vice versa. There are 12
possible outcomes of this action (Table 2), and each event is expected to occur with equal
probability.

Twelve Equally Likely Outcomes of Rolling a Die and Flipping a Penny

Rolling Die Flipping Penny

D1 PH
 EU Test Material

Rolling Die Flipping Penny

D1 PT

D2 PH

D2 PT

D3 PH

D3 PT

D4 PH

D4 PT

D5 PH

D5 PT

D6 PH

D6 PT

Table .2

Of the 12 possible outcomes, the die has a 2/12 (or 1/6) probability of rolling a two, and the
penny has a 6/12 (or 1/2) probability of coming up heads. By the product rule, the probability
that you will obtain the combined outcome 2 and heads is: (D2) x (PH) = (1/6) x (1/2) or 1/12
(Table.3). Notice the word “and” in the description of the probability. The “and” is a signal to
apply the product rule. For example, consider how the product rule is applied to the dihybrid
cross: the probability of having both dominant traits (for example, yellow and round) in the
F2 progeny is the product of the probabilities of having the dominant trait for each
characteristic, as shown here:

3/4× 3/4 = 9/16

On the other hand, the sum rule of probability is applied when considering two mutually
exclusive outcomes that can come about by more than one pathway. The sum rule states that
the probability of the occurrence of one event or the other event, of two mutually exclusive
events, is the sum of their individual probabilities. Notice the word “or” in the description of
 EU Test Material

the probability. The “or” indicates that you should apply the sum rule. In this case, let’s
imagine you are flipping a penny (P) and a quarter (Q). What is the probability of one coin
coming up heads and one coin coming up tails? This outcome can be achieved by two cases:
the penny may be heads (PH) and the quarter may be tails (QT), or the quarter may be heads
(QH) and the penny may be tails (PT). Either case fulfills the outcome. By the sum rule, we
calculate the probability of obtaining one head and one tail as [(PH) × (QT)] + [(QH) × (PT)] =
[(1/2) × (1/2)] + [(1/2) × (1/2)] = 1/2 (Table .3). You should also notice that we used the
product rule to calculate the probability of PH and QT, and also the probability of PT and QH,
before we summed them. Again, the sum rule can be applied to show the probability of
having at least one dominant trait in the F2 generation of a dihybrid cross:

(1/4×3/4)+(3/4×1/4)=3/16+3/16=6/16=3/8

The Product Rule and Sum Rule

Product Rule Sum Rule

For independent events A and B, the probability (P) For mutually exclusive events A and B, the probability
of them both occurring (A and B) is (PA × PB) (P) that at least one occurs (A or B) is (PA + PB)

Table .3

To use probability laws in practice, we must work with large sample sizes because small
sample sizes are prone to deviations caused by chance. The large quantities of pea plants that
Mendel examined allowed him to calculate the probabilities of the traits appearing in his
F2 generation. As you will learn, this discovery meant that when parental traits were known,
the offspring’s traits could be predicted accurately even before fertilization.
 EU Test Material

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain the relationship between genotypes and phenotypes in dominant and recessive
gene systems
 Develop a Punnett square to calculate the expected proportions of genotypes and
phenotypes in a monohybrid cross
 Explain the purpose and methods of a test cross
 Identify non-Mendelian inheritance patterns such as incomplete dominance,
codominance, recessive lethals, multiple alleles, and sex linkage

Physical characteristics are expressed through genes carried on chromosomes. The genetic
makeup of peas consists of two similar, or homologous, copies of each chromosome, one
from each parent. Each pair of homologous chromosomes has the same linear order of genes.
In other words, peas are diploid organisms in that they have two copies of each chromosome.
The same is true for many other plants and for virtually all animals. Diploid organisms
produce haploid gametes, which contain one copy of each homologous chromosome that
unite at fertilization to create a diploid zygote.

For cases in which a single gene controls a single characteristic, a diploid organism has two
genetic copies that may or may not encode the same version of that characteristic. Gene
variants that arise by mutation and exist at the same relative locations on homologous
chromosomes are called alleles. Mendel examined the inheritance of genes with just two
allele forms, but it is common to encounter more than two alleles for any given gene in a
natural population.

Phenotypes and Genotypes

Two alleles for a given gene in a diploid organism are expressed and interact to produce
physical characteristics. The observable traits expressed by an organism are referred to as
its phenotype. An organism’s underlying genetic makeup, consisting of both physically
visible and non-expressed alleles, is called its genotype. Mendel’s hybridization experiments
demonstrate the difference between phenotype and genotype. When true-breeding plants in
which one parent had yellow pods and one had green pods were cross-fertilized, all of the
F1 hybrid offspring had yellow pods. That is, the hybrid offspring were phenotypically
identical to the true-breeding parent with yellow pods. However, we know that the allele
donated by the parent with green pods was not simply lost because it reappeared in some of
the F2 offspring. Therefore, the F1 plants must have been genotypically different from the
parent with yellow pods.

The P1 plants that Mendel used in his experiments were each homozygous for the trait he was
studying. Diploid organisms that are homozygous at a given gene, or locus, have two
identical alleles for that gene on their homologous chromosomes. Mendel’s parental pea
plants always bred true because both of the gametes produced carried the same trait. When
P1 plants with contrasting traits were cross-fertilized, all of the offspring
were heterozygous for the contrasting trait, meaning that their genotype reflected that they
had different alleles for the gene being examined.
 EU Test Material

Dominant and Recessive Alleles

Our discussion of homozygous and heterozygous organisms brings us to why the

F1 heterozygous offspring were identical to one of the parents, rather than expressing both
alleles. In all seven pea-plant characteristics, one of the two contrasting alleles was dominant,
and the other was recessive. Mendel called the dominant allele the expressed unit factor; the
recessive allele was referred to as the latent unit factor. We now know that these so-called
unit factors are actually genes on homologous chromosome pairs. For a gene that is expressed
in a dominant and recessive pattern, homozygous dominant and heterozygous organisms will
look identical (that is, they will have different genotypes but the same phenotype). The traits
of the recessive allele will only be observed in homozygous recessive individuals (Table .4).

Human Inheritance in Dominant and Recessive Patterns

Dominant Traits Recessive Traits

Achondroplasia Albinism

Brachydactyly Cystic fibrosis

Huntington’s disease Duchenne muscular dystrophy

Marfan syndrome Galactosemia

Neurofibromatosis Phenylketonuria

Widow’s peak Sickle-cell anemia

Wooly hair Tay-Sachs disease

Table .4

Several conventions exist for referring to genes and alleles. For the purposes of this chapter,
we will abbreviate genes using the first letter of the gene’s corresponding dominant trait. For
example, violet is the dominant trait for a pea plant’s flower color, so the flower-color gene
would be abbreviated as V (note that it is customary to italicize gene designations).
Furthermore, we will use uppercase and lowercase letters to represent dominant and recessive
alleles, respectively. Therefore, we would refer to the genotype of a homozygous dominant
pea plant with violet flowers as VV, a homozygous recessive pea plant with white flowers
as vv, and a heterozygous pea plant with violet flowers as Vv.

The Punnett Square Approach for a Monohybrid Cross

 EU Test Material

When fertilization occurs between two true-breeding parents that differ in only one
characteristic, the process is called a monohybrid cross, and the resulting offspring are
monohybrids. Mendel performed seven monohybrid crosses involving contrasting traits for
each characteristic. On the basis of his results in F1 and F2 generations, Mendel postulated
that each parent in the monohybrid cross contributed one of two paired unit factors to each
offspring, and every possible combination of unit factors was equally likely.

To demonstrate a monohybrid cross, consider the case of true-breeding pea plants with
yellow versus green pea seeds. The dominant seed color is yellow; therefore, the parental
genotypes were YY for the plants with yellow seeds and yy for the plants with green seeds,
respectively. A Punnett square, devised by the British geneticist Reginald Punnett, can be
drawn that applies the rules of probability to predict the possible outcomes of a genetic cross
or mating and their expected frequencies. To prepare a Punnett square, all possible
combinations of the parental alleles are listed along the top (for one parent) and side (for the
other parent) of a grid, representing their meiotic segregation into haploid gametes. Then the
combinations of egg and sperm are made in the boxes in the table to show which alleles are
combining. Each box then represents the diploid genotype of a zygote, or fertilized egg, that
could result from this mating. Because each possibility is equally likely, genotypic ratios can
be determined from a Punnett square. If the pattern of inheritance (dominant or recessive) is
known, the phenotypic ratios can be inferred as well. For a monohybrid cross of two true-
breeding parents, each parent contributes one type of allele. In this case, only one genotype is
possible. All offspring are Yy and have yellow seeds (Figure .4).
 EU Test Material

Figure .4 In the P generation, pea plants that are true-breeding for the dominant yellow
phenotype are crossed with plants with the recessive green phenotype. This cross produces
F1 heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the
genotypes of the F2 generation.

A self-cross of one of the Yy heterozygous offspring can be represented in a 2 × 2 Punnett

square because each parent can donate one of two different alleles. Therefore, the offspring
can potentially have one of four allele combinations: YY, Yy, yY, or yy (Figure 4). Notice that
there are two ways to obtain the Yy genotype: a Y from the egg and a y from the sperm, or
a y from the egg and a Y from the sperm. Both of these possibilities must be counted. Recall
that Mendel’s pea-plant characteristics behaved in the same way in reciprocal crosses.
Therefore, the two possible heterozygous combinations produce offspring that are
genotypically and phenotypically identical despite their dominant and recessive alleles
deriving from different parents. They are grouped together. Because fertilization is a random
event, we expect each combination to be equally likely and for the offspring to exhibit a ratio
of YY:Yy:yy genotypes of 1:2:1 (Figure 4). Furthermore, because the YY and Yy offspring have
yellow seeds and are phenotypically identical, applying the sum rule of probability, we
expect the offspring to exhibit a phenotypic ratio of 3 yellow:1 green. Indeed, working with
large sample sizes, Mendel observed approximately this ratio in every F2 generation resulting
from crosses for individual traits.

Mendel validated these results by performing an F3 cross in which he self-crossed the

dominant- and recessive-expressing F2 plants. When he self-crossed the plants expressing
green seeds, all of the offspring had green seeds, confirming that all green seeds had
homozygous genotypes of yy. When he self-crossed the F2 plants expressing yellow seeds, he
found that one-third of the plants bred true, and two-thirds of the plants segregated at a 3:1
ratio of yellow:green seeds. In this case, the true-breeding plants had homozygous (YY)
genotypes, whereas the segregating plants corresponded to the heterozygous (Yy) genotype.
When these plants self-fertilized, the outcome was just like the F1 self-fertilizing cross.

The Test Cross Distinguishes the Dominant Phenotype

Beyond predicting the offspring of a cross between known homozygous or heterozygous

parents, Mendel also developed a way to determine whether an organism that expressed a
dominant trait was a heterozygote or a homozygote. Called the test cross, this technique is
still used by plant and animal breeders. In a test cross, the dominant-expressing organism is
crossed with an organism that is homozygous recessive for the same characteristic. If the
dominant-expressing organism is a homozygote, then all F1 offspring will be heterozygotes
expressing the dominant trait (Figure 5). Alternatively, if the dominant expressing organism
is a heterozygote, the F1 offspring will exhibit a 1:1 ratio of heterozygotes and recessive
homozygotes (Figure .5). The test cross further validates Mendel’s postulate that pairs of unit
factors segregate equally.

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .5 A test cross can be performed to determine whether an organism expressing a

dominant trait is a homozygote or a heterozygote.

In pea plants, round peas (R) are dominant to wrinkled peas (r). You do a test cross between a
pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round
peas. You end up with three plants, all which have round peas. From this data, can you tell if
the round pea parent plant is homozygous dominant or heterozygous? If the round pea parent
plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all
be round?

Many human diseases are genetically inherited. A healthy person in a family in which some
members suffer from a recessive genetic disorder may want to know if they have the disease-
causing gene and what risk exists of passing the disorder on to their offspring. Of course,
doing a test cross in humans is unethical and impractical. Instead, geneticists use pedigree
analysis to study the inheritance pattern of human genetic diseases (Figure .6).

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .6 Alkaptonuria is a recessive genetic disorder in which two amino acids,

phenylalanine and tyrosine, are not properly metabolized. Affected individuals may have
darkened skin and brown urine, and may suffer joint damage and other complications. In this
pedigree, individuals with the disorder are indicated in blue and have the genotype aa.
Unaffected individuals are indicated in yellow and have the genotype AA or Aa. Note that it is
often possible to determine a person’s genotype from the genotype of their offspring. For
example, if neither parent has the disorder but their child does, they must be heterozygous.
Two individuals on the pedigree have an unaffected phenotype but unknown genotype.
Because they do not have the disorder, they must have at least one normal allele, so their
genotype gets the “A?” designation.

What are the genotypes of the individuals labeled 1, 2, and 3?

Alternatives to Dominance and Recessiveness

Mendel’s experiments with pea plants suggested that: (1) two “units” or alleles exist for every
gene; (2) alleles maintain their integrity in each generation (no blending); and (3) in the
presence of the dominant allele, the recessive allele is hidden and makes no contribution to
the phenotype. Therefore, recessive alleles can be “carried” and not expressed by individuals.
Such heterozygous individuals are sometimes referred to as “carriers.” Further genetic studies
in other plants and animals have shown that much more complexity exists, but that the
fundamental principles of Mendelian genetics still hold true. In the sections to follow, we
consider some of the extensions of Mendelism. If Mendel had chosen an experimental system
that exhibited these genetic complexities, it’s possible that he would not have understood
what his results meant.

Incomplete Dominance

Mendel’s results, that traits are inherited as dominant and recessive pairs, contradicted the
view at that time that offspring exhibited a blend of their parents’ traits. However, the
heterozygote phenotype occasionally does appear to be intermediate between the two parents.
 EU Test Material

For example, in the snapdragon, Antirrhinum majus (Figure .7), a cross between a
homozygous parent with white flowers (CWCW) and a homozygous parent with red flowers
(CRCR) will produce offspring with pink flowers (CRCW). (Note that different genotypic
abbreviations are used for Mendelian extensions to distinguish these patterns from simple
dominance and recessiveness.) This pattern of inheritance is described as incomplete
dominance, denoting the expression of two contrasting alleles such that the individual
displays an intermediate phenotype. The allele for red flowers is incompletely dominant over
the allele for white flowers. However, the results of a heterozygote self-cross can still be
predicted, just as with Mendelian dominant and recessive crosses. In this case, the genotypic
ratio would be 1 CRCR:2 CRCW:1 CWCW, and the phenotypic ratio would be 1:2:1 for
red:pink:white.

Figure .7 These pink flowers of a heterozygote snapdragon result from incomplete

dominance. (credit: “storebukkebruse”/Flickr)

Codominance

A variation on incomplete dominance is codominance, in which both alleles for the same
characteristic are simultaneously expressed in the heterozygote. An example of codominance
is the MN blood groups of humans. The M and N alleles are expressed in the form of an M or
N antigen present on the surface of red blood cells. Homozygotes (LMLM and LNLN) express
either the M or the N allele, and heterozygotes (LMLN) express both alleles equally. In a self-
cross between heterozygotes expressing a codominant trait, the three possible offspring
genotypes are phenotypically distinct. However, the 1:2:1 genotypic ratio characteristic of a
Mendelian monohybrid cross still applies.
 EU Test Material

Multiple Alleles

Mendel implied that only two alleles, one dominant and one recessive, could exist for a given
gene. We now know that this is an oversimplification. Although individual humans (and all
diploid organisms) can only have two alleles for a given gene, multiple alleles may exist at
the population level such that many combinations of two alleles are observed. Note that when
many alleles exist for the same gene, the convention is to denote the most common phenotype
or genotype among wild animals as the wild type (often abbreviated “+”); this is considered
the standard or norm. All other phenotypes or genotypes are considered variants of this
standard, meaning that they deviate from the wild type. The variant may be recessive or
dominant to the wild-type allele.

An example of multiple alleles is coat color in rabbits (Figure 8). Here, four alleles exist for
the c gene. The wild-type version, C+C+, is expressed as brown fur. The chinchilla
phenotype, cchcch, is expressed as black-tipped white fur. The Himalayan phenotype, chch, has
black fur on the extremities and white fur elsewhere. Finally, the albino, or “colorless”
phenotype, cc, is expressed as white fur. In cases of multiple alleles, dominance hierarchies
can exist. In this case, the wild-type allele is dominant over all the others, chinchilla is
incompletely dominant over Himalayan and albino, and Himalayan is dominant over albino.
This hierarchy, or allelic series, was revealed by observing the phenotypes of each possible
heterozygote offspring.

Figure .8 Four different alleles exist for the rabbit coat color (C) gene.

The complete dominance of a wild-type phenotype over all other mutants often occurs as an
effect of “dosage” of a specific gene product, such that the wild-type allele supplies the
correct amount of gene product whereas the mutant alleles cannot. For the allelic series in
rabbits, the wild-type allele may supply a given dosage of fur pigment, whereas the mutants
supply a lesser dosage or none at all. Interestingly, the Himalayan phenotype is the result of
an allele that produces a temperature-sensitive gene product that only produces pigment in
the cooler extremities of the rabbit’s body.
 EU Test Material

Alternatively, one mutant allele can be dominant over all other phenotypes, including the
wild type. This may occur when the mutant allele somehow interferes with the genetic
message so that even a heterozygote with one wild-type allele copy expresses the mutant
phenotype. One way in which the mutant allele can interfere is by enhancing the function of
the wild-type gene product or changing its distribution in the body. One example of this is
the Antennapedia mutation in Drosophila (Figure .9). In this case, the mutant allele expands
the distribution of the gene product, and as a result, the Antennapedia heterozygote develops
legs on its head where its antennae should be.

Figure .9 As seen in comparing the wild-type Drosophila (left) and the Antennapedia mutant
(right), the Antennapedia mutant has legs on its head in place of antennae.

EVOLUTION CONNECTION

Evolution Connection

Multiple Alleles Confer Drug Resistance in the Malaria ParasiteMalaria is a parasitic

disease in humans that is transmitted by infected female mosquitoes, including Anopheles
gambiae (Figure .10a), and is characterized by cyclic high fevers, chills, flu-like symptoms,
and severe anemia. Plasmodium falciparum and P. vivax are the most common causative
agents of malaria, and P. falciparum is the most deadly (Figure .10b). When promptly and
correctly treated, P. falciparum malaria has a mortality rate of 0.1 percent. However, in some
parts of the world, the parasite has evolved resistance to commonly used malaria treatments,
so the most effective malarial treatments can vary by geographic region.

Figure .10 The (a) Anopheles gambiae, or African malaria mosquito, acts as a vector in the
transmission to humans of the malaria-causing parasite (b) Plasmodium falciparum, here
 EU Test Material

visualized using false-color transmission electron microscopy. (credit a: James D. Gathany;

credit b: Ute Frevert; false color by Margaret Shear; scale-bar data from Matt Russell)

In Southeast Asia, Africa, and South America, P. falciparum has developed resistance to the
anti-malarial drugs chloroquine, mefloquine, and sulfadoxine-pyrimethamine. P. falciparum,
which is haploid during the life stage in which it is infectious to humans, has evolved
multiple drug-resistant mutant alleles of the dhps gene. Varying degrees of sulfadoxine
resistance are associated with each of these alleles. Being haploid, P. falciparum needs only
one drug-resistant allele to express this trait.

In Southeast Asia, different sulfadoxine-resistant alleles of the dhps gene are localized to
different geographic regions. This is a common evolutionary phenomenon that occurs
because drug-resistant mutants arise in a population and interbreed with other P.
falciparum isolates in close proximity. Sulfadoxine-resistant parasites cause considerable
human hardship in regions where this drug is widely used as an over-the-counter malaria
remedy. As is common with pathogens that multiply to large numbers within an infection
cycle, P. falciparum evolves relatively rapidly (over a decade or so) in response to the
selective pressure of commonly used anti-malarial drugs. For this reason, scientists must
constantly work to develop new drugs or drug combinations to combat the worldwide malaria
burden.2

X-Linked Traits

In humans, as well as in many other animals and some plants, the sex of the individual is
determined by sex chromosomes. The sex chromosomes are one pair of non-homologous
chromosomes. Until now, we have only considered inheritance patterns among non-sex
chromosomes, or autosomes. In addition to 22 homologous pairs of autosomes, human
females have a homologous pair of X chromosomes, whereas human males have an XY
chromosome pair. Although the Y chromosome contains a small region of similarity to the X
chromosome so that they can pair during meiosis, the Y chromosome is much shorter and
contains many fewer genes. In fact, when Nettie Stevens discovered that the X and Y
chromosomes were the determinants of sex, she differentiated them only by size. (Note that
in this case and in the description below, the terms X and Y chromosome were not used at the
time.) When a gene being examined is present on the X chromosome, but not on the Y
chromosome, it is said to be X-linked.

Eye color in Drosophila was one of the first X-linked traits to be identified. Thomas Hunt
Morgan mapped this trait to what became known as the X chromosome in 1910. Like
humans, Drosophila males have an XY chromosome pair, and females are XX. In flies, the
wild-type eye color is red (XW) and it is dominant to white eye color (Xw) (Figure .11).
Because of the location of the eye-color gene, reciprocal crosses do not produce the same
offspring ratios. Males are said to be hemizygous, because they have only one allele for any
X-linked characteristic. Hemizygosity makes the descriptions of dominance and
recessiveness irrelevant for XY males. Drosophila males lack a second allele copy on the Y
chromosome; that is, their genotype can only be XWY or XwY. In contrast, females have two
allele copies of this gene and can be XWXW, XWXw, or XwXw.
 EU Test Material

Figure .11 In Drosophila, several genes determine eye color. The genes for white and
vermilion eye colors are located on the X chromosome. Others are located on the autosomes.
Clockwise from top left are brown, cinnabar, sepia, vermilion, white, and red. Red eye color
is wild-type and is dominant to white eye color.

In an X-linked cross, the genotypes of F1 and F2 offspring depend on whether the recessive
trait was expressed by the male or the female in the P1 generation. With regard
to Drosophila eye color, when the P1 male expresses the white-eye phenotype and the female
is homozygous red-eyed, all members of the F1 generation exhibit red eyes (Figure 12). The
F1 females are heterozygous (XWXw), and the males are all XWY, having received their X
chromosome from the homozygous dominant P1 female and their Y chromosome from the
P1 male. A subsequent cross between the XWXw female and the XWY male would produce
only red-eyed females (with XWXW or XWXw genotypes) and both red- and white-eyed males
(with XWY or XwY genotypes). Now, consider a cross between a homozygous white-eyed
female and a male with red eyes. The F1 generation would exhibit only heterozygous red-
eyed females (XWXw) and only white-eyed males (XwY). Half of the F2 females would be red-
eyed (XWXw) and half would be white-eyed (XwXw). Similarly, half of the F2 males would be
red-eyed (XWY) and half would be white-eyed (XwY).

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure 12 Punnett square analysis is used to determine the ratio of offspring from a cross
between a red-eyed male fruit fly and a white-eyed female fruit fly.

What ratio of offspring would result from a cross between a white-eyed male and a female
that is heterozygous for red eye color?

Discoveries in fruit fly genetics can be applied to human genetics. When a female parent is
homozygous for a recessive X-linked trait, she will pass the trait on to 100 percent of her
offspring. Her male offspring are, therefore, destined to express the trait, as they will inherit
their father's Y chromosome. In humans, the alleles for certain conditions (some forms of
color blindness, hemophilia, and muscular dystrophy) are X-linked. Females who are
heterozygous for these diseases are said to be carriers and may not exhibit any phenotypic
effects. These females will pass the disease to half of their sons and will pass carrier status to
half of their daughters; therefore, recessive X-linked traits appear more frequently in males
than females.

In some groups of organisms with sex chromosomes, the sex with the non-homologous sex
chromosomes is the female rather than the male. This is the case for all birds. In this case,
sex-linked traits will be more likely to appear in the female, in which they are hemizygous.

Human Sex-linked Disorders

Sex-linkage studies in Morgan’s laboratory provided the fundamentals for understanding X-

linked recessive disorders in humans, which include red-green color blindness, and Types A
and B hemophilia. Because human males need to inherit only one recessive mutant X allele to
be affected, X-linked disorders are disproportionately observed in males. Females must
 EU Test Material

inherit recessive X-linked alleles from both of their parents in order to express the trait. When
they inherit one recessive X-linked mutant allele and one dominant X-linked wild-type allele,
they are carriers of the trait and are typically unaffected. Carrier females can manifest mild
forms of the trait due to the inactivation of the dominant allele located on one of the X
chromosomes. However, female carriers can contribute the trait to their male children,
resulting in the male exhibiting the trait, or they can contribute the recessive allele to their
female children, resulting in the children being carriers of the trait (Figure .13). Although
some Y-linked recessive disorders exist, typically they are associated with infertility in males
and are therefore not transmitted to subsequent generations.

Figure .13 The male offspring of a person who is a carrier of a recessive X-linked disorder
will have a 50 percent chance of being affected. A female will not be affected, but she will
have a 50 percent chance of being a carrier like the female parent.

LINK TO LEARNING

Link to Learning

Watch this video to learn more about sex-linked traits.

Lethality

A large proportion of genes in an individual’s genome are essential for survival.

Occasionally, a nonfunctional allele for an essential gene can arise by mutation and be
transmitted in a population as long as individuals with this allele also have a wild-type,
functional copy. The wild-type allele functions at a capacity sufficient to sustain life and is
therefore considered to be dominant over the nonfunctional allele. However, consider two
 EU Test Material

heterozygous parents that have a genotype of wild-type/nonfunctional mutant for a

hypothetical essential gene. In one quarter of their offspring, we would expect to observe
individuals that are homozygous recessive for the nonfunctional allele. Because the gene is
essential, these individuals might fail to develop past fertilization, die in utero, or die later in
life, depending on what life stage requires this gene. An inheritance pattern in which an allele
is only lethal in the homozygous form and in which the heterozygote may be normal or have
some altered nonlethal phenotype is referred to as recessive lethal.

For crosses between heterozygous individuals with a recessive lethal allele that causes death
before birth when homozygous, only wild-type homozygotes and heterozygotes would be
observed. The genotypic ratio would therefore be 2:1. In other instances, the recessive lethal
allele might also exhibit a dominant (but not lethal) phenotype in the heterozygote. For
instance, the recessive lethal Curly allele in Drosophila affects wing shape in the
heterozygote form but is lethal in the homozygote.

A single copy of the wild-type allele is not always sufficient for normal functioning or even
survival. The dominant lethal inheritance pattern is one in which an allele is lethal both in
the homozygote and the heterozygote; this allele can only be transmitted if the lethality
phenotype occurs after reproductive age. Individuals with mutations that result in dominant
lethal alleles fail to survive even in the heterozygote form. Dominant lethal alleles are very
rare because, as you might expect, the allele only lasts one generation and is not transmitted.
However, just as the recessive lethal allele might not immediately manifest the phenotype of
death, dominant lethal alleles also might not be expressed until adulthood. Once the
individual reaches reproductive age, the allele may be unknowingly passed on, resulting in a
delayed death in both generations. An example of this in humans is Huntington’s disease, in
which the nervous system gradually wastes away (Figure .14). People who are heterozygous
for the dominant Huntington allele (Hh) will inevitably develop the fatal disease. However,
the onset of Huntington’s disease may not occur until age 40, at which point the afflicted
persons may have already passed the allele to 50 percent of their offspring.
 EU Test Material

Figure .14 The neuron in the center of this micrograph (yellow) has nuclear inclusions
characteristic of Huntington’s disease (orange area in the center of the neuron). Huntington’s
disease occurs when an abnormal dominant allele for the Huntington gene is present. (credit:
Dr. Steven Finkbeiner, Gladstone Institute of Neurological Disease, The Taube-Koret Center
for Huntington's Disease Research, and the University of California San
Francisco/Wikimedia)

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain Mendel’s law of segregation and independent assortment in terms of genetics

and the events of meiosis
 Use the forked-line method and the probability rules to calculate the probability of
genotypes and phenotypes from multiple gene crosses
 Explain the effect of linkage and recombination on gamete genotypes
 Explain the phenotypic outcomes of epistatic effects between genes

Mendel generalized the results of his pea-plant experiments into four postulates, some of
which are sometimes called “laws,” that describe the basis of dominant and recessive
inheritance in diploid organisms. As you have learned, more complex extensions of
Mendelism exist that do not exhibit the same F2 phenotypic ratios (3:1). Nevertheless, these
laws summarize the basics of classical genetics.

Pairs of Unit Factors, or Genes

Mendel proposed first that paired unit factors of heredity were transmitted faithfully from
generation to generation by the dissociation and reassociation of paired factors during
gametogenesis and fertilization, respectively. After he crossed peas with contrasting traits and
found that the recessive trait resurfaced in the F2 generation, Mendel deduced that hereditary
factors must be inherited as discrete units. This finding contradicted the belief at that time
that parental traits were blended in the offspring.

Alleles Can Be Dominant or Recessive

Mendel’s law of dominance states that in a heterozygote, one trait will conceal the presence
of another trait for the same characteristic. Rather than both alleles contributing to a
phenotype, the dominant allele will be expressed exclusively. The recessive allele will remain
“latent” but will be transmitted to offspring by the same manner in which the dominant allele
is transmitted. The recessive trait will only be expressed by offspring that have two copies of
this allele (Figure 15), and these offspring will breed true when self-crossed.

Since Mendel’s experiments with pea plants, researchers have found that the law of
dominance does not always hold true. Instead, several different patterns of inheritance have
been found to exist.
 EU Test Material

Figure .15 The child in the photo expresses albinism, a recessive trait.

Equal Segregation of Alleles

Observing that true-breeding pea plants with contrasting traits gave rise to F1 generations that
all expressed the dominant trait and F2 generations that expressed the dominant and recessive
traits in a 3:1 ratio, Mendel proposed the law of segregation. This law states that paired unit
factors (genes) must segregate equally into gametes such that offspring have an equal
likelihood of inheriting either factor. For the F2 generation of a monohybrid cross, the
following three possible combinations of genotypes could result: homozygous dominant,
heterozygous, or homozygous recessive. Because heterozygotes could arise from two
different pathways (receiving one dominant and one recessive allele from either parent), and
because heterozygotes and homozygous dominant individuals are phenotypically identical,
the law supports Mendel’s observed 3:1 phenotypic ratio. The equal segregation of alleles is
the reason we can apply the Punnett square to accurately predict the offspring of parents with
known genotypes. The physical basis of Mendel’s law of segregation is the first division of
meiosis, in which the homologous chromosomes with their different versions of each gene
are segregated into daughter nuclei. The role of the meiotic segregation of chromosomes in
sexual reproduction was not understood by the scientific community during Mendel’s
lifetime.

Independent Assortment

Mendel’s law of independent assortment states that genes do not influence each other with
regard to the sorting of alleles into gametes, and every possible combination of alleles for
every gene is equally likely to occur. The independent assortment of genes can be illustrated
by the dihybrid cross, a cross between two true-breeding parents that express different traits
for two characteristics. Consider the characteristics of seed color and seed texture for two pea
plants, one that has green, wrinkled seeds (yyrr) and another that has yellow, round seeds
(YYRR). Because each parent is homozygous, the law of segregation indicates that the
gametes for the green/wrinkled plant all are yr, and the gametes for the yellow/round plant
are all YR. Therefore, the F1 generation of offspring all are YyRr (Figure .16).
 EU Test Material

VISUAL CONNECTION

Visual Connection

Figure .16 This dihybrid cross of pea plants involves the genes for seed color and texture.

In pea plants, round seed shape (R) is dominant to wrinkled seed shape (r) and yellow peas
(Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a
cross between RrYY and rrYy pea plants? How many squares do you need to do a Punnett
square analysis of this cross?

For the F2 generation, the law of segregation requires that each gamete receive either
an R allele or an r allele along with either a Y allele or a y allele. The law of independent
assortment states that a gamete into which an r allele sorted would be equally likely to
contain either a Y allele or a y allele. Thus, there are four equally likely gametes that can be
formed when the YyRr heterozygote is self-crossed, as follows: YR, Yr, yR, and yr. Arranging
these gametes along the top and left of a 4 × 4 Punnett square (Figure 16) gives us 16 equally
likely genotypic combinations. From these genotypes, we infer a phenotypic ratio of 9
round/yellow:3 round/green:3 wrinkled/yellow:1 wrinkled/green (Figure .16). These are the
offspring ratios we would expect, assuming we performed the crosses with a large enough
sample size.

Because of independent assortment and dominance, the 9:3:3:1 dihybrid phenotypic ratio can
be collapsed into two 3:1 ratios, characteristic of any monohybrid cross that follows a
dominant and recessive pattern. Ignoring seed color and considering only seed texture in the
above dihybrid cross, we would expect that three quarters of the F2 generation offspring
would be round, and one quarter would be wrinkled. Similarly, isolating only seed color, we
would assume that three quarters of the F2 offspring would be yellow and one quarter would
be green. The sorting of alleles for texture and color are independent events, so we can apply
the product rule. Therefore, the proportion of round and yellow F2 offspring is expected to be
(3/4) × (3/4) = 9/16, and the proportion of wrinkled and green offspring is expected to be
 EU Test Material

(1/4) × (1/4) = 1/16. These proportions are identical to those obtained using a Punnett square.
Round, green and wrinkled, yellow offspring can also be calculated using the product rule, as
each of these genotypes includes one dominant and one recessive phenotype. Therefore, the
proportion of each is calculated as (3/4) × (1/4) = 3/16.

The law of independent assortment also indicates that a cross between yellow, wrinkled
(YYrr) and green, round (yyRR) parents would yield the same F1 and F2 offspring as in
the YYRR x yyrr cross.

The physical basis for the law of independent assortment also lies in meiosis I, in which the
different homologous pairs line up in random orientations. Each gamete can contain any
combination of paternal and maternal chromosomes (and therefore the genes on them)
because the orientation of tetrads on the metaphase plane is random.

Forked-Line Method

When more than two genes are being considered, the Punnett-square method becomes
unwieldy. For instance, examining a cross involving four genes would require a 16 × 16 grid
containing 256 boxes. It would be extremely cumbersome to manually enter each genotype.
For more complex crosses, the forked-line and probability methods are preferred.

To prepare a forked-line diagram for a cross between F1 heterozygotes resulting from a cross
between AABBCC and aabbcc parents, we first create rows equal to the number of genes
being considered, and then segregate the alleles in each row on forked lines according to the
probabilities for individual monohybrid crosses (Figure .17). We then multiply the values
along each forked path to obtain the F2 offspring probabilities. Note that this process is a
diagrammatic version of the product rule. The values along each forked pathway can be
multiplied because each gene assorts independently. For a trihybrid cross, the F2 phenotypic
ratio is 27:9:9:9:3:3:3:1.

Figure .17 The forked-line method can be used to analyze a trihybrid cross. Here, the
probability for color in the F2 generation occupies the top row (3 yellow:1 green). The
probability for shape occupies the second row (3 round: 1 wrinkled), and the probability for
height occupies the third row (3 tall:1 dwarf). The probability for each possible combination
of traits is calculated by multiplying the probability for each individual trait. Thus, the
probability of F2 offspring having yellow, round, and tall traits is 3 × 3 × 3, or 27.
 EU Test Material

Probability Method

While the forked-line method is a diagrammatic approach to keeping track of probabilities in

a cross, the probability method gives the proportions of offspring expected to exhibit each
phenotype (or genotype) without the added visual assistance. Both methods make use of the
product rule and consider the alleles for each gene separately. Earlier, we examined the
phenotypic proportions for a trihybrid cross using the forked-line method; now we will use
the probability method to examine the genotypic proportions for a cross with even more
genes.

For a trihybrid cross, writing out the forked-line method is tedious, albeit not as tedious as
using the Punnett-square method. To fully demonstrate the power of the probability method,
however, we can consider specific genetic calculations. For instance, for a tetrahybrid cross
between individuals that are heterozygotes for all four genes, and in which all four genes are
sorting independently and in a dominant and recessive pattern, what proportion of the
offspring will be expected to be homozygous recessive for all four alleles? Rather than
writing out every possible genotype, we can use the probability method. We know that for
each gene, the fraction of homozygous recessive offspring will be 1/4. Therefore, multiplying
this fraction for each of the four genes, (1/4) × (1/4) × (1/4) × (1/4), we determine that 1/256
of the offspring will be quadruply homozygous recessive.

For the same tetrahybrid cross, what is the expected proportion of offspring that have the
dominant phenotype at all four loci? We can answer this question using phenotypic
proportions, but let’s do it the hard way—using genotypic proportions. The question asks for
the proportion of offspring that are 1) homozygous dominant at A or heterozygous at A, and
2) homozygous at B or heterozygous at B, and so on. Noting the “or” and “and” in each
circumstance makes clear where to apply the sum and product rules. The probability of a
homozygous dominant at A is 1/4 and the probability of a heterozygote at A is 1/2. The
probability of the homozygote or the heterozygote is 1/4 + 1/2 = 3/4 using the sum rule. The
same probability can be obtained in the same way for each of the other genes, so that the
probability of a dominant phenotype at A and B and C and D is, using the product rule, equal
to 3/4 × 3/4 × 3/4 × 3/4, or 81/256. If you are ever unsure about how to combine
probabilities, returning to the forked-line method should make it clear.

Rules for Multihybrid Fertilization

 EU Test Material

Predicting the genotypes and phenotypes of offspring from given crosses is the best way to
test your knowledge of Mendelian genetics. Given a multihybrid cross that obeys
independent assortment and follows a dominant and recessive pattern, several generalized
rules exist; you can use these rules to check your results as you work through genetics
calculations (Table .5). To apply these rules, first you must determine n, the number of
heterozygous gene pairs (the number of genes segregating two alleles each). For example, a
cross between AaBb and AaBb heterozygotes has an n of 2. In contrast, a cross
between AABb and AABb has an n of 1 because A is not heterozygous.

Number of
Heterozygous Gene
General Rule Pairs

Number of different F1 gametes 2n

Number of different F2 genotypes 3n

Given dominant and recessive inheritance, the

2n
number of different F2 phenotypes

General Rules for Multihybrid Crosses

Table .5

Linked Genes Violate the Law of Independent Assortment

Although all of Mendel’s pea characteristics behaved according to the law of independent
assortment, we now know that some allele combinations are not inherited independently of
each other. Genes that are located on separate non-homologous chromosomes will always
sort independently. However, each chromosome contains hundreds or thousands of genes,
organized linearly on chromosomes like beads on a string. The segregation of alleles into
gametes can be influenced by linkage, in which genes that are located physically close to
each other on the same chromosome are more likely to be inherited as a pair. However,
because of the process of recombination, or “crossover,” it is possible for two genes on the
same chromosome to behave independently, or as if they are not linked. To understand this,
let’s consider the biological basis of gene linkage and recombination.

Homologous chromosomes possess the same genes in the same linear order. The alleles may
differ on homologous chromosome pairs, but the genes to which they correspond do not. In
preparation for the first division of meiosis, homologous chromosomes replicate and synapse.
Like genes on the homologs align with each other. At this stage, segments of homologous
chromosomes exchange linear segments of genetic material (Figure .18). This process is
called recombination, or crossover, and it is a common genetic process. Because the genes
are aligned during recombination, the gene order is not altered. Instead, the result of
recombination is that maternal and paternal alleles are combined onto the same chromosome.
 EU Test Material

Across a given chromosome, several recombination events may occur, causing extensive
shuffling of alleles.

Figure .18 The process of crossover, or recombination, occurs when two homologous
chromosomes align during meiosis and exchange a segment of genetic material. Here, the
alleles for gene C were exchanged. The result is two recombinant and two non-recombinant
chromosomes.

When two genes are located in close proximity on the same chromosome, they are considered
linked, and their alleles tend to be transmitted through meiosis together. To exemplify this,
imagine a dihybrid cross involving flower color and plant height in which the genes are next
to each other on the chromosome. If one homologous chromosome has alleles for tall plants
and red flowers, and the other chromosome has genes for short plants and yellow flowers,
then when the gametes are formed, the tall and red alleles will go together into a gamete and
the short and yellow alleles will go into other gametes. These are called the parental
genotypes because they have been inherited intact from the parents of the individual
producing gametes. But unlike if the genes were on different chromosomes, there will be no
gametes with tall and yellow alleles and no gametes with short and red alleles. If you create
the Punnett square with these gametes, you will see that the classical Mendelian prediction of
a 9:3:3:1 outcome of a dihybrid cross would not apply. As the distance between two genes
increases, the probability of one or more crossovers between them increases, and the genes
behave more like they are on separate chromosomes. Geneticists have used the proportion of
recombinant gametes (the ones not like the parents) as a measure of how far apart genes are
on a chromosome. Using this information, they have constructed elaborate maps of genes on
chromosomes for well-studied organisms, including humans.
 EU Test Material

Mendel’s seminal publication makes no mention of linkage, and many researchers have
questioned whether he encountered linkage but chose not to publish those crosses out of
concern that they would invalidate his independent assortment postulate. The garden pea has
seven pairs of chromosomes, and some have suggested that his choice of seven characteristics
was not a coincidence. However, even if the genes he examined were not located on separate
chromosomes, it is possible that he simply did not observe linkage because of the extensive
shuffling effects of recombination.

SCIENTIFIC METHOD CONNECTION

Scientific Method Connection

Testing the Hypothesis of Independent AssortmentTo better appreciate the amount of

labor and ingenuity that went into Mendel’s experiments, proceed through one of Mendel’s
dihybrid crosses.

Question: What will be the offspring of a dihybrid cross?

Background: Consider that pea plants mature in one growing season, and you have access to
a large garden in which you can cultivate thousands of pea plants. There are several true-
breeding plants with the following pairs of traits: tall plants with inflated pods, and dwarf
plants with constricted pods. Before the plants have matured, you remove the pollen-
producing organs from the tall/inflated plants in your crosses to prevent self-fertilization.
Upon plant maturation, the plants are manually crossed by transferring pollen from the
dwarf/constricted plants to the stigmata of the tall/inflated plants.

Hypothesis: Both trait pairs will sort independently according to Mendelian laws. When the
true-breeding parents are crossed, all of the F1 offspring are tall and have inflated pods, which
indicates that the tall and inflated traits are dominant over the dwarf and constricted traits,
respectively. A self-cross of the F1 heterozygotes results in 2,000 F2 progeny.

Test the hypothesis: Because each trait pair sorts independently, the ratios of tall:dwarf and
inflated:constricted are each expected to be 3:1. The tall/dwarf trait pair is called T/t, and the
inflated/constricted trait pair is designated I/i. Each member of the F1 generation therefore has
a genotype of TtIi. Construct a grid analogous to Figure 12.16, in which you cross
two TtIi individuals. Each individual can donate four combinations of two traits: TI, Ti, tI,
or ti, meaning that there are 16 possibilities of offspring genotypes. Because
the T and I alleles are dominant, any individual having one or two of those alleles will
express the tall or inflated phenotypes, respectively, regardless if they also have a t or i allele.
Only individuals that are tt or ii will express the dwarf and constricted alleles, respectively.
As shown in Figure 12.19, you predict that you will observe the following offspring
proportions: tall/inflated:tall/constricted:dwarf/inflated:dwarf/constricted in a 9:3:3:1 ratio.
Notice from the grid that when considering the tall/dwarf and inflated/constricted trait pairs
in isolation, they are each inherited in 3:1 ratios.
 EU Test Material

Figure .19 This figure shows all possible combinations of offspring resulting from a dihybrid
cross of pea plants that are heterozygous for the tall/dwarf and inflated/constricted alleles.

Test the hypothesis: You cross the dwarf and tall plants and then self-cross the offspring.
For best results, this is repeated with hundreds or even thousands of pea plants. What special
precautions should be taken in the crosses and in growing the plants?

Analyze your data: You observe the following plant phenotypes in the F2 generation: 2706
tall/inflated, 930 tall/constricted, 888 dwarf/inflated, and 300 dwarf/constricted. Reduce these
findings to a ratio and determine if they are consistent with Mendelian laws.

Form a conclusion: Were the results close to the expected 9:3:3:1 phenotypic ratio? Do the
results support the prediction? What might be observed if far fewer plants were used, given
that alleles segregate randomly into gametes? Try to imagine growing that many pea plants,
and consider the potential for experimental error. For instance, what would happen if it was
extremely windy one day?

Epistasis

Mendel’s studies in pea plants implied that the sum of an individual’s phenotype was
controlled by genes (or as he called them, unit factors), such that every characteristic was
distinctly and completely controlled by a single gene. In fact, single observable
characteristics are almost always under the influence of multiple genes (each with two or
more alleles) acting in unison. For example, at least eight genes contribute to eye color in
humans.

In some cases, several genes can contribute to aspects of a common phenotype without their
gene products ever directly interacting. In the case of organ development, for instance, genes
may be expressed sequentially, with each gene adding to the complexity and specificity of the
organ. Genes may function in complementary or synergistic fashions, such that two or more
genes need to be expressed simultaneously to affect a phenotype. Genes may also oppose
each other, with one gene modifying the expression of another.
 EU Test Material

In epistasis, the interaction between genes is antagonistic, such that one gene masks or
interferes with the expression of another. “Epistasis” is a word composed of Greek roots that
mean “standing upon.” The alleles that are being masked or silenced are said to be hypostatic
to the epistatic alleles that are doing the masking. Often the biochemical basis of epistasis is a
gene pathway in which the expression of one gene is dependent on the function of a gene that
precedes or follows it in the pathway.

An example of epistasis is pigmentation in mice. The wild-type coat color, agouti (AA), is
dominant to solid-colored fur (aa). However, a separate gene (C) is necessary for pigment
production. A mouse with a recessive c allele at this locus is unable to produce pigment and
is albino regardless of the allele present at locus A (Figure .20). Therefore, the
genotypes AAcc, Aacc, and aacc all produce the same albino phenotype. A cross between
heterozygotes for both genes (AaCc x AaCc) would generate offspring with a phenotypic
ratio of 9 agouti:3 solid color:4 albino (Figure .20). In this case, the C gene is epistatic to
the A gene.

Figure .20 In mice, the mottled agouti coat color (A) is dominant to a solid coloration, such
as black or gray. A gene at a separate locus (C) is responsible for pigment production. The
recessive c allele does not produce pigment, and a mouse with the homozygous
recessive cc genotype is albino regardless of the allele present at the A locus. Thus,
the C gene is epistatic to the A gene.

Epistasis can also occur when a dominant allele masks expression at a separate gene. Fruit
color in summer squash is expressed in this way. Homozygous recessive expression of
the W gene (ww) coupled with homozygous dominant or heterozygous expression of
the Y gene (YY or Yy) generates yellow fruit, and the wwyy genotype produces green fruit.
 EU Test Material

However, if a dominant copy of the W gene is present in the homozygous or heterozygous

form, the summer squash will produce white fruit regardless of the Y alleles. A cross between
white heterozygotes for both genes (WwYy × WwYy) would produce offspring with a
phenotypic ratio of 12 white:3 yellow:1 green.

Finally, epistasis can be reciprocal such that either gene, when present in the dominant (or
recessive) form, expresses the same phenotype. In the shepherd’s purse plant (Capsella
bursa-pastoris), the characteristic of seed shape is controlled by two genes in a dominant
epistatic relationship. When the genes A and B are both homozygous recessive (aabb), the
seeds are ovoid. If the dominant allele for either of these genes is present, the result is
triangular seeds. That is, every possible genotype other than aabb results in triangular seeds,
and a cross between heterozygotes for both genes (AaBb x AaBb) would yield offspring with
a phenotypic ratio of 15 triangular:1 ovoid.

As you work through genetics problems, keep in mind that any single characteristic that
results in a phenotypic ratio that totals 16 is typical of a two-gene interaction. Recall the
phenotypic inheritance pattern for Mendel’s dihybrid cross, which considered two
noninteracting genes—9:3:3:1. Similarly, we would expect interacting gene pairs to also
exhibit ratios expressed as 16 parts. Note that we are assuming the interacting genes are not
linked; they are still assorting independently into gametes.

Chapter Outline
 EU Test Material

.1 Chromosomal Theory and Genetic Linkage

.2 Chromosomal Basis of Inherited Disorders

The gene is the physical unit of inheritance, and genes are arranged in a linear order on
chromosomes. Chromosome behavior and interaction during meiosis explain, at a cellular
level, inheritance patterns that we observe in populations. Genetic disorders involving
alterations in chromosome number or structure may have dramatic effects and can prevent a
fertilized egg from developing.

Learning Objectives

By the end of this section, you will be able to do the following:

 Discuss Sutton’s Chromosomal Theory of Inheritance

 Describe genetic linkage
 Explain the process of homologous recombination, or crossing over
 Describe chromosome creation
 Calculate the distances between three genes on a chromosome using a three-point test
cross

Long before scientists visualized chromosomes under a microscope, the father of modern
genetics, Gregor Mendel, began studying heredity in 1843. With improved microscopic
techniques during the late 1800s, cell biologists could stain and visualize subcellular
structures with dyes and observe their actions during cell division and meiosis. With each
mitotic division, chromosomes replicated, condensed from an amorphous (no constant shape)
nuclear mass into distinct X-shaped bodies (pairs of identical sister chromatids), and migrated
to separate cellular poles.

Chromosomal Theory of Inheritance

The speculation that chromosomes might be the key to understanding heredity led several
scientists to examine Mendel’s publications and reevaluate his model in terms of
chromosome behavior during mitosis and meiosis. In 1902, Theodor Boveri observed that
proper sea urchin embryonic development does not occur unless chromosomes are present.
That same year, Walter Sutton observed chromosome separation into daughter cells during
meiosis (Figure .2). Together, these observations led to the Chromosomal Theory of
Inheritance, which identified chromosomes as the genetic material responsible for
Mendelian inheritance.
 EU Test Material

Figure .2 (a) Walter Sutton and (b) Theodor Boveri developed the Chromosomal Theory of
Inheritance, which states that chromosomes carry the unit of heredity (genes). Eleanor
Carothers (c), was the first to provide physical evidence supporting the theory.

The Chromosomal Theory of Inheritance was consistent with Mendel’s laws, which the
following observations supported:

 During meiosis, homologous chromosome pairs migrate as discrete structures that are
independent of other chromosome pairs.
 Chromosome sorting from each homologous pair into pre-gametes appears to be
random.
 Each parent synthesizes gametes that contain only half their chromosomal
complement.
 Even though male and female gametes (sperm and egg) differ in size and morphology,
they have the same number of chromosomes, suggesting equal genetic contributions
from each parent.
 The gametic chromosomes combine during fertilization to produce offspring with the
same chromosome number as their parents.

Despite the lack of direct evidence that chromosomes carry traits, the compelling correlation
between chromosome behavior during meiosis and Mendel's abstract laws led scientists to
propose the Chromosomal Theory of Inheritance. Critics pointed out that individuals had far
more independently segregating traits than they had chromosomes. About ten years after the
theory was proposed, Eleanor Carothers was the first to discover physical evidence
supporting it; she observed independent chromosome assortment in grasshoppers. Then, after
several years of carrying out crosses with the fruit fly, Drosophila melanogaster, Thomas
Hunt Morgan provided additional experimental evidence to support the Chromosomal Theory
of Inheritance.

Genetic Linkage and Distances

Mendel’s work suggested that traits are inherited independently of each other. Morgan
identified a 1:1 correspondence between a segregating trait and the X chromosome,
suggesting that random chromosome segregation was the physical basis of Mendel’s model.
This also demonstrated that linked genes disrupt Mendel’s predicted outcomes. That each
 EU Test Material

chromosome can carry many linked genes explains how individuals can have many more
traits than they have chromosomes. However, researchers in Morgan’s laboratory suggested
that alleles positioned on the same chromosome were not always inherited together. During
meiosis, linked genes somehow became unlinked.

Homologous Recombination

In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact
with each other and may exchange segments—prior to the first meiotic division. He
suggested that alleles become unlinked and chromosomes physically exchange segments. As
chromosomes condensed and paired with their homologs, they appeared to interact at distinct
points. Janssen suggested that these points corresponded to regions in which chromosome
segments exchanged. We now know that the pairing and interaction between homologous
chromosomes, or synapsis, does more than simply organize the homologs for migration to
separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal
physical exchanges at their arms in homologous recombination, or more simply, “crossing
over.”

To better understand the type of experimental results that researchers were obtaining at this
time, consider a heterozygous individual that inherited dominant maternal alleles for two
genes on the same chromosome (such as A and B) and two recessive paternal alleles for those
same genes (such as a and b). If the genes are linked, one would expect this individual to
produce gametes that are either AB or ab with a 1:1 ratio. If the genes are unlinked, the
individual should produce AB, Ab, aB, and ab gametes with equal frequencies, according to
the Mendelian concept of independent assortment. Because they correspond to new allele
combinations, the genotypes Ab and aB are nonparental types that result from homologous
recombination during meiosis. Parental types are progeny that exhibit the same allelic
combination as their parents. Morgan and his colleagues, however, found that when they test
crossed such heterozygous individuals to a homozygous recessive parent (AaBb × aabb), both
parental and nonparental cases occurred. For example, 950 offspring might be recovered that
were either AaBb or aabb, but 50 offspring would also result that were either Aabb or aaBb.
These results suggested that linkage occurred most often, but a significant minority of
offspring were the products of recombination.

One of the experiments in Morgan's lab involving the crosses of flies for two traits, body
color (gray or black) and wing shape (normal and vestigial), demonstrated the recombination
events that lead to the development of nonparental phenotypes (Figure .3).

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .3 This figure shows unlinked and linked gene inheritance patterns. In (a), two genes
are located on different chromosomes so independent assortment occurs during meiosis. The
offspring have an equal chance of being the parental type (inheriting the same combination of
traits as the parents) or a nonparental type (inheriting a different combination of traits than
the parents). In (b), two genes are very close together on the same chromosome so that no
crossing over occurs between them. Therefore, the genes are always inherited together and all
 EU Test Material

the offspring are the parental type. In (c), two genes are far apart on the chromosome such
that crossing over occurs during every meiotic event. The recombination frequency will be
the same as if the genes were on separate chromosomes. (d) The actual recombination
frequency of fruit fly wing length and body color that Thomas Morgan observed in 1912 was
17 percent. A crossover frequency between 0 percent and 50 percent indicates that the genes
are on the same chromosome and crossover sometimes occurs.

In a test cross for two characteristics such as the one here, can the recombinant offspring's
predicted frequency be 60 percent? Why or why not?

Genetic Maps

Janssen did not have the technology to demonstrate crossing over so it remained an abstract
idea that scientists did not widely believe. Scientists thought chiasmata were a variation on
synapsis and could not understand how chromosomes could break and rejoin. Yet, the data
were clear that linkage did not always occur. Ultimately, it took a young undergraduate
student and an “all-nighter” to mathematically elucidate the linkage and recombination
problem.

In 1913, Alfred Sturtevant, a student in Morgan’s laboratory, gathered results from

researchers in the laboratory, and took them home one night to mull them over. By the next
morning, he had created the first “chromosome map,” a linear representation of gene order
and relative distance on a chromosome (Figure .4).

VISUAL CONNECTION

Visual Connection

Figure .4 This genetic map orders Drosophila genes on the basis of recombination
frequency.

Which of the following statements is true?

a. Recombination of the body color and red/cinnabar eye alleles will occur more
frequently than recombination of the alleles for wing length and aristae length.
b. Recombination of the body color and aristae length alleles will occur more frequently
than recombination of red/brown eye alleles and the aristae length alleles.
 EU Test Material

c. Recombination of the gray/black body color and long/short aristae alleles will not
occur.
d. Recombination of the red/brown eye and long/short aristae alleles will occur more
frequently than recombination of the alleles for wing length and body color.

As (Figure .4) shows, by using recombination frequency to predict genetic distance, we can
infer the relative gene order on chromosome 2. The values represent map distances in
centimorgans (cM), which correspond to recombination frequencies (in percent). Therefore,
the genes for body color and wing size were 65.5 − 48.5 = 17 cM apart, indicating that the
maternal and paternal alleles for these genes recombine in 17 percent of offspring, on
average.

To construct a chromosome map, Sturtevant assumed that genes were ordered serially on
threadlike chromosomes. He also assumed that the incidence of recombination between two
homologous chromosomes could occur with equal likelihood anywhere along the
chromosome's length. Operating under these assumptions, Sturtevant postulated that alleles
that were far apart on a chromosome were more likely to dissociate during meiosis simply
because there was a larger region over which recombination could occur. Conversely, alleles
that were close to each other on the chromosome were likely to be inherited together. The
average number of crossovers between two alleles—that is, their recombination
frequency—correlated with their genetic distance from each other, relative to the locations of
other genes on that chromosome. Considering the example cross
between AaBb and aabb above, we could calculate the recombination's frequency as 50/1000
= 0.05. That is, the likelihood of a crossover between genes A/a and B/b was 0.05, or 5
percent. Such a result would indicate that the genes were definitively linked, but that they
were far enough apart for crossovers to occasionally occur. Sturtevant divided his genetic
map into map units, or centimorgans (cM), in which a 0.01 recombination frequency
corresponds to 1 cM.

By representing alleles in a linear map, Sturtevant suggested that genes can range from
linking perfectly (recombination frequency = 0) to unlinking perfectly (recombination
frequency = 0.5) when genes are on different chromosomes or genes separate very far apart
on the same chromosome. Perfectly unlinked genes correspond to the frequencies Mendel
predicted to assort independently in a dihybrid cross. A 0.5 recombination frequency
indicates that 50 percent of offspring are recombinants and the other 50 percent are parental
types. That is, every type of allele combination is represented with equal frequency. This
representation allowed Sturtevant to additively calculate distances between several genes on
the same chromosome. However, as the genetic distances approached 0.50, his predictions
became less accurate because it was not clear whether the genes were very far apart on the
same or on different chromosomes.

In 1931, Barbara McClintock and Harriet Creighton demonstrated the crossover of

homologous chromosomes in corn plants. Weeks later, Curt Stern demonstrated
microscopically homologous recombination in Drosophila. Stern observed several X-linked
phenotypes that were associated with a structurally unusual and dissimilar X chromosome
pair in which one X was missing a small terminal segment, and the other X was fused to a
piece of the Y chromosome. By crossing flies, observing their offspring, and then visualizing
the offspring’s chromosomes, Stern demonstrated that every time the offspring allele
combination deviated from either of the parental combinations, there was a corresponding
exchange of an X chromosome segment. Using mutant flies with structurally distinct X
 EU Test Material

chromosomes was the key to observing the products of recombination because DNA
sequencing and other molecular tools were not yet available. We now know that homologous
chromosomes regularly exchange segments in meiosis by reciprocally breaking and rejoining
their DNA at precise locations. Aurora Ruiz-Herrera, for example, studies the occurrence of
genetic breakpoints at locations in the chromosomes known as fragile sites. By identifying
chromosomal fragile sites that are shared between humans and other primates, Ruiz-Herrera
has provided a deeper understanding of mammalian and specifically human evolution.

Mendel’s Mapped Traits

Homologous recombination is a common genetic process, yet Mendel never observed it. Had
he investigated both linked and unlinked genes, it would have been much more difficult for
him to create a unified model of his data on the basis of probabilistic calculations.
Researchers who have since mapped the seven traits that Mendel investigated onto a pea
plant genome's seven chromosomes have confirmed that all the genes he examined are either
on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some
have suggested that Mendel was enormously lucky to select only unlinked genes; whereas,
others question whether Mendel discarded any data suggesting linkage. In any case, Mendel
consistently observed independent assortment because he examined genes that were
effectively unlinked.

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe how a karyogram is created

 Explain how nondisjunction leads to disorders in chromosome number
 Compare disorders that aneuploidy causes
 Describe how errors in chromosome structure occur through inversions and
translocations

Inherited disorders can arise when chromosomes behave abnormally during meiosis. We can
divide chromosome disorders into two categories: abnormalities in chromosome number and
chromosomal structural rearrangements. Because even small chromosome segments can span
many genes, chromosomal disorders are characteristically dramatic and often fatal.

Chromosome Identification

Chromosome isolation and microscopic observation forms the basis of cytogenetics and is the
primary method by which clinicians detect chromosomal abnormalities in humans.
A karyotype is the number and appearance of chromosomes, and includes their length,
banding pattern, and centromere position. To obtain a view of an individual’s karyotype,
cytologists photograph the chromosomes and then cut and paste each chromosome into a
chart, or karyogram. Another name is an ideogram (Figure .5).
 EU Test Material

Figure .5 This karyotype is of a female human. Notice that homologous chromosomes are the
same size, and have the same centromere positions and banding patterns. A human male
would have an XY chromosome pair instead of the XX pair. (credit: Andreas Blozer et al)

In a given species, we can identify chromosomes by their number, size, centromere position,
and banding pattern. In a human karyotype, autosomes or “body chromosomes” (all of the
non–sex chromosomes) are generally organized in approximate order of size from largest
(chromosome 1) to smallest (chromosome 22). The X and Y chromosomes are not
autosomes. However, chromosome 21 is actually shorter than chromosome 22. Researchers
discovered this after naming Down syndrome as trisomy 21, reflecting how this disorder
results from possessing one extra chromosome 21 (three total). Not wanting to change the
name of this important disorder, scientists retained the numbering of chromosome 21 despite
describing it having the shortest set of chromosomes. We may designate the chromosome
“arms” projecting from either end of the centromere as short or long, depending on their
relative lengths. We abbreviate the short arm p (for “petite”); whereas, we abbreviate the long
arm q (because it follows “p” alphabetically). Numbers further subdivide and denote each
arm. Using this naming system, we can describe chromosome locations consistently in the
scientific literature.

CAREER CONNECTION

Career Connection

Geneticists Use Karyograms to Identify Chromosomal AberrationsAlthough we refer to

Mendel as the “father of modern genetics,” he performed his experiments with none of the
tools that the geneticists of today routinely employ. One such powerful cytological technique
is karyotyping, a method in which geneticists can identify traits characterized by
chromosomal abnormalities from a single cell. To observe an individual’s karyotype, a
geneticist first collects a person’s cells (like white blood cells) from a blood sample or other
tissue. In the laboratory, the geneticist stimulates the isolated cells to begin actively dividing.
The geneticist then applies the chemical colchicine to cells to arrest condensed chromosomes
in metaphase. The geneticist then induces swelling in the cells using a hypotonic solution so
the chromosomes spread apart. Finally, the geneticist preserves the sample in a fixative and
applies it to a slide.

The geneticist then stains chromosomes with one of several dyes to better visualize each
chromosome pair's distinct and reproducible banding patterns. Following staining, the
geneticist views the chromosomes using bright-field microscopy. A common stain choice is
the Giemsa stain. Giemsa staining results in approximately 400–800 bands (of tightly coiled
DNA and condensed proteins) arranged along all 23 chromosome pairs. An experienced
 EU Test Material

geneticist can identify each band. In addition to the banding patterns, geneticists further
identify chromosomes on the basis of size and centromere location. To obtain the classic
depiction of the karyotype in which homologous chromosome pairs align in numerical order
from longest to shortest, the geneticist obtains a digital image, identifies each chromosome,
and manually arranges the chromosomes into this pattern (Figure .5).

At its most basic, the karyogram may reveal genetic abnormalities in which an individual has
too many or too few chromosomes per cell. Examples of this are Down Syndrome, which one
identifies by a third copy of chromosome 21, and Turner Syndrome, which is characterized
by the presence of only one X chromosome in females instead of the normal two. Geneticists
can also identify large DNA deletions or insertions. For instance, geneticists can identify
Jacobsen Syndrome—which involves distinctive facial features as well as heart and bleeding
defects—by a deletion on chromosome 11. Finally, the karyotype can
pinpoint translocations, which occur when a segment of genetic material breaks from one
chromosome and reattaches to another chromosome or to a different part of the same
chromosome. Translocations are implicated in certain cancers, including chronic
myelogenous leukemia.

During Mendel’s lifetime, inheritance was an abstract concept that one could only infer by
performing crosses and observing the traits that offspring expressed. By observing a
karyogram, today’s geneticists can actually visualize an individual's chromosomal
composition to confirm or predict genetic abnormalities in offspring, even before birth.

Chromosome Number Disorders

Of all of the chromosomal disorders, chromosome number abnormalities are the most
obviously identifiable from a karyogram. Chromosome number disorders include duplicating
or losing entire chromosomes, as well as changes in the number of complete sets of
chromosomes. They are caused by nondisjunction, which occurs when homologous
chromosome pairs or sister chromatids fail to separate during meiosis. Misaligned or
incomplete synapsis, or a spindle apparatus dysfunction that facilitates chromosome
migration, can cause nondisjunction. The risk of nondisjunction occurring increases with the
parents' age.

Nondisjunction can occur during either meiosis I or II, with differing results (Figure .6). If
homologous chromosomes fail to separate during meiosis I, the result is two gametes that
lack that particular chromosome and two gametes with two chromosome copies. If sister
chromatids fail to separate during meiosis II, the result is one gamete that lacks that
chromosome, two normal gametes with one chromosome copy, and one gamete with two
chromosome copies.

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .6 Nondisjunction occurs when homologous chromosomes or sister chromatids fail to

separate during meiosis, resulting in an abnormal chromosome number. Nondisjunction may
occur during meiosis I or meiosis II. Credit: Rao, A. and Tag, A. Department of Biology,
Texas A&M University.

Which of the following statements about nondisjunction is true?

a. Nondisjunction only results in gametes with n+1 or n–1 chromosomes.

b. Nondisjunction occurring during meiosis II results in 50 percent normal gametes.
c. Nondisjunction during meiosis I results in 50 percent normal gametes.
d. Nondisjunction always results in four different kinds of gametes.

Aneuploidy

Scientists call an individual with the appropriate number of chromosomes for their
species euploid. In humans, euploidy corresponds to 22 pairs of autosomes and one pair of
sex chromosomes. An individual with an error in chromosome number is described
as aneuploid, a term that includes monosomy (losing one chromosome) or trisomy (gaining
an extraneous chromosome). Monosomic human zygotes missing any one copy of an
autosome invariably fail to develop to birth because they lack essential genes. This
underscores the importance of “gene dosage” in humans. Most autosomal trisomies also fail
to develop to birth; however, duplications of some smaller chromosomes (13, 15, 18, 21, or
22) can result in offspring that survive for several weeks to many years. Trisomic individuals
suffer from a different type of genetic imbalance: an excess in gene dose. Individuals with an
extra chromosome may synthesize an abundance of the gene products, which that
chromosome encodes. This extra dose (150 percent) of specific genes can lead to a number of
functional challenges and often precludes development. The most common trisomy among
 EU Test Material

viable births is that of chromosome 21, which corresponds to Down Syndrome. Short stature
and stunted digits, facial distinctions that include a broad skull and large tongue, and
significant developmental delays characterize individuals with this inherited disorder. We can
correlate the incidence of Down syndrome with maternal age. Older people are more likely to
become pregnant with fetuses carrying the trisomy 21 genotype (Figure .7).

Figure .7 The incidence of having a fetus with trisomy 21 increases dramatically with
maternal age.

Polyploidy

We call an individual with more than the correct number of chromosome sets (two for diploid
species) polyploid. For instance, fertilizing an abnormal diploid egg with a normal haploid
sperm would yield a triploid zygote. Polyploid animals are extremely rare, with only a few
examples among the flatworms, crustaceans, amphibians, fish, and lizards. Polyploid animals
are sterile because meiosis cannot proceed normally and instead produces mostly aneuploid
daughter cells that cannot yield viable zygotes. Rarely, polyploid animals can reproduce
asexually by haplodiploidy, in which an unfertilized egg divides mitotically to produce
offspring. In contrast, polyploidy is very common in the plant kingdom, and polyploid plants
tend to be larger and more robust than euploids of their species (Figure .8).
 EU Test Material

Figure .8 As with many polyploid plants, this triploid orange daylily (Hemerocallis fulva) is
particularly large and robust, and grows flowers with triple the number of petals of its diploid
counterparts. (credit: Steve Karg)

Sex Chromosome Nondisjunction in Humans

Humans display dramatic deleterious effects with autosomal trisomies and monosomies.
Therefore, it may seem counterintuitive that human females and males can function normally,
despite carrying different numbers of the X chromosome. Rather than a gain or loss of
autosomes, variations in the number of sex chromosomes occur with relatively mild effects.
In part, this happens because of the molecular process X inactivation. Early in development,
when female mammalian embryos consist of just a few thousand cells (relative to trillions in
the newborn), one X chromosome in each cell inactivates by tightly condensing into a
quiescent (dormant) structure, or a Barr body. The chance that an X chromosome (maternally
or paternally derived) inactivates in each cell is random, but once this occurs, all cells derived
from that one will have the same inactive X chromosome or Barr body. By this process,
females compensate for their double genetic dose of X chromosome. In so-called
“tortoiseshell” cats, we observe embryonic X inactivation as color variegation (Figure .9).
Females that are heterozygous for an X-linked coat color gene will express one of two
different coat colors over different regions of their body, corresponding to whichever X
chromosome inactivates in that region's embryonic cell progenitor.

Figure .9 In cats, the gene for coat color is located on the X chromosome. In female cats'
embryonic development, one of the two X chromosomes randomly inactivates in each cell,
resulting in a tortoiseshell pattern if the cat has two different alleles for coat color. Male cats,
having only one X chromosome, never exhibit a tortoiseshell coat color. (credit: Michael
Bodega)
 EU Test Material

An individual carrying an abnormal number of X chromosomes will inactivate all but one X
chromosome in each of her cells. However, even inactivated X chromosomes continue to
express a few genes, and X chromosomes must reactivate for the proper maturation of female
ovaries. As a result, X-chromosomal abnormalities typically occur with mild intellectual and
physical disorders or disabilities, as well as sterility. If the X chromosome is absent
altogether, the individual will not develop in utero.

Scientists have identified and characterized several errors in sex chromosome number.
Individuals with three X chromosomes, triplo-X, are phenotypically female but express
developmental delays and reduced fertility. The XXY genotype, corresponding to one type of
Klinefelter syndrome, corresponds to phenotypically male individuals with small testes,
enlarged breasts, and reduced body hair. More complex types of Klinefelter syndrome exist
in which the individual has as many as five X chromosomes. In all types, every X
chromosome except one undergoes inactivation to compensate for the excess genetic dosage.
We see this as several Barr bodies in each cell nucleus. Turner syndrome, characterized as an
X0 genotype (i.e., only a single sex chromosome), corresponds to a phenotypically female
individual with short stature, webbed skin in the neck region, hearing and cardiac
impairments, and sterility.

Duplications and Deletions

In addition to losing or gaining an entire chromosome, a chromosomal segment may

duplicate or lose itself. Duplications and deletions often produce offspring that survive but
exhibit abnormalities. Duplicated chromosomal segments may fuse to existing chromosomes
or may be free in the nucleus. Cri-du-chat (from the French for “cry of the cat”) is a
syndrome that occurs with nervous system abnormalities and identifiable physical features
that result from a deletion of most 5p (the small arm of chromosome 5) (Figure .10). Infants
with this genotype emit a characteristic high-pitched cry on which the disorder’s name is
based.

Figure .10 This figure shows an individual with cri-du-chat syndrome at two, four, nine, and
12 years of age. (credit: Paola Cerruti Mainardi)

Chromosomal Structural Rearrangements

Cytologists have characterized numerous structural rearrangements in chromosomes, but

chromosome inversions and translocations are the most common. We can identify both
 EU Test Material

during meiosis by the adaptive pairing of rearranged chromosomes with their former
homologs to maintain appropriate gene alignment. If the genes on two homologs are not
oriented correctly, a recombination event could result in losing genes from one chromosome
and gaining genes on the other. This would produce aneuploid gametes.

Chromosome Inversions

A chromosome inversion is the detachment, 180° rotation, and reinsertion of part of a

chromosome. Inversions may occur in nature as a result of mechanical shear, or from
transposable elements' action (special DNA sequences capable of facilitating rearranging
chromosome segments with the help of enzymes that cut and paste DNA sequences). Unless
they disrupt a gene sequence, inversions only change gene orientation and are likely to have
more mild effects than aneuploid errors. However, altered gene orientation can result in
functional changes because regulators of gene expression could move out of position with
respect to their targets, causing aberrant levels of gene products.

An inversion can be pericentric and include the centromere, or paracentric and occur
outside the centromere (Figure 11). A pericentric inversion that is asymmetric about the
centromere can change the chromosome arms' relative lengths, making these inversions
easily identifiable.

Figure .11 Pericentric inversions include the centromere, and paracentric inversions do not.
A pericentric inversion can change the chromosome arms' relative lengths. A paracentric
inversion cannot.

When one homologous chromosome undergoes an inversion but the other does not, the
individual is an inversion heterozygote. To maintain point-for-point synapsis during meiosis,
one homolog must form a loop, and the other homolog must mold around it. Although this
topology can ensure that the genes correctly align, it also forces the homologs to stretch and
can occur with imprecise synapsis regions (Figure 12).
 EU Test Material

Figure .12 When one chromosome undergoes an inversion but the other does not, one
chromosome must form an inverted loop to retain point-for-point interaction during synapsis.
This inversion pairing is essential to maintaining gene alignment during meiosis and to allow
for recombination.

EVOLUTION CONNECTION

Evolution Connection

The Chromosome 18 InversionNot all chromosomes' structural rearrangements produce

nonviable, impaired, or infertile individuals. In rare instances, such a change can result in
new species evolving. In fact, a pericentric inversion in chromosome 18 appears to have
contributed to human evolution. This inversion is not present in our closest genetic relatives,
the chimpanzees. Humans and chimpanzees differ cytogenetically by pericentric inversions
on several chromosomes and by the fusion of two separate chromosomes in chimpanzees that
correspond to chromosome two in humans.

Scientists believe the pericentric chromosome 18 inversion occurred in early humans

following their divergence from a common ancestor with chimpanzees approximately five
million years ago. Researchers characterizing this inversion have suggested that
approximately 19,000 nucleotide bases were duplicated on 18p, and the duplicated region
inverted and reinserted on chromosome 18 of an ancestral human.

A comparison of human and chimpanzee genes in the region of this inversion indicates that
two genes—ROCK1 and USP14—that are adjacent on chimpanzee chromosome 17 (which
corresponds to human chromosome 18) are more distantly positioned on human chromosome
18. This suggests that one of the inversion breakpoints occurred between these two genes.
Interestingly, humans and chimpanzees express USP14 at distinct levels in specific cell types,
including cortical cells and fibroblasts. Perhaps the chromosome 18 inversion in an ancestral
human repositioned specific genes and reset their expression levels in a useful way. Because
both ROCK1 and USP14 encode cellular enzymes, a change in their expression could alter
cellular function. We do not know how this inversion contributed to hominid evolution, but it
appears to be a significant factor in the divergence of humans from other primates.1
 EU Test Material

Translocations

A translocation occurs when a chromosome segment dissociates and reattaches to a

different, nonhomologous chromosome. Translocations can be benign or have devastating
effects depending on how the positions of genes are altered with respect to regulatory
sequences. Notably, specific translocations have occurred with several cancers and with
schizophrenia. Reciprocal translocations result from exchanging chromosome segments
between two nonhomologous chromosomes such that there is no genetic information gain or
loss (Figure 13).

Figure 13 A reciprocal translocation occurs when a DNA segment transfers from one
chromosome to another, nonhomologous chromosome. (credit: modification of work by
National Human Genome Research/USA)

Chapter Outline

.1 Historical Basis of Modern Understanding

 EU Test Material

.2 DNA Structure and Sequencing

.3 Basics of DNA Replication

.4 DNA Replication in Prokaryotes

.5 DNA Replication in Eukaryotes

.6 DNA Repair

The three letters “DNA” have now become synonymous with crime solving and genetic
testing. DNA can be retrieved from hair, blood, or saliva. Each person’s DNA is unique, and
it is possible to detect differences between individuals within a species on the basis of these
unique features.

DNA analysis has many practical applications beyond forensics. In humans, DNA testing is
applied to numerous uses: determining paternity, tracing genealogy, identifying pathogens,
archeological research, tracing disease outbreaks, and studying human migration patterns. In
the medical field, DNA is used in diagnostics, new vaccine development, and cancer therapy.
It is now possible to determine predisposition to diseases by looking at genes.

Each human cell has 23 pairs of chromosomes: one set of chromosomes is inherited from the
female parent and the other set is inherited from the male parent. There is also a
mitochondrial genome, inherited exclusively from the female parent, which can be involved
in inherited genetic disorders. On each chromosome, there are thousands of genes that are
responsible for determining the genotype and phenotype of the individual. A gene is defined
as a sequence of DNA that codes for a functional product. The human haploid genome
contains 3 billion base pairs and has between 20,000 and 25,000 functional genes.

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain transformation of DNA

 Describe the key experiments that helped identify that DNA is the genetic material
 State and explain Chargaff’s rules

Our current understanding of DNA began with the discovery of nucleic acids followed by the
development of the double-helix model. In the 1860s, Friedrich Miescher (Figure 2), a
physician by profession, isolated phosphate-rich chemicals from white blood cells
(leukocytes). He named these chemicals (which would eventually be known as
DNA) nuclein because they were isolated from the nuclei of the cells.
 EU Test Material

Figure 2 Friedrich Miescher (1844–1895) discovered nucleic acids.

A half century later, in 1928, British bacteriologist Frederick Griffith reported the first
demonstration of bacterial transformation—a process in which external DNA is taken up by
a cell, thereby changing its morphology and physiology. Griffith conducted his experiments
with Streptococcus pneumoniae, a bacterium that causes pneumonia. Griffith worked with
two strains of this bacterium called rough (R) and smooth (S). (The two cell types were called
“rough” and “smooth” after the appearance of their colonies grown on a nutrient agar plate.)

The R strain is non-pathogenic (does not cause disease). The S strain is pathogenic (disease-
causing), and has a capsule outside its cell wall. The capsule allows the cell to escape the
immune responses of the host mouse.

When Griffith injected the living S strain into mice, they died from pneumonia. In contrast,
when Griffith injected the live R strain into mice, they survived. In another experiment, when
he injected mice with the heat-killed S strain, they also survived. This experiment showed
that the capsule alone was not the cause of death. In a third set of experiments, a mixture of
live R strain and heat-killed S strain were injected into mice, and—to his surprise—the mice
died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was
recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith
concluded that something had passed from the heat-killed S strain into the live R strain and
transformed it into the pathogenic S strain. He called this the transforming principle (Figure
3). These experiments are now known as Griffith's transformation experiments.
 EU Test Material

Figure .3 Two strains of S. pneumoniae were used in Griffith’s transformation experiments.

The R strain is non-pathogenic, whereas the S strain is pathogenic and causes death. When
Griffith injected a mouse with the heat-killed S strain and a live R strain, the mouse died. The
S strain was recovered from the dead mouse. Griffith concluded that something had passed
from the heat-killed S strain to the R strain, transforming the R strain into the S strain in the
process. Credit: Rao, A., Ryan, K. and Tag, A. Department of Biology, Texas A&M
University.

Scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in
exploring this transforming principle further. They isolated the S strain from the dead mice
and isolated the proteins and nucleic acids (RNA and DNA) as these were possible candidates
for the molecule of heredity. They used enzymes that specifically degraded each component
and then used each mixture separately to transform the R strain. They found that when DNA
was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all
of the other combinations were able to transform the bacteria. This led them to conclude that
DNA was the transforming principle.

CAREER CONNECTION

Career Connection

Forensic ScientistForensic Scientists used DNA analysis evidence for the first time to solve
an immigration case. The story started with a teenage boy returning to London from Ghana to
 EU Test Material

be with his mother. Immigration authorities at the airport were suspicious of him, thinking
that he was traveling on a forged passport. After much persuasion, he was allowed to go live
with his mother, but the immigration authorities did not drop the case against him. All types
of evidence, including photographs, were provided to the authorities, but deportation
proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester
University in the United Kingdom had invented a technique known as DNA fingerprinting.
The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from
the mother and three of her children, as well as an unrelated mother, and compared the
samples with the boy’s DNA. Because the biological father was not in the picture, DNA from
the three children was compared with the boy’s DNA. He found a match in the boy’s DNA
for both the mother and his three siblings. He concluded that the boy was indeed the mother’s
son.

Forensic scientists analyze many items, including documents, handwriting, firearms, and
biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and
compare it with a database of DNA profiles of known criminals. Analysis includes DNA
isolation, sequencing, and sequence analysis. Forensic scientists are expected to appear at
court hearings to present their findings. They are usually employed in crime labs of city and
state government agencies. Geneticists experimenting with DNA techniques also work for
scientific and research organizations, pharmaceutical industries, and college and university
labs. Students wishing to pursue a career as a forensic scientist should have at least a
bachelor's degree in chemistry, biology, or physics, and preferably some experience working
in a laboratory.

Although the experiments of Avery, McCarty and McLeod had demonstrated that DNA was
the informational component transferred during transformation, DNA was still considered to
be too simple a molecule to carry biological information. Proteins, with their 20 different
amino acids, were regarded as more likely candidates. The decisive experiment, conducted by
Martha Chase and Alfred Hershey in 1952, provided confirmatory evidence that DNA was
indeed the genetic material and not proteins. Chase and Hershey were studying
a bacteriophage—a virus that infects bacteria. Viruses typically have a simple structure: a
protein coat, called the capsid, and a nucleic acid core that contains the genetic material
(either DNA or RNA). The bacteriophage infects the host bacterial cell by attaching to its
surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple
copies of itself using the host machinery, and eventually the host cell bursts, releasing a large
number of bacteriophages. Hershey and Chase selected radioactive elements that would
specifically distinguish the protein from the DNA in infected cells. They labeled one batch of
phage with radioactive sulfur, 35S, to label the protein coat. Another batch of phage were
labeled with radioactive phosphorus, 32P. Because phosphorous is found in DNA, but not
protein, the DNA and not the protein would be tagged with radioactive phosphorus. Likewise,
sulfur is absent from DNA, but present in several amino acids such as methionine and
cysteine.

Each batch of phage was allowed to infect the cells separately. After infection, the phage
bacterial suspension was put in a blender, which caused the phage coat to detach from the
host cell. Cells exposed long enough for infection to occur were then examined to see which
of the two radioactive molecules had entered the cell. The phage and bacterial suspension
was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet,
whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage
labeled with 35S, the supernatant contained the radioactively labeled phage, whereas no
 EU Test Material

radioactivity was detected in the pellet. In the tube that contained the phage labeled with 32P,
the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no
radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the
phage DNA that was injected into the cell and carried information to produce more phage
particles, thus providing evidence that DNA was the genetic material and not proteins (Figure
.4).

Figure .4 In Hershey and Chase's experiments, bacteria were infected with phage
radiolabeled with either 35S, which labels protein, or 32P, which labels DNA. Only 32P entered
the bacterial cells, indicating that DNA is the genetic material.

Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in
different species and found that the amounts of adenine, thymine, guanine, and cytosine were
not found in equal quantities, and that relative concentrations of the four nucleotide bases
varied from species to species, but not within tissues of the same individual or between
individuals of the same species. He also discovered something unexpected: That the amount
of adenine equaled the amount of thymine, and the amount of cytosine equaled the amount of
guanine (that is, A = T and G = C). Different species had equal amounts of purines (A+G)
and pyrimidines (T + C), but different ratios of A+T to G+C. These observations became
known as Chargaff’s rules. Chargaff's findings proved immensely useful when Watson and
Crick were getting ready to propose their DNA double helix model! You can see after
reading the past few pages how science builds upon previous discoveries, sometimes in a
slow and laborious process.
 EU Test Material

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe the structure of DNA

 Explain the Sanger method of DNA sequencing
 Discuss the similarities and differences between eukaryotic and prokaryotic DNA

The building blocks of DNA are nucleotides. The important components of the nucleotide are
a nitrogenous (nitrogen-bearing) base, a 5-carbon sugar (pentose), and a phosphate group
(Figure .5). The nucleotide is named depending on the nitrogenous base. The nitrogenous
base can be a purine such as adenine (A) and guanine (G), or a pyrimidine such as cytosine
(C) and thymine (T).

VISUAL CONNECTION

Visual Connection

Figure .5 The purines have a double ring structure with a six-membered ring fused to a five-
membered ring. Pyrimidines are smaller in size; they have a single six-membered ring
structure.

The images above illustrate the five bases of DNA and RNA. Examine the images and
explain why these are called “nitrogenous bases.” How are the purines different from the
pyrimidines? How is one purine or pyrimidine different from another, e.g., adenine from
guanine? How is a nucleoside different from a nucleotide?

The purines have a double ring structure with a six-membered ring fused to a five-membered
ring. Pyrimidines are smaller in size; they have a single six-membered ring structure.

The sugar is deoxyribose in DNA and ribose in RNA. The carbon atoms of the five-carbon
sugar are numbered 1', 2', 3', 4', and 5' (1' is read as “one prime”). The phosphate, which
makes DNA and RNA acidic, is connected to the 5' carbon of the sugar by the formation of
an ester linkage between phosphoric acid and the 5'-OH group (an ester is an acid + an
alcohol). In DNA nucleotides, the 3' carbon of the sugar deoxyribose is attached to a
hydroxyl (OH) group. In RNA nucleotides, the 2' carbon of the sugar ribose also contains a
hydroxyl group. The base is attached to the 1'carbon of the sugar.
 EU Test Material

The nucleotides combine with each other to produce phosphodiester bonds. The phosphate
residue attached to the 5' carbon of the sugar of one nucleotide forms a second ester linkage
with the hydroxyl group of the 3' carbon of the sugar of the next nucleotide, thereby forming
a 5'-3' phosphodiester bond. In a polynucleotide, one end of the chain has a free 5' phosphate,
and the other end has a free 3'-OH. These are called the 5' and 3' ends of the chain.

In the 1950s, Francis Crick and James Watson worked together to determine the structure of
DNA at the University of Cambridge, England. Other scientists like Linus Pauling and
Maurice Wilkins were also actively exploring this field. Pauling previously had discovered
the secondary structure of proteins using X-ray crystallography. In Wilkins’ lab, researcher
Rosalind Franklin was using X-ray diffraction methods to understand the structure of DNA.
Watson and Crick were able to piece together the puzzle of the DNA molecule on the basis of
Franklin's data because Crick had also studied X-ray diffraction (Figure .6). In 1962, James
Watson, Francis Crick, and Maurice Wilkins were awarded the Nobel Prize in Medicine.
Unfortunately, by then Franklin had died, and Nobel prizes are not awarded posthumously.

Figure .6 The X-ray diffraction pattern of DNA, which helped to elucidate its double-helix
structure.

Watson and Crick proposed that DNA is made up of two strands that are twisted around each
other to form a right-handed helix. Base pairing takes place between a purine and pyrimidine
on opposite strands, so that A pairs with T, and G pairs with C (suggested by Chargaff's
Rules). Thus, adenine and thymine are complementary base pairs, and cytosine and guanine
are also complementary base pairs. The base pairs are stabilized by hydrogen bonds: adenine
and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen
bonds. The two strands are anti-parallel in nature; that is, the 3' end of one strand faces the 5'
end of the other strand. The sugar and phosphate of the nucleotides form the backbone of the
structure, whereas the nitrogenous bases are stacked inside, like the rungs of a ladder. Each
base pair is separated from the next base pair by a distance of 0.34 nm, and each turn of the
helix measures 3.4 nm. Therefore, 10 base pairs are present per turn of the helix. The
 EU Test Material

diameter of the DNA double-helix is 2 nm, and it is uniform throughout. Only the pairing
between a purine and pyrimidine and the antiparallel orientation of the two DNA strands can
explain the uniform diameter. The twisting of the two strands around each other results in the
formation of uniformly spaced major and minor grooves (Figure .7).

Figure .7 DNA has (a) a double helix structure and (b) phosphodiester bonds; the dotted lines
between Thymine and Adenine and Guanine and Cytosine represent hydrogen bonds. The (c)
major and minor grooves are binding sites for DNA binding proteins during processes such as
transcription (the copying of RNA from DNA) and replication.

DNA Sequencing Techniques

Until the 1990s, the sequencing of DNA (reading the sequence of DNA) was a relatively
expensive and long process. Using radiolabeled nucleotides also compounded the problem
through safety concerns. With currently available technology and automated machines, the
process is cheaper, safer, and can be completed in a matter of hours. Fred Sanger developed
the sequencing method used for the human genome sequencing project, which is widely used
today (Figure .8).

The sequencing method is known as the dideoxy chain termination method. The method is
based on the use of chain terminators, the dideoxynucleotides (ddNTPs). The ddNTPSs differ
from the deoxynucleotides by the lack of a free 3' OH group on the five-carbon sugar. If a
ddNTP is added to a growing DNA strand, the chain cannot be extended any further because
the free 3' OH group needed to add another nucleotide is not available. By using a
predetermined ratio of deoxynucleotides to dideoxynucleotides, it is possible to generate
DNA fragments of different sizes.

Figure .8 In Frederick Sanger's dideoxy chain termination method, dye-labeled

dideoxynucleotides are used to generate DNA fragments that terminate at different points.
 EU Test Material

The DNA is separated by capillary electrophoresis (not defined) on the basis of size, and
from the order of fragments formed, the DNA sequence can be read. The DNA sequence
readout is shown on an electropherogram (not defined) that is generated by a laser scanner.

The DNA sample to be sequenced is denatured (separated into two strands by heating it to
high temperatures). The DNA is divided into four tubes in which a primer, DNA polymerase,
and all four nucleoside triphosphates (A, T, G, and C) are added. In addition, limited
quantities of one of the four dideoxynucleoside triphosphates (ddCTP, ddATP, ddGTP, and
ddTTP) are added to each tube respectively. The tubes are labeled as A, T, G, and C
according to the ddNTP added. For detection purposes, each of the four dideoxynucleotides
carries a different fluorescent label. Chain elongation continues until a fluorescent dideoxy
nucleotide is incorporated, after which no further elongation takes place. After the reaction is
over, electrophoresis is performed. Even a difference in length of a single base can be
detected. The sequence is read from a laser scanner that detects the fluorescent marker of
each fragment. For his work on DNA sequencing, Sanger received a Nobel Prize in
Chemistry in 1980.

Gel electrophoresis is a technique used to separate DNA fragments of different sizes.

Usually the gel is made of a chemical called agarose (a polysaccharide polymer extracted
from seaweed that is high in galactose residues). Agarose powder is added to a buffer and
heated. After cooling, the gel solution is poured into a casting tray. Once the gel has
solidified, the DNA is loaded on the gel and electric current is applied. The DNA has a net
negative charge and moves from the negative electrode toward the positive electrode. The
electric current is applied for sufficient time to let the DNA separate according to size; the
smallest fragments will be farthest from the well (where the DNA was loaded), and the
heavier molecular weight fragments will be closest to the well. Once the DNA is separated,
the gel is stained with a DNA-specific dye for viewing it (Figure .9).

Figure .9 DNA can be separated on the basis of size using gel electrophoresis. (credit: James
Jacob, Tompkins Cortland Community College)

EVOLUTION CONNECTION

Evolution Connection

Neanderthal Genome: How Are We Related?The first draft sequence of the Neanderthal
genome was recently published by Richard E. Green et al. in 2010.1 Neanderthals are the
 EU Test Material

closest ancestors of present-day humans. They were known to have lived in Europe and
Western Asia (and now, perhaps, in Northern Africa) before they disappeared from fossil
records approximately 30,000 years ago. Green’s team studied almost 40,000-year-old fossil
remains that were selected from sites across the world. Extremely sophisticated means of
sample preparation and DNA sequencing were employed because of the fragile nature of the
bones and heavy microbial contamination. In their study, the scientists were able to sequence
some four billion base pairs. The Neanderthal sequence was compared with that of present-
day humans from across the world. After comparing the sequences, the researchers found that
the Neanderthal genome had 2 to 3 percent greater similarity to people living outside Africa
than to people in Africa. While current theories have suggested that all present-day humans
can be traced to a small ancestral population in Africa, the data from the Neanderthal genome
suggest some interbreeding between Neanderthals and early modern humans.

Green and his colleagues also discovered DNA segments among people in Europe and Asia
that are more similar to Neanderthal sequences than to other contemporary human sequences.
Another interesting observation was that Neanderthals are as closely related to people from
Papua New Guinea as to those from China or France. This is surprising because Neanderthal
fossil remains have been located only in Europe and West Asia. Most likely, genetic
exchange took place between Neanderthals and modern humans as modern humans emerged
out of Africa, before the divergence of Europeans, East Asians, and Papua New Guineans.

Several genes seem to have undergone changes from Neanderthals during the evolution of
present-day humans. These genes are involved in cranial structure, metabolism, skin
morphology, and cognitive development. One of the genes that is of particular interest
is RUNX2, which is different in modern day humans and Neanderthals. This gene is
responsible for the prominent frontal bone, bell-shaped rib cage, and dental differences seen
in Neanderthals. It is speculated that an evolutionary change in RUNX2 was important in the
origin of modern-day humans, and this affected the cranium and the upper body.

DNA Packaging in Cells

Prokaryotes are much simpler than eukaryotes in many of their features (Figure 10). Most
prokaryotes contain a single, circular chromosome that is found in an area of the cytoplasm
called the nucleoid region.

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .10 A eukaryote contains a well-defined nucleus, whereas in prokaryotes, the

chromosome lies in the cytoplasm in an area called the nucleoid.

In eukaryotic cells, DNA and RNA synthesis occur in a separate compartment from protein
synthesis. In prokaryotic cells, both processes occur together. What advantages might there
be to separating the processes? What advantages might there be to having them occur
together?

The size of the genome in one of the most well-studied prokaryotes, E.coli, is 4.6 million
base pairs (approximately 1.1 mm, if cut and stretched out). So how does this fit inside a
small bacterial cell? The DNA is twisted by what is known as supercoiling. Supercoiling
suggests that DNA is either “under-wound” (less than one turn of the helix per 10 base pairs)
or “over-wound” (more than 1 turn per 10 base pairs) from its normal relaxed state. Some
proteins are known to be involved in the supercoiling; other proteins and enzymes such as
DNA gyrase help in maintaining the supercoiled structure.

Eukaryotes, whose chromosomes each consist of a linear DNA molecule, employ a different
type of packing strategy to fit their DNA inside the nucleus (Figure .11). At the most basic
level, DNA is wrapped around proteins known as histones to form structures called
nucleosomes. The histones are evolutionarily conserved proteins that are rich in basic amino
acids and form an octamer composed of two molecules of each of four different histones.
Their composition and properties are important to understanding gene expression, and were
partially uncovered based on research by Marie M. Daly and Alfred E. Mirsky in the early
1950s. The DNA (remember, it is negatively charged because of the phosphate groups) is
wrapped tightly around the histone core. This nucleosome is linked to the next one with the
help of a linker DNA. This is also known as the “beads on a string” structure. With the help of
a fifth histone, a string of nucleosomes is further compacted into a 30-nm fiber, which is the
diameter of the structure. Metaphase chromosomes are even further condensed by association
with scaffolding proteins. At the metaphase stage, the chromosomes are at their most
compact, approximately 700 nm in width.

In interphase, eukaryotic chromosomes have two distinct regions that can be distinguished by
staining. The tightly packaged region is known as heterochromatin, and the less dense region
is known as euchromatin. Heterochromatin usually contains genes that are not expressed, and
is found in the regions of the centromere and telomeres. The euchromatin usually contains
 EU Test Material

genes that are transcribed, with DNA packaged around nucleosomes but not further
compacted.

Figure .11 These figures illustrate the compaction of the eukaryotic chromosome.

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain how the structure of DNA reveals the replication process

 Describe the Meselson and Stahl experiments

The elucidation of the structure of the double helix provided a hint as to how DNA divides
and makes copies of itself. In their 1953 paper, Watson and Crick penned an incredible
understatement: "It has not escaped our notice that the specific pairing we have postulated
immediately suggests a possible copying mechanism for the genetic material." With specific
base pairs, the sequence of one DNA strand can be predicted from its complement. The
double-helix model suggests that the two strands of the double helix separate during
replication, and each strand serves as a template from which the new complementary strand is
copied. What was not clear was how the replication took place. There were three models
suggested (Figure 12): conservative, semi-conservative, and dispersive.
 EU Test Material

Figure .12 The three suggested models of DNA replication. Gray indicates the original DNA
strands, and blue indicates newly synthesized DNA.

In conservative replication, the parental DNA remains together, and the newly formed
daughter strands are together. The semi-conservative method suggests that each of the two
parental DNA strands acts as a template for new DNA to be synthesized; after replication,
each double-stranded DNA includes one parental or “old” strand and one “new” strand. In the
dispersive model, both copies of DNA have double-stranded segments of parental DNA and
newly synthesized DNA interspersed.

Meselson and Stahl were interested in understanding how DNA replicates. They grew E.
coli for several generations in a medium containing a “heavy” isotope of nitrogen (15N),
which gets incorporated into nitrogenous bases, and eventually into the DNA (Figure .13).
 EU Test Material

Figure .13 Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (15N)
then in 14N. DNA grown in 15N (red band) is heavier than DNA grown in 14N (orange band),
and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA
grown in 15N is switched to media containing 14N, after one round of cell division the DNA
sediments halfway between the 15N and 14N levels, indicating that it now contains fifty
percent 14N. In subsequent cell divisions, an increasing amount of DNA contains 14N only.
These data support the semi-conservative replication model. (credit: modification of work by
Mariana Ruiz Villareal)

The E. coli culture was then placed into medium containing 14N and allowed to grow for
several generations. After each of the first few generations, the cells were harvested and the
DNA was isolated, then centrifuged at high speeds in an ultracentrifuge. During the
centrifugation, the DNA was loaded into a gradient (typically a solution of salt such as
cesium chloride or sucrose) and spun at high speeds of 50,000 to 60,000 rpm. Under these
circumstances, the DNA will form a band according to its buoyant density: the density within
the gradient at which it floats. DNA grown in 15N will form a band at a higher density
position (i.e., farther down the centrifuge tube) than that grown in 14N. Meselson and Stahl
noted that after one generation of growth in 14N after they had been shifted from 15N, the
single band observed was intermediate in position in between DNA of cells grown
exclusively in 15N and 14N. This suggested either a semi-conservative or dispersive mode of
replication. The DNA harvested from cells grown for two generations in 14N formed two
bands: one DNA band was at the intermediate position between 15N and 14N, and the other
corresponded to the band of 14N DNA. These results could only be explained if DNA
replicates in a semi-conservative manner. And for this reason, therefore, the other two models
were ruled out.

During DNA replication, each of the two strands that make up the double helix serves as a
template from which new strands are copied. The new strands will be complementary to the
parental or “old” strands. When two daughter DNA copies are formed, they have the same
sequence and are divided equally into the two daughter cells.

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain the process of DNA replication in prokaryotes

 Discuss the role of different enzymes and proteins in supporting this process

DNA replication has been well studied in prokaryotes primarily because of the small size of
the genome and because of the large variety of mutants that are available. E. coli has 4.6
million base pairs in a single circular chromosome and all of it gets replicated in
approximately 42 minutes, starting from a single site along the chromosome and proceeding
around the circle in both directions. This means that approximately 1000 nucleotides are
added per second. Thus, the process is quite rapid and occurs without many mistakes.

DNA replication employs a large number of structural proteins and enzymes, each of which
plays a critical role during the process. One of the key players is the enzyme DNA
polymerase, also known as DNA pol, which adds nucleotides one-by-one to the growing
DNA chain that is complementary to the template strand. The addition of nucleotides requires
 EU Test Material

energy; this energy is obtained from the nucleoside triphosphates ATP, GTP, TTP and CTP.
Like ATP, the other NTPs (nucleoside triphosphates) are high-energy molecules that can
serve both as the source of DNA nucleotides and the source of energy to drive the
polymerization. When the bond between the phosphates is “broken,” the energy released is
used to form the phosphodiester bond between the incoming nucleotide and the growing
chain. In prokaryotes, three main types of polymerases are known: DNA pol I, DNA pol II,
and DNA pol III. It is now known that DNA pol III is the enzyme required for DNA
synthesis; DNA pol I is an important accessory enzyme in DNA replication, and along with
DNA pol II, is primarily required for repair.

How does the replication machinery know where to begin? It turns out that there are specific
nucleotide sequences called origins of replication where replication begins. In E. coli, which
has a single origin of replication on its one chromosome (as do most prokaryotes), this origin
of replication is approximately 245 base pairs long and is rich in AT sequences. The origin of
replication is recognized by certain proteins that bind to this site. An enzyme
called helicase unwinds the DNA by breaking the hydrogen bonds between the nitrogenous
base pairs. ATP hydrolysis is required for this process. As the DNA opens up, Y-shaped
structures called replication forks are formed. Two replication forks are formed at the origin
of replication and these get extended bi-directionally as replication proceeds. Single-strand
binding proteins coat the single strands of DNA near the replication fork to prevent the
single-stranded DNA from winding back into a double helix.

DNA polymerase has two important restrictions: it is able to add nucleotides only in the 5' to
3' direction (a new DNA strand can be only extended in this direction). It also requires a free
3'-OH group to which it can add nucleotides by forming a phosphodiester bond between the
3'-OH end and the 5' phosphate of the next nucleotide. This essentially means that it cannot
add nucleotides if a free 3'-OH group is not available. Then how does it add the first
nucleotide? The problem is solved with the help of a primer that provides the free 3'-OH end.
Another enzyme, RNA primase, synthesizes an RNA segment that is about five to ten
nucleotides long and complementary to the template DNA. Because this sequence primes the
DNA synthesis, it is appropriately called the primer. DNA polymerase can now extend this
RNA primer, adding nucleotides one-by-one that are complementary to the template strand
(Figure 14).

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure 14 First Components of DNA Replication. As DNA replication begins, DNA

Helicase, a large enzyme, separates the two strands of DNA so that they can act as templates
for replication. Single-strand binding proteins bind to each strand to stabilize and prevent
them from reforming the double helix. Primase, an RNA polymerase, binds to the single
stranded DNA and synthesizes a short RNA primer in the 5’ to 3’ direction that is antiparallel
to the parental strand. This RNA primer allows for DNA polymerase to begin replicating the
DNA. Topoisomerase binds to the double helix upstream of the replication fork to prevent
additional coiling by making small cuts in one of the DNA strands. Credit: Rao, A., Ryan, K.
Fletcher, S. and Tag, A. Department of Biology, Texas A&M University.

Question: You isolate a cell strain in which the joining of Okazaki fragments is impaired and
suspect that a mutation has occurred in an enzyme found at the replication fork. Which
enzyme is most likely to be mutated?

The replication fork moves at the rate of 1000 nucleotides per second. Topoisomerase
prevents the over-winding of the DNA double helix ahead of the replication fork as the DNA
is opening up; it does so by causing temporary nicks in the DNA helix and then resealing it.
Because DNA polymerase can only extend in the 5' to 3' direction, and because the DNA
double helix is antiparallel, there is a slight problem at the replication fork. The two template
DNA strands have opposing orientations: one strand is in the 5' to 3' direction and the other is
oriented in the 3' to 5' direction. Only one new DNA strand, the one that is complementary to
the 3' to 5' parental DNA strand, can be synthesized continuously towards the replication
fork. This continuously synthesized strand is known as the leading strand. The other strand,
complementary to the 5' to 3' parental DNA, is extended away from the replication fork, in
small fragments known as Okazaki fragments, each requiring a primer to start the synthesis.
New primer segments are laid down in the direction of the replication fork, but each pointing
away from it. (Okazaki fragments are named after the Japanese scientist who first discovered
them. The strand with the Okazaki fragments is known as the lagging strand.)
 EU Test Material

The leading strand can be extended from a single primer, whereas the lagging strand needs a
new primer for each of the short Okazaki fragments. The overall direction of the lagging
strand will be 3' to 5', and that of the leading strand 5' to 3'. A protein called the sliding
clamp holds the DNA polymerase in place as it continues to add nucleotides. The sliding
clamp is a ring-shaped protein that binds to the DNA and holds the polymerase in place. As
synthesis proceeds, the RNA primers are replaced by DNA. The primers are removed by the
exonuclease activity of DNA pol I, which uses DNA behind the RNA as its own primer and
fills in the gaps left by removal of the RNA nucleotides by the addition of DNA nucleotides.
The nicks that remain between the newly synthesized DNA (that replaced the RNA primer)
and the previously synthesized DNA are sealed by the enzyme DNA ligase, which catalyzes
the formation of phosphodiester linkages between the 3'-OH end of one nucleotide and the 5'
phosphate end of the other fragment.

Once the chromosome has been completely replicated, the two DNA copies move into two
different cells during cell division.

The process of DNA replication can be summarized as follows:

1. DNA unwinds at the origin of replication.

2. Helicase opens up the DNA-forming replication forks; these are extended
bidirectionally.
3. Single-strand binding proteins coat the DNA around the replication fork to prevent
rewinding of the DNA.
4. Topoisomerase binds at the region ahead of the replication fork to prevent
supercoiling.
5. Primase synthesizes RNA primers complementary to the DNA strand.
6. DNA polymerase III starts adding nucleotides to the 3'-OH end of the primer.
7. Elongation of both the lagging and the leading strand continues.
8. RNA primers are removed by exonuclease activity.
9. Gaps are filled by DNA pol I by adding dNTPs.
10. The gap between the two DNA fragments is sealed by DNA ligase, which helps in the
formation of phosphodiester bonds.

Table 1 summarizes the enzymes involved in prokaryotic DNA replication and the functions
of each.
 EU Test Material

Learning Objectives

By the end of this section, you will be able to do the following:

 Discuss the similarities and differences between DNA replication in eukaryotes and
prokaryotes
 State the role of telomerase in DNA replication

Eukaryotic genomes are much more complex and larger in size than prokaryotic genomes.
Eukaryotes also have a number of different linear chromosomes. The human genome has 3
billion base pairs per haploid set of chromosomes, and 6 billion base pairs are replicated
during the S phase of the cell cycle. There are multiple origins of replication on each
eukaryotic chromosome; humans can have up to 100,000 origins of replication across the
genome. The rate of replication is approximately 100 nucleotides per second, much slower
than prokaryotic replication. In yeast, which is a eukaryote, special sequences known as
autonomously replicating sequences (ARS) are found on the chromosomes. These are
equivalent to the origin of replication in E. coli.

The number of DNA polymerases in eukaryotes is much more than in prokaryotes: 14 are
known, of which five are known to have major roles during replication and have been well
studied. They are known as pol α, pol β, pol γ, pol δ, and pol ε.

The essential steps of replication are the same as in prokaryotes. Before replication can start,
the DNA has to be made available as a template. Eukaryotic DNA is bound to basic proteins
known as histones to form structures called nucleosomes. Histones must be removed and then
replaced during the replication process, which helps to account for the lower replication rate
 EU Test Material

in eukaryotes. The chromatin (the complex between DNA and proteins) may undergo some
chemical modifications, so that the DNA may be able to slide off the proteins or be accessible
to the enzymes of the DNA replication machinery. At the origin of replication, a pre-
replication complex is made with other initiator proteins. Helicase and other proteins are then
recruited to start the replication process (Table .2).

Difference between Prokaryotic and Eukaryotic Replication

Property Prokaryotes Eukaryotes

Origin of replication Single Multiple

Rate of replication 1000 nucleotides/s 50 to 100 nucleotides/s

DNA polymerase types 5 14

Telomerase Not present Present

RNA primer removal DNA pol I RNase H

Strand elongation DNA pol III Pol α, pol δ, pol ε

Sliding clamp Sliding clamp PCNA

Table 14.2

A helicase using the energy from ATP hydrolysis opens up the DNA helix. Replication forks
are formed at each replication origin as the DNA unwinds. The opening of the double helix
causes over-winding, or supercoiling, in the DNA ahead of the replication fork. These are
resolved with the action of topoisomerases. Primers are formed by the enzyme primase, and
using the primer, DNA pol can start synthesis. Three major DNA polymerases are then
involved: α, δ and ε. DNA pol α adds a short (20 to 30 nucleotides) DNA fragment to the
RNA primer on both strands, and then hands off to a second polymerase. While the leading
strand is continuously synthesized by the enzyme pol ε, the lagging strand is synthesized by
pol δ. A sliding clamp protein known as PCNA (proliferating cell nuclear antigen) holds the
DNA pol in place so that it does not slide off the DNA. As pol δ runs into the primer RNA on
the lagging strand, it displaces it from the DNA template. The displaced primer RNA is then
removed by RNase H (AKA flap endonuclease) and replaced with DNA nucleotides. The
Okazaki fragments in the lagging strand are joined after the replacement of the RNA primers
with DNA. The gaps that remain are sealed by DNA ligase, which forms the phosphodiester
bond.

Telomere replication
 EU Test Material

Unlike prokaryotic chromosomes, eukaryotic chromosomes are linear. As you’ve learned, the
enzyme DNA pol can add nucleotides only in the 5' to 3' direction. In the leading strand,
synthesis continues until the end of the chromosome is reached. On the lagging strand, DNA
is synthesized in short stretches, each of which is initiated by a separate primer. When the
replication fork reaches the end of the linear chromosome, there is no way to replace the
primer on the 5’ end of the lagging strand. The DNA at the ends of the chromosome thus
remains unpaired, and over time these ends, called telomeres, may get progressively shorter
as cells continue to divide.

Telomeres comprise repetitive sequences that code for no particular gene. In humans, a six-
base-pair sequence, TTAGGG, is repeated 100 to 1000 times in the telomere regions. In a
way, these telomeres protect the genes from getting deleted as cells continue to divide. The
telomeres are added to the ends of chromosomes by a separate enzyme, telomerase (Figure
16), whose discovery helped in the understanding of how these repetitive chromosome ends
are maintained. The telomerase enzyme contains a catalytic part and a built-in RNA
template. It attaches to the end of the chromosome, and DNA nucleotides complementary to
the RNA template are added on the 3' end of the DNA strand. Once the 3' end of the lagging
strand template is sufficiently elongated, DNA polymerase can add the nucleotides
complementary to the ends of the chromosomes. Thus, the ends of the chromosomes are
replicated.

Figure .15 The ends of linear chromosomes are maintained by the action of the telomerase
enzyme. Credit: Rao, A. and Fletcher, S. Department of Biology, Texas A&M University.

Telomerase is typically active in germ cells and adult stem cells. It is not active in adult
somatic cells. For their discovery of telomerase and its action, Elizabeth Blackburn, Carol W.
Greider, and Jack W. Szostak (Figure 16) received the Nobel Prize for Medicine and
Physiology in 2009. Later research using HeLa cells (obtained from Henrietta Lacks)
 EU Test Material

confirmed that telomerase is present in human cells. And in 2001, researchers including
Diane L. Wright found that telomerase is necessary for cells in human embryos to rapidly
proliferate.

Figure .16 Elizabeth Blackburn, 2009 Nobel Laureate, is one of the scientists who discovered
how telomerase works. (credit: US Embassy Sweden)

Telomerase and Aging

Cells that undergo cell division continue to have their telomeres shortened because most
somatic cells do not make telomerase. This essentially means that telomere shortening is
associated with aging. With the advent of modern medicine, preventative health care, and
healthier lifestyles, the human life span has increased, and there is an increasing demand for
people to look younger and have a better quality of life as they grow older.

In 2010, scientists found that telomerase can reverse some age-related conditions in mice.
This may have potential in regenerative medicine.2 Telomerase-deficient mice were used in
these studies; these mice have tissue atrophy, stem cell depletion, organ system failure, and
impaired tissue injury responses. Telomerase reactivation in these mice caused extension of
telomeres, reduced DNA damage, reversed neurodegeneration, and improved the function of
the testes, spleen, and intestines. Thus, telomere reactivation may have potential for treating
age-related diseases in humans.

Cancer is characterized by uncontrolled cell division of abnormal cells. The cells accumulate
mutations, proliferate uncontrollably, and can migrate to different parts of the body through a
process called metastasis. Scientists have observed that cancerous cells have considerably
shortened telomeres and that telomerase is active in these cells. Interestingly, only after the
telomeres were shortened in the cancer cells did the telomerase become active. If the action
of telomerase in these cells can be inhibited by drugs during cancer therapy, then the
cancerous cells could potentially be stopped from further division.

Learning Objectives
 EU Test Material

By the end of this section, you will be able to do the following:

 Discuss the different types of mutations in DNA

 Explain DNA repair mechanisms

DNA replication is a highly accurate process, but mistakes can occasionally occur, such as a
DNA polymerase inserting a wrong base. Uncorrected mistakes may sometimes lead to
serious consequences, such as cancer. Repair mechanisms correct the mistakes. In rare cases,
mistakes are not corrected, leading to mutations; in other cases, repair enzymes are
themselves mutated or defective.

Most of the mistakes during DNA replication are promptly corrected by the proofreading
ability of DNA polymerase itself. (Figure 17). In proofreading, the DNA pol reads the newly
added base before adding the next one, so a correction can be made. The polymerase checks
whether the newly added base has paired correctly with the base in the template strand. If it is
the right base, the next nucleotide is added. If an incorrect base has been added, the enzyme
makes a cut at the phosphodiester bond and releases the wrong nucleotide. This is performed
by the 3' exonuclease action of DNA pol. Once the incorrect nucleotide has been removed, it
can be replaced by the correct one.

Figure .17 Proofreading by DNA polymerase corrects errors during replication.

Some errors are not corrected during replication, but are instead corrected after replication is
completed; this type of repair is known as mismatch repair (Figure .18). Specific repair
enzymes recognize the mispaired nucleotide and excise part of the strand that contains it; the
excised region is then resynthesized. If the mismatch remains uncorrected, it may lead to
more permanent damage when the mismatched DNA is replicated. How do mismatch repair
enzymes recognize which of the two bases is the incorrect one? In E. coli, after replication,
the nitrogenous base adenine acquires a methyl group; the parental DNA strand will have
methyl groups, whereas the newly synthesized strand lacks them. Thus, DNA polymerase is
able to remove the wrongly incorporated bases from the newly synthesized, non-methylated
strand. In eukaryotes, the mechanism is not very well understood, but it is believed to involve
recognition of unsealed nicks in the new strand, as well as a short-term continuing association
of some of the replication proteins with the new daughter strand after replication has
completed.
 EU Test Material

Figure .18 In mismatch repair, the incorrectly added base is detected after replication. The
mismatch repair proteins detect this base and remove it from the newly synthesized strand by
nuclease action. The gap is now filled with the correctly paired base.

Another type of repair mechanism, nucleotide excision repair, is similar to mismatch repair,
except that it is used to remove damaged bases rather than mismatched ones. The repair
enzymes replace abnormal bases by making a cut on both the 3' and 5' ends of the damaged
base (Figure 19). The segment of DNA is removed and replaced with the correctly paired
nucleotides by the action of DNA pol. Once the bases are filled in, the remaining gap is
sealed with a phosphodiester linkage catalyzed by DNA ligase. This repair mechanism is
often employed when UV exposure causes the formation of pyrimidine dimers.
 EU Test Material

Figure 19 Nucleotide excision repairs thymine dimers. When exposed to UV light, thymines
lying adjacent to each other can form thymine dimers. In normal cells, they are excised and
replaced. Credit: Rao, A., Fletcher, S. and Tag, A. Department of Biology, Texas A&M
University.

A well-studied example of mistakes not being corrected is seen in people suffering from
xeroderma pigmentosa (Figure .20). Affected individuals have skin that is highly sensitive to
UV rays from the sun. When individuals are exposed to UV light, pyrimidine dimers,
especially those of thymine, are formed; people with xeroderma pigmentosa are not able to
repair the damage. These are not repaired because of a defect in the nucleotide excision repair
enzymes, whereas in normal individuals, the thymine dimers are excised and the defect is
corrected. The thymine dimers distort the structure of the DNA double helix, and this may
cause problems during DNA replication. People with xeroderma pigmentosa may have a
higher risk of contracting skin cancer than those who don't have the condition.
 EU Test Material

Figure .20 Xeroderma pigmentosa is a condition in which thymine dimerization from

exposure to UV light is not repaired. Exposure to sunlight results in skin lesions. (credit:
James Halpern et al.)

Errors during DNA replication are not the only reason why mutations arise in
DNA. Mutations, variations in the nucleotide sequence of a genome, can also occur because
of damage to DNA. Such mutations may be of two types: induced or spontaneous. Induced
mutations are those that result from an exposure to chemicals, UV rays, x-rays, or some
other environmental agent. For example, Charlotte Auerbach and J.M Robson discovered the
mutation-inducing effects of mustard gas. Spontaneous mutations occur without any
exposure to any environmental agent; they are a result of natural reactions taking place within
the body.

Mutations may have a wide range of effects. Point mutations are those mutations that affect a
single base pair. The most common nucleotide mutations are substitutions, in which one base
is replaced by another. These substitutions can be of two types, either transitions or
transversions. Transition substitution refers to a purine or pyrimidine being replaced by a
base of the same kind; for example, a purine such as adenine may be replaced by the purine
guanine. Transversion substitution refers to a purine being replaced by a pyrimidine, or
vice versa; for example, cytosine, a pyrimidine, is replaced by adenine, a purine. Some point
mutations are not detectable in the final product; these are known as silent mutations. Silent
mutations are usually due to a substitution in the third base of a codon, which often represents
the same amino acid as the original codon. Other point mutations can result in the
replacement of one amino acid by another, which may alter the function of the protein. Point
mutations that generate a stop codon can terminate a protein early.

Some mutations can result in an increased number of copies of the same codon. These are
called trinucleotide repeat expansions and result in repeated regions of the same amino acid.
Mutations can also be the result of the addition of a base, known as an insertion, or the
removal of a base, also known as deletion. If an insertion or deletion results in the alteration
of the translational reading frame (a frameshift mutation), the resultant protein is usually
nonfunctional. Sometimes a piece of DNA from one chromosome may get translocated to
another chromosome or to another region of the same chromosome; this is also known as
translocation. These mutation types are shown in Figure .21.
 EU Test Material

VISUAL CONNECTION

Visual Connection

Figure .21 Mutations can lead to changes in the protein sequence encoded by the DNA.

A frameshift mutation that results in the insertion of three nucleotides is often less deleterious
than a mutation that results in the insertion of one nucleotide. Why?

Mutations in repair genes have been known to cause cancer. Many mutated repair genes have
been implicated in certain forms of pancreatic cancer, colon cancer, and colorectal cancer.
Mutations can affect either somatic cells or germ cells. If many mutations accumulate in a
somatic cell, they may lead to problems such as the uncontrolled cell division observed in
cancer. If a mutation takes place in germ cells, the mutation will be passed on to the next
generation, as in the case of hemophilia and xeroderma pigmentosa.
 EU Test Material

Figure .1 Genes, which are carried on (a) chromosomes, are linearly organized instructions
for making the RNA and protein molecules that are necessary for all of the processes of life.
The (b) interleukin-2 protein and (c) alpha-2u-globulin protein are just two examples of the
array of different molecular structures that are encoded by genes. (credit “chromosome:
National Human Genome Research Institute; credit “interleukin-2”: Ramin Herati/Created
from PDB 1M47 and rendered with Pymol; credit “alpha-2u-globulin”: Darren
Logan/rendered with AISMIG)

Chapter Outline

.1 The Genetic Code

.2 Prokaryotic Transcription

.3 Eukaryotic Transcription

.4 RNA Processing in Eukaryotes

.5 Ribosomes and Protein Synthesis

Since the rediscovery of Mendel’s work in 1900, the definition of the gene has progressed
from an abstract unit of heredity to a tangible molecular entity capable of replication,
expression, and mutation (Figure 1). Genes are composed of DNA and are linearly arranged
on chromosomes. Genes specify the sequences of amino acids, which are the building blocks
of proteins. In turn, proteins are responsible for orchestrating nearly every function of the
cell. Both genes and the proteins they encode are absolutely essential to life as we know it.

Learning Objectives

By the end of this section, you will be able to do the following:

 Explain the “central dogma” of DNA-protein synthesis

 Describe the genetic code and how the nucleotide sequence prescribes the amino acid
and the protein sequence
 EU Test Material

The cellular process of transcription generates messenger RNA (mRNA), a mobile molecular
copy of one or more genes with an alphabet of A, C, G, and uracil (U). Translation of the
mRNA template on ribosomes converts nucleotide-based genetic information into a protein
product. That is the central dogma of DNA-protein synthesis. Protein sequences consist of 20
commonly occurring amino acids; therefore, it can be said that the protein alphabet consists
of 20 “letters” (Figure .2). Different amino acids have different chemistries (such as acidic
versus basic, or polar and nonpolar) and different structural constraints. Variation in amino
acid sequence is responsible for the enormous variation in protein structure and function.

Figure .2 Structures of the 20 amino acids found in proteins are shown. Each amino acid is
composed of an amino group (NH+3NH3+), a carboxyl group (COO-), and a side chain
(blue). The side chain may be nonpolar, polar, or charged, as well as large or small. It is the
variety of amino acid side chains that gives rise to the incredible variation of protein structure
and function.

The Central Dogma: DNA Encodes RNA; RNA Encodes Protein

The flow of genetic information in cells from DNA to mRNA to protein is described by
the central dogma (Figure .3), which states that genes specify the sequence of mRNAs,
which in turn specify the sequence of amino acids making up all proteins. The decoding of
one molecule to another is performed by specific proteins and RNAs. Because the
information stored in DNA is so central to cellular function, it makes intuitive sense that the
cell would make mRNA copies of this information for protein synthesis, while keeping the
 EU Test Material

DNA itself intact and protected. The copying of DNA to RNA is relatively straightforward,
with one nucleotide being added to the mRNA strand for every nucleotide read in the DNA
strand. The translation to protein is a bit more complex because three mRNA nucleotides
correspond to one amino acid in the polypeptide sequence. However, the translation to
protein is still systematic and colinear, such that nucleotides 1 to 3 correspond to amino acid
1, nucleotides 4 to 6 correspond to amino acid 2, and so on.
 EU Test Material

Figure .3 Instructions on DNA are transcribed onto messenger RNA. Ribosomes are able to
read the genetic information inscribed on a strand of messenger RNA and use this
information to string amino acids together into a protein.

The Genetic Code Is Degenerate and Universal

Each amino acid is defined by a three-nucleotide sequence called the triplet codon. Given the
different numbers of “letters” in the mRNA and protein “alphabets,” scientists theorized that
single amino acids must be represented by combinations of nucleotides. Nucleotide doublets
would not be sufficient to specify every amino acid because there are only 16 possible two-
nucleotide combinations (42). In contrast, there are 64 possible nucleotide triplets (43), which
is far more than the number of amino acids. Scientists theorized that amino acids were
encoded by nucleotide triplets and that the genetic code was “degenerate.” In other words, a
given amino acid could be encoded by more than one nucleotide triplet. This was later
confirmed experimentally: Francis Crick and Sydney Brenner used the chemical mutagen
proflavin to insert one, two, or three nucleotides into the gene of a virus. When one or two
nucleotides were inserted, the normal proteins were not produced. When three nucleotides
were inserted, the protein was synthesized and functional. This demonstrated that the amino
acids must be specified by groups of three nucleotides. These nucleotide triplets are
called codons. The insertion of one or two nucleotides completely changed the
triplet reading frame, thereby altering the message for every subsequent amino acid (Figure
.5). Though insertion of three nucleotides caused an extra amino acid to be inserted during
translation, the integrity of the rest of the protein was maintained.

Scientists painstakingly solved the genetic code by translating synthetic mRNAs in vitro and
sequencing the proteins they specified (Figure .4).

Figure .4 This figure shows the genetic code for translating each nucleotide triplet in mRNA
into an amino acid or a termination signal in a protein. (credit: modification of work by NIH)
 EU Test Material

In addition to codons that instruct the addition of a specific amino acid to a polypeptide
chain, three of the 64 codons terminate protein synthesis and release the polypeptide from the
translation machinery. These triplets are called nonsense codons, or stop codons. Another
codon, AUG, also has a special function. In addition to specifying the amino acid methionine,
it also serves as the start codon to initiate translation. The reading frame for translation is set
by the AUG start codon near the 5' end of the mRNA. Following the start codon, the mRNA
is read in groups of three until a stop codon is encountered.

The arrangement of the coding table reveals the structure of the code. There are sixteen
"blocks" of codons, each specified by the first and second nucleotides of the codons within
the block, e.g., the "AC*" block that corresponds to the amino acid threonine (Thr). Some
blocks are divided into a pyrimidine half, in which the codon ends with U or C, and a purine
half, in which the codon ends with A or G. Some amino acids get a whole block of four
codons, like alanine (Ala), threonine (Thr) and proline (Pro). Some get the pyrimidine half of
their block, like histidine (His) and asparagine (Asn). Others get the purine half of their
block, like glutamate (Glu) and lysine (Lys). Note that some amino acids get a block and a
half-block for a total of six codons.

The specification of a single amino acid by multiple similar codons is called "degeneracy."
Degeneracy is believed to be a cellular mechanism to reduce the negative impact of random
mutations. Codons that specify the same amino acid typically only differ by one nucleotide.
In addition, amino acids with chemically similar side chains are encoded by similar codons.
For example, aspartate (Asp) and glutamate (Glu), which occupy the GA* block, are both
negatively charged. This nuance of the genetic code ensures that a single-nucleotide
substitution mutation might specify the same amino acid but have no effect or specify a
similar amino acid, preventing the protein from being rendered completely nonfunctional.

The genetic code is nearly universal. With a few minor exceptions, virtually all species use
the same genetic code for protein synthesis. Conservation of codons means that a purified
mRNA encoding the globin protein in horses could be transferred to a tulip cell, and the tulip
would synthesize horse globin. That there is only one genetic code is powerful evidence that
all of life on Earth shares a common origin, especially considering that there are about
1084 possible combinations of 20 amino acids and 64 triplet codons.

Figure .5 The deletion of two nucleotides shifts the reading frame of an mRNA and changes
the entire protein message, creating a nonfunctional protein or terminating protein synthesis
altogether.

SCIENTIFIC METHOD CONNECTION

Scientific Method Connection

Which Has More DNA: A Kiwi or a Strawberry?

 EU Test Material

Figure .6 Do you think that a kiwi or a strawberry has more DNA per fruit? (credit “kiwi”:
"Kelbv"/Flickr; credit: “strawberry”: Alisdair McDiarmid)

Question: Would a kiwi and strawberry that are approximately the same size (Figure .6) also
have approximately the same amount of DNA?

Background: Genes are carried on chromosomes and are made of DNA. All mammals are
diploid, meaning they have two copies of each chromosome. However, not all plants are
diploid. The common strawberry is octoploid (8n) and the cultivated kiwi is hexaploid (6n).
Research the total number of chromosomes in the cells of each of these fruits and think about
how this might correspond to the amount of DNA in these fruits’ cell nuclei. What other
factors might contribute to the total amount of DNA in a single fruit? Read about the
technique of DNA isolation to understand how each step in the isolation protocol helps
liberate and precipitate DNA.

Hypothesis: Hypothesize whether you would be able to detect a difference in DNA quantity
from similarly sized strawberries and kiwis. Which fruit do you think would yield more
DNA?

Test your hypothesis: Isolate the DNA from a strawberry and a kiwi that are similarly sized.
Perform the experiment in at least triplicate for each fruit

1. Prepare a bottle of DNA extraction buffer from 900 mL water, 50 mL dish detergent,
and two teaspoons of table salt. Mix by inversion (cap it and turn it upside down a few
times).
2. Grind a strawberry and a kiwi by hand in a plastic bag, or using a mortar and pestle,
or with a metal bowl and the end of a blunt instrument. Grind for at least two minutes
per fruit.
3. Add 10 mL of the DNA extraction buffer to each fruit, and mix well for at least one
minute.
4. Remove cellular debris by filtering each fruit mixture through cheesecloth or porous
cloth and into a funnel placed in a test tube or an appropriate container.
5. Pour ice-cold ethanol or isopropanol (rubbing alcohol) into the test tube. You should
observe white, precipitated DNA.
6. Gather the DNA from each fruit by winding it around separate glass rods.

Record your observations: Because you are not quantitatively measuring DNA volume, you
can record for each trial whether the two fruits produced the same or different amounts of
 EU Test Material

DNA as observed by eye. If one or the other fruit produced noticeably more DNA, record this
as well. Determine whether your observations are consistent with several pieces of each fruit.

Analyze your data: Did you notice an obvious difference in the amount of DNA produced
by each fruit? Were your results reproducible?

Draw a conclusion: Given what you know about the number of chromosomes in each fruit,
can you conclude that chromosome number necessarily correlates to DNA amount? Can you
identify any drawbacks to this procedure? If you had access to a laboratory, how could you
standardize your comparison and make it more quantitative?

Learning Objectives

By the end of this section, you will be able to do the following:

 List the different steps in prokaryotic transcription

 Discuss the role of promoters in prokaryotic transcription
 Describe how and when transcription is terminated

The prokaryotes, which include Bacteria and Archaea, are mostly single-celled organisms
that, by definition, lack membrane-bound nuclei and other organelles. A bacterial
chromosome is a closed circle that, unlike eukaryotic chromosomes, is not organized around
histone proteins. The central region of the cell in which prokaryotic DNA resides is called the
nucleoid region. In addition, prokaryotes often have abundant plasmids, which are shorter,
circular DNA molecules that may only contain one or a few genes. Plasmids can be
transferred independently of the bacterial chromosome during cell division and often carry
traits such as those involved with antibiotic resistance.

Transcription in prokaryotes (and in eukaryotes) requires the DNA double helix to partially
unwind in the region of mRNA synthesis. The region of unwinding is called a transcription
bubble. Transcription always proceeds from the same DNA strand for each gene, which is
called the template strand. The mRNA product is complementary to the template strand and
is almost identical to the other DNA strand, called the nontemplate strand, or the coding
strand. The only nucleotide difference is that in mRNA, all of the T nucleotides are replaced
with U nucleotides (Figure .7). In an RNA double helix, A can bind U via two hydrogen
bonds, just as in A–T pairing in a DNA double helix.
 EU Test Material

Figure .7 Messenger RNA is a copy of protein-coding information in the coding strand of

DNA, with the substitution of U in the RNA for T in the coding sequence. However, new
RNA nucleotides base pair with the nucleotides of the template strand. RNA is synthesized in
its 5'-3' direction, using the enzyme RNA polymerase. As the template is read, the DNA
unwinds ahead of the polymerase and then rewinds behind it.

The nucleotide pair in the DNA double helix that corresponds to the site from which the first
5' mRNA nucleotide is transcribed is called the +1 site, or the initiation site. Nucleotides
preceding the initiation site are denoted with a “-” and are designated upstream nucleotides.
Conversely, nucleotides following the initiation site are denoted with “+” numbering and are
called downstream nucleotides.

Initiation of Transcription in Prokaryotes

Prokaryotes do not have membrane-enclosed nuclei. Therefore, the processes of transcription,

translation, and mRNA degradation can all occur simultaneously. The intracellular level of a
bacterial protein can quickly be amplified by multiple transcription and translation events that
occur concurrently on the same DNA template. Prokaryotic genomes are very compact, and
prokaryotic transcripts often cover more than one gene or cistron (a coding sequence for a
single protein). Polycistronic mRNAs are then translated to produce more than one kind of
protein.

Our discussion here will exemplify transcription by describing this process in Escherichia
coli, a well-studied eubacterial species. Although some differences exist between
transcription in E. coli and transcription in archaea, an understanding of E. coli transcription
can be applied to virtually all bacterial species.

Prokaryotic RNA Polymerase

Prokaryotes use the same RNA polymerase to transcribe all of their genes. In E. coli, the
polymerase is composed of five polypeptide subunits, two of which are identical. Four of
these subunits, denoted α, α, β, and β', comprise the polymerase core enzyme. These subunits
assemble every time a gene is transcribed, and they disassemble once transcription is
complete. Each subunit has a unique role; the two α-subunits are necessary to assemble the
polymerase on the DNA; the β-subunit binds to the ribonucleoside triphosphate that will
become part of the nascent mRNA molecule; and the β' subunit binds the DNA template
strand. The fifth subunit, σ, is involved only in transcription initiation. It confers
transcriptional specificity such that the polymerase begins to synthesize mRNA from an
appropriate initiation site. Without σ, the core enzyme would transcribe from random sites
and would produce mRNA molecules that specified protein gibberish. The polymerase
comprised of all five subunits is called the holoenzyme.

Prokaryotic Promoters

A promoter is a DNA sequence onto which the transcription machinery, including RNA
polymerase, binds and initiates transcription. In most cases, promoters exist upstream of the
genes they regulate. The specific sequence of a promoter is very important because it
determines whether the corresponding gene is transcribed all the time, some of the time, or
infrequently. Although promoters vary among prokaryotic genomes, a few elements are
 EU Test Material

evolutionarily conserved in many species. At the -10 and -35 regions upstream of the
initiation site, there are two promoter consensus sequences, or regions that are similar across
all promoters and across various bacterial species (Figure .8). The -10 sequence, called the -
10 region, has the consensus sequence TATAAT. The -35 sequence has the consensus
sequence TTGACA. These consensus sequences are recognized and bound by σ. Once this
interaction is made, the subunits of the core enzyme bind to the site. The A–T-rich -10 region
facilitates unwinding of the DNA template, and several phosphodiester bonds are made. The
transcription initiation phase ends with the production of abortive transcripts, which are
polymers of approximately 10 nucleotides that are made and released.

Figure .8 The σ subunit of prokaryotic RNA polymerase recognizes consensus sequences

found in the promoter region upstream of the transcription start site. The σ subunit dissociates
from the polymerase after transcription has been initiated.

Elongation and Termination in Prokaryotes

The transcription elongation phase begins with the release of the σ subunit from the
polymerase. The dissociation of σ allows the core enzyme to proceed along the DNA
template, synthesizing mRNA in the 5' to 3' direction at a rate of approximately 40
nucleotides per second. As elongation proceeds, the DNA is continuously unwound ahead of
the core enzyme and rewound behind it. The base pairing between DNA and RNA is not
stable enough to maintain the stability of the mRNA synthesis components. Instead, the RNA
polymerase acts as a stable linker between the DNA template and the nascent RNA strands to
ensure that elongation is not interrupted prematurely.

Prokaryotic Termination Signals

Once a gene is transcribed, the prokaryotic polymerase needs to be instructed to dissociate

from the DNA template and liberate the newly made mRNA. Depending on the gene being
transcribed, there are two kinds of termination signals. One is protein-based and the other is
RNA-based. Rho-dependent termination is controlled by the rho protein, which tracks
along behind the polymerase on the growing mRNA chain. Near the end of the gene, the
polymerase encounters a run of G nucleotides on the DNA template and it stalls. As a result,
the rho protein collides with the polymerase. The interaction with rho releases the mRNA
from the transcription bubble.

Rho-independent termination is controlled by specific sequences in the DNA template

strand. As the polymerase nears the end of the gene being transcribed, it encounters a region
rich in C–G nucleotides. The mRNA folds back on itself, and the complementary C–G
nucleotides bind together. The result is a stable hairpin that causes the polymerase to stall as
 EU Test Material

soon as it begins to transcribe a region rich in A–T nucleotides. The complementary U–A
region of the mRNA transcript forms only a weak interaction with the template DNA. This,
coupled with the stalled polymerase, induces enough instability for the core enzyme to break
away and liberate the new mRNA transcript.

Upon termination, the process of transcription is complete. By the time termination occurs,
the prokaryotic transcript would already have been used to begin synthesis of numerous
copies of the encoded protein because these processes can occur concurrently. The
unification of transcription, translation, and even mRNA degradation is possible because all
of these processes occur in the same 5' to 3' direction, and because there is no membranous
compartmentalization in the prokaryotic cell (Figure .9). In contrast, the presence of a nucleus
in eukaryotic cells precludes simultaneous transcription and translation.

Figure .9 Multiple polymerases can transcribe a single bacterial gene while numerous
ribosomes concurrently translate the mRNA transcripts into polypeptides. In this way, a
specific protein can rapidly reach a high concentration in the bacterial cell.

Learning Objectives

By the end of this section, you will be able to do the following:

 List the steps in eukaryotic transcription

 Discuss the role of RNA polymerases in transcription
 Compare and contrast the three RNA polymerases
 Explain the significance of transcription factors

Prokaryotes and eukaryotes perform fundamentally the same process of transcription, with a
few key differences. The most important difference between prokaryote and eukaryote
transcription is due to the latter’s membrane-bound nucleus and organelles. With the genes
bound in a nucleus, the eukaryotic cell must be able to transport its mRNA to the cytoplasm
and must protect its mRNA from degrading before it is translated. Eukaryotes also employ
three different polymerases that each transcribe a different subset of genes. Eukaryotic
mRNAs are usually monogenic, meaning that they specify a single protein.

Initiation of Transcription in Eukaryotes

Unlike the prokaryotic polymerase that can bind to a DNA template on its own, eukaryotes
require several other proteins, called transcription factors, to first bind to the promoter region
and then to help recruit the appropriate polymerase.
 EU Test Material

The Three Eukaryotic RNA Polymerases

The features of eukaryotic mRNA synthesis are markedly more complex than those of
prokaryotes. Instead of a single polymerase comprising five subunits, the eukaryotes have
three polymerases that are each made up of 10 subunits or more. Each eukaryotic polymerase
also requires a distinct set of transcription factors to bring it to the DNA template.

RNA polymerase I is located in the nucleolus, a specialized nuclear substructure in which

ribosomal RNA (rRNA) is transcribed, processed, and assembled into ribosomes (Table.1).
The rRNA molecules are considered structural RNAs because they have a cellular role but
are not translated into protein. The rRNAs are components of the ribosome and are essential
to the process of translation. RNA polymerase I synthesizes all of the rRNAs from the
tandemly duplicated set of 18S, 5.8S, and 28S ribosomal genes. (Note that the “S”
designation applies to “Svedberg” units, a nonadditive value that characterizes the speed at
which a particle sediments during centrifugation.)

Locations, Products, and Sensitivities of the Three Eukaryotic RNA Polymerases

Table .1

RNA polymerase II is located in the nucleus and synthesizes all protein-coding nuclear pre-
mRNAs. Eukaryotic pre-mRNAs undergo extensive processing after transcription but before
translation. For clarity, this module’s discussion of transcription and translation in eukaryotes
will use the term “mRNAs” to describe only the mature, processed molecules that are ready
to be translated. RNA polymerase II is responsible for transcribing the overwhelming
majority of eukaryotic genes.

RNA polymerase III is also located in the nucleus. This polymerase transcribes a variety of
structural RNAs that includes the 5S pre-rRNA, transfer pre-RNAs (pre-tRNAs), and small
nuclear pre-RNAs. The tRNAs have a critical role in translation; they serve as the “adaptor
molecules” between the mRNA template and the growing polypeptide chain. Small nuclear
RNAs have a variety of functions, including “splicing” pre-mRNAs and regulating
transcription factors.

A scientist characterizing a new gene can determine which polymerase transcribes it by

testing whether the gene is expressed in the presence of α-amanitin, an oligopeptide toxin
produced by the fly agaric toadstool mushroom and other species of Amanita. Interestingly,
the α-amanitin affects the three polymerases very differently (Table 1). RNA polymerase I is
completely insensitive to α-amanitin, meaning that the polymerase can transcribe DNA in
 EU Test Material

vitro in the presence of this poison. RNA polymerase III is moderately sensitive to the toxin.
In contrast, RNA polymerase II is extremely sensitive to α-amanitin. The toxin prevents the
enzyme from progressing down the DNA, and thus inhibits transcription. Knowing the
transcribing polymerase can provide clues as to the general function of the gene being
studied. Because RNA polymerase II transcribes the vast majority of genes, we will focus on
this polymerase in our subsequent discussions about eukaryotic transcription factors and
promoters.

RNA Polymerase II Promoters and Transcription Factors

Eukaryotic promoters are much larger and more intricate than prokaryotic promoters.
However, both have a sequence similar to the -10 sequence of prokaryotes. In eukaryotes,
this sequence is called the TATA box, and has the consensus sequence TATAAA on the
coding strand. It is located at -25 to -35 bases relative to the initiation (+1) site (Figure 10).
This sequence is not identical to the E. coli -10 box, but it conserves the A–T rich element.
The thermostability of A–T bonds is low and this helps the DNA template to locally unwind
in preparation for transcription.

Instead of the simple σ factor that helps bind the prokaryotic RNA polymerase to its
promoter, eukaryotes assemble a complex of transcription factors required to recruit RNA
polymerase II to a protein coding gene. Transcription factors that bind to the promoter are
called basal transcription factors. These basal factors are all called TFII (for Transcription
Factor/polymerase II) plus an additional letter (A-J). The core complex is TFIID, which
includes a TATA-binding protein (TBP). The other transcription factors systematically fall
into place on the DNA template, with each one further stabilizing the pre-initiation complex
and contributing to the recruitment of RNA polymerase II.

VISUAL CONNECTION
 EU Test Material

Visual Connection

Figure .10 A generalized promoter of a gene transcribed by RNA polymerase II is shown.

Transcription factors recognize the promoter. RNA polymerase II then binds and forms the
transcription initiation complex.

A scientist splices a eukaryotic promoter in front of a bacterial gene and inserts the gene in a
bacterial chromosome. Would you expect the bacteria to transcribe the gene?

Some eukaryotic promoters also have a conserved CAAT box (GGCCAATCT) at

approximately -80. Further upstream of the TATA box, eukaryotic promoters may also
contain one or more GC-rich boxes (GGCG) or octamer boxes (ATTTGCAT). These
elements bind cellular factors that increase the efficiency of transcription initiation and are
often identified in more “active” genes that are constantly being expressed by the cell.

Basal transcription factors are crucial in the formation of a preinitiation complex on the DNA
template that subsequently recruits RNA polymerase II for transcription initiation. The
complexity of eukaryotic transcription does not end with the polymerases and promoters. An
army of other transcription factors, which bind to upstream enhancers and silencers, also help
to regulate the frequency with which pre-mRNA is synthesized from a gene. Enhancers and
silencers affect the efficiency of transcription but are not necessary for transcription to
proceed.

Promoter Structures for RNA Polymerases I and III

The processes of bringing RNA polymerases I and III to the DNA template involve slightly
less complex collections of transcription factors, but the general theme is the same.
 EU Test Material

The conserved promoter elements for genes transcribed by polymerases I and III differ from
those transcribed by RNA polymerase II. RNA polymerase I transcribes genes that have two
GC-rich promoter sequences in the -45 to +20 region. These sequences alone are sufficient
for transcription initiation to occur, but promoters with additional sequences in the region
from -180 to -105 upstream of the initiation site will further enhance initiation. Genes that are
transcribed by RNA polymerase III have upstream promoters or promoters that occur within
the genes themselves.

Eukaryotic transcription is a tightly regulated process that requires a variety of proteins to

interact with each other and with the DNA strand. Although the process of transcription in
eukaryotes involves a greater metabolic investment than in prokaryotes, it ensures that the
cell transcribes precisely the pre-mRNAs that it needs for protein synthesis.

EVOLUTION CONNECTION

Evolution Connection

The Evolution of PromotersThe evolution of genes may be a familiar concept. Mutations

can occur in genes during DNA replication, and the result may or may not be beneficial to the
cell. By altering an enzyme, structural protein, or some other factor, the process of mutation
can transform functions or physical features. However, eukaryotic promoters and other gene
regulatory sequences may evolve as well. For instance, consider a gene that, over many
generations, becomes more valuable to the cell. Maybe the gene encodes a structural protein
that the cell needs to synthesize in abundance for a certain function. If this is the case, it
would be beneficial to the cell for that gene’s promoter to recruit transcription factors more
efficiently and increase gene expression.

Scientists examining the evolution of promoter sequences have reported varying results. In
part, this is because it is difficult to infer exactly where a eukaryotic promoter begins and
ends. Some promoters occur within genes; others are located very far upstream, or even
downstream, of the genes they are regulating. However, when researchers limited their
examination to human core promoter sequences that were defined experimentally as
sequences that bind the preinitiation complex, they found that promoters evolve even faster
than protein-coding genes.

It is still unclear how promoter evolution might correspond to the evolution of humans or
other complex organisms. However, the evolution of a promoter to effectively make more or
less of a given gene product is an intriguing alternative to the evolution of the genes
themselves.1

Eukaryotic Elongation and Termination

Following the formation of the preinitiation complex, the polymerase is released from the
other transcription factors, and elongation is allowed to proceed as it does in prokaryotes with
the polymerase synthesizing pre-mRNA in the 5' to 3' direction. As discussed previously,
RNA polymerase II transcribes the major share of eukaryotic genes, so in this section we will
focus on how this polymerase accomplishes elongation and termination.

Although the enzymatic process of elongation is essentially the same in eukaryotes and
prokaryotes, the DNA template is considerably more complex. When eukaryotic cells are not
 EU Test Material

dividing, their genes exist as a diffuse mass of DNA and proteins called chromatin. The DNA
is tightly packaged around charged histone proteins at repeated intervals. These DNA–histone
complexes, collectively called nucleosomes, are regularly spaced and include 146 nucleotides
of DNA wound around eight histones like thread around a spool.

For polynucleotide synthesis to occur, the transcription machinery needs to move histones out
of the way every time it encounters a nucleosome. This is accomplished by a special protein
complex called FACT, which stands for “facilitates chromatin transcription.” This complex
pulls histones away from the DNA template as the polymerase moves along it. Once the pre-
mRNA is synthesized, the FACT complex replaces the histones to recreate the nucleosomes.

The termination of transcription is different for the different polymerases. Unlike in

prokaryotes, elongation by RNA polymerase II in eukaryotes takes place 1,000 to 2,000
nucleotides beyond the end of the gene being transcribed. This pre-mRNA tail is
subsequently removed by cleavage during mRNA processing. On the other hand, RNA
polymerases I and III require termination signals. Genes transcribed by RNA polymerase I
contain a specific 18-nucleotide sequence that is recognized by a termination protein. The
process of termination in RNA polymerase III involves an mRNA hairpin similar to rho-
independent termination of transcription in prokaryotes.
 EU Test Material

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe the different steps in RNA processing

 Understand the significance of exons, introns, and splicing for mRNAs
 Explain how tRNAs and rRNAs are processed

After transcription, eukaryotic pre-mRNAs must undergo several processing steps before
they can be translated. Eukaryotic (and prokaryotic) tRNAs and rRNAs also undergo
processing before they can function as components in the protein-synthesis machinery.

mRNA Processing

The eukaryotic pre-mRNA undergoes extensive processing before it is ready to be translated.

Eukaryotic protein-coding sequences are not continuous, as they are in prokaryotes. The
coding sequences (exons) are interrupted by noncoding introns, which must be removed to
make a translatable mRNA. The additional steps involved in eukaryotic mRNA maturation
also create a molecule with a much longer half-life than a prokaryotic mRNA. Eukaryotic
mRNAs last for several hours, whereas the typical E. coli mRNA lasts no more than five
seconds.

Pre-mRNAs are first coated in RNA-stabilizing proteins; these protect the pre-mRNA from
degradation while it is processed and exported out of the nucleus. The three most important
steps of pre-mRNA processing are the addition of stabilizing and signaling factors at the 5'
and 3' ends of the molecule, and the removal of the introns (Figure 11). In rare cases, the
mRNA transcript can be “edited” after it is transcribed.

Figure .11 Eukaryotic pre-mRNA processing. In addition to 5’ Cap and 3’ Poly-A Tail
addition, introns must be precisely removed and exons joined to generate a functional
mRNA. Nucleotides upstream (towards the 5’cap) of the translation START codon are part of
the 5’ untranslated region (5’ UTR). Nucleotides downstream (towards 3’end) of the STOP
codon form the 3’ UTR. Both 5’ and 3’ UTRs are important for regulating translation
 EU Test Material

initiation and mRNA stability. Credit: Rao, A., Ryan, K. Fletcher, S. and Tag, A. Department
of Biology, Texas A&M University.

EVOLUTION CONNECTION

Evolution Connection

RNA Editing in TrypanosomesThe trypanosomes are a group of protozoa that include the
pathogen Trypanosoma brucei, which causes nagana in cattle and sleeping sickness in
humans throughout great areas of Africa (Figure 12). The trypanosome is carried by biting
flies in the genus Glossina (commonly called tsetse flies). Trypanosomes, and virtually all
other eukaryotes, have organelles called mitochondria that supply the cell with chemical
energy. Mitochondria are organelles that express their own DNA and are believed to be the
remnants of a symbiotic relationship between a eukaryote and an engulfed prokaryote. The
mitochondrial DNA of trypanosomes exhibit an interesting exception to the central dogma:
their pre-mRNAs do not have the correct information to specify a functional protein. Usually,
this is because the mRNA is missing several U nucleotides. The cell performs an additional
RNA processing step called RNA editing to remedy this.

Figure .12 Trypanosoma brucei is the causative agent of sleeping sickness in humans. The
mRNAs of this pathogen must be modified by the addition of nucleotides before protein
synthesis can occur. (credit: modification of work by Torsten Ochsenreiter)

Other genes in the mitochondrial genome encode 40- to 80-nucleotide guide RNAs. One or
more of these molecules interacts by complementary base pairing with some of the
nucleotides in the pre-mRNA transcript. However, the guide RNA has more A nucleotides
than the pre-mRNA has U nucleotides with which to bind. In these regions, the guide RNA
loops out. The 3' ends of guide RNAs have a long poly-U tail, and these U bases are inserted
in regions of the pre-mRNA transcript at which the guide RNAs are looped. This process is
entirely mediated by RNA molecules. That is, guide RNAs—rather than proteins—serve as
the catalysts in RNA editing.

RNA editing is not just a phenomenon of trypanosomes. In the mitochondria of some plants,
almost all pre-mRNAs are edited. RNA editing has also been identified in mammals such as
rats, rabbits, and even humans. What could be the evolutionary reason for this additional step
 EU Test Material

in pre-mRNA processing? One possibility is that the mitochondria, being remnants of ancient
prokaryotes, have an equally ancient RNA-based method for regulating gene expression. In
support of this hypothesis, edits made to pre-mRNAs differ depending on cellular conditions.
Although speculative, the process of RNA editing may be a holdover from a primordial time
when RNA molecules, instead of proteins, were responsible for catalyzing reactions.

5' Capping

While the pre-mRNA is still being synthesized, a 7-methylguanosine cap is added to the 5'
end of the growing transcript by a phosphate linkage. This functional group protects the
nascent mRNA from degradation. In addition, factors involved in protein synthesis recognize
the cap to help initiate translation by ribosomes.

3' Poly-A Tail

Once elongation is complete, the pre-mRNA is cleaved by an endonuclease between an

AAUAAA consensus sequence and a GU-rich sequence, leaving the AAUAAA sequence on
the pre-mRNA. An enzyme called poly-A polymerase then adds a string of approximately
200 A residues, called the poly-A tail. This modification further protects the pre-mRNA from
degradation and is also the binding site for a protein necessary for exporting the processed
mRNA to the cytoplasm.

Pre-mRNA Splicing

Eukaryotic genes are composed of exons, which correspond to protein-coding sequences (ex-
on signifies that they are expressed), and intervening sequences called introns (int-ron
denotes their intervening role), which may be involved in gene regulation but are removed
from the pre-mRNA during processing. Intron sequences in mRNA do not encode functional
proteins.

The discovery of introns came as a surprise to researchers in the 1970s who expected that
pre-mRNAs would specify protein sequences without further processing, as they had
observed in prokaryotes. The genes of higher eukaryotes very often contain one or more
introns. These regions may correspond to regulatory sequences; however, the biological
significance of having many introns or having very long introns in a gene is unclear. It is
possible that introns slow down gene expression because it takes longer to transcribe pre-
mRNAs with lots of introns. Alternatively, introns may be nonfunctional sequence remnants
left over from the fusion of ancient genes throughout the course of evolution. This is
supported by the fact that separate exons often encode separate protein subunits or domains.
For the most part, the sequences of introns can be mutated without ultimately affecting the
protein product.

All of a pre-mRNA’s introns must be completely and precisely removed before protein
synthesis. If the process errs by even a single nucleotide, the reading frame of the rejoined
exons would shift, and the resulting protein would be dysfunctional. The process of removing
introns and reconnecting exons is called splicing (Figure 13). Introns are removed and
degraded while the pre-mRNA is still in the nucleus. Splicing occurs by a sequence-specific
mechanism that ensures introns will be removed and exons rejoined with the accuracy and
precision of a single nucleotide. Although the intron itself is noncoding, the beginning and
end of each intron is marked with specific nucleotides: GU at the 5' end and AG at the 3' end
 EU Test Material

of the intron. The splicing of pre-mRNAs is conducted by complexes of proteins and RNA
molecules called spliceosomes.

VISUAL CONNECTION

Visual Connection

Figure .13 Pre-mRNA splicing involves the precise removal of introns from the primary
RNA transcript. The splicing process is catalyzed by protein complexes called spliceosomes
that are composed of proteins and RNA molecules called small nuclear RNAs (snRNAs).
Spliceosomes recognize sequences at the 5' and 3' end of the intron. Rao, A. and Ryan, K.
Department of Biology, Texas A&M University.

Errors in splicing are implicated in cancers and other human diseases. What kinds of
mutations might lead to splicing errors? Think of different possible outcomes if splicing
errors occur.

Note that more than 70 individual introns can be present, and each has to undergo the process
of splicing—in addition to 5' capping and the addition of a poly-A tail—just to generate a
single, translatable mRNA molecule.

Processing of tRNAs and rRNAs

The tRNAs and rRNAs are structural molecules that have roles in protein synthesis; however,
these RNAs are not themselves translated. Pre-rRNAs are transcribed, processed, and
assembled into ribosomes in the nucleolus. Pre-tRNAs are transcribed and processed in the
nucleus and then released into the cytoplasm where they are linked to free amino acids for
protein synthesis.

Most of the tRNAs and rRNAs in eukaryotes and prokaryotes are first transcribed as a long
precursor molecule that spans multiple rRNAs or tRNAs. Enzymes then cleave the precursors
into subunits corresponding to each structural RNA. Some of the bases of pre-rRNAs
are methylated; that is, a –CH3 methyl functional group is added for stability. Pre-tRNA
 EU Test Material

molecules also undergo methylation. As with pre-mRNAs, subunit excision occurs in

eukaryotic pre-RNAs destined to become tRNAs or rRNAs.

Mature rRNAs make up approximately 50 percent of each ribosome. Some of a ribosome’s

RNA molecules are purely structural, whereas others have catalytic or binding activities.
Mature tRNAs take on a three-dimensional structure through local regions of base pairing
stabilized by intramolecular hydrogen bonding. The tRNA folds to position the amino acid
binding site at one end and the anticodon at the other end (Figure 14). The anticodon is a
three-nucleotide sequence in a tRNA that interacts with an mRNA codon through
complementary base pairing.

Figure .14 This is a space-filling model of a tRNA molecule that adds the amino acid
phenylalanine to a growing polypeptide chain. The anticodon AAG binds the Codon UUC on
the mRNA. The amino acid phenylalanine is attached to the other end of the tRNA.
 EU Test Material

Learning Objectives

By the end of this section, you will be able to do the following:

 Describe the different steps in protein synthesis

 Discuss the role of ribosomes in protein synthesis

The synthesis of proteins consumes more of a cell’s energy than any other metabolic process.
In turn, proteins account for more mass than any other component of living organisms (with
the exception of water), and proteins perform virtually every function of a cell. The process
of translation, or protein synthesis, involves the decoding of an mRNA message into a
polypeptide product. Amino acids are covalently strung together by interlinking peptide
bonds in lengths ranging from approximately 50 to more than 1000 amino acid residues. Each
individual amino acid has an amino group (NH2) and a carboxyl (COOH) group.
Polypeptides are formed when the amino group of one amino acid forms an amide (i.e.,
peptide) bond with the carboxyl group of another amino acid (Figure 15). This reaction is
catalyzed by ribosomes and generates one water molecule.

Figure 15 A peptide bond links the carboxyl end of one amino acid with the amino end of
another, producing one water molecule during the process. For simplicity in this image, only
the functional groups involved in the peptide bond are shown. The R and R' designations
refer to the rest of each amino acid structure.

The Protein Synthesis Machinery

In addition to the mRNA template, many molecules and macromolecules contribute to the
process of translation. The composition of each component may vary across species; for
example, ribosomes may consist of different numbers of rRNAs and polypeptides depending
on the organism. However, the general structures and functions of the protein synthesis
machinery are comparable from bacteria to human cells. Translation requires the input of an
mRNA template, ribosomes, tRNAs, and various enzymatic factors. (Note: A ribosome can
be thought of as an enzyme whose amino acid binding sites are specified by mRNA.)
 EU Test Material

Ribosomes

Even before an mRNA is translated, a cell must invest energy to build each of its ribosomes.
In E. coli, there are between 10,000 and 70,000 ribosomes present in each cell at any given
time. A ribosome is a complex macromolecule composed of structural and catalytic rRNAs,
and many distinct polypeptides. In eukaryotes, the nucleolus is completely specialized for the
synthesis and assembly of rRNAs.

Ribosomes exist in the cytoplasm of prokaryotes and in the cytoplasm and rough
endoplasmic reticulum of eukaryotes. Mitochondria and chloroplasts also have their own
ribosomes in the matrix and stroma, which look more similar to prokaryotic ribosomes (and
have similar drug sensitivities) than the ribosomes just outside their outer membranes in the
cytoplasm. Ribosomes dissociate into large and small subunits when they are not
synthesizing proteins and reassociate during the initiation of translation. In E. coli, the small
subunit is described as 30S, and the large subunit is 50S, for a total of 70S (recall that
Svedberg units are not additive). Mammalian ribosomes have a small 40S subunit and a large
60S subunit, for a total of 80S. The small subunit is responsible for binding the mRNA
template, whereas the large subunit sequentially binds tRNAs. Each mRNA molecule is
simultaneously translated by many ribosomes, all synthesizing protein in the same direction:
reading the mRNA from 5' to 3' and synthesizing the polypeptide from the N terminus to the
C terminus. The complete mRNA/poly-ribosome structure is called a polysome.

tRNAs

The tRNAs are structural RNA molecules that were transcribed from genes by RNA
polymerase III. Depending on the species, 40 to 60 types of tRNAs exist in the cytoplasm.
Transfer RNAs serve as adaptor molecules. Each tRNA carries a specific amino acid and
recognizes one or more of the mRNA codons that define the order of amino acids in a
protein. Aminoacyl-tRNAs bind to the ribosome and add the corresponding amino acid to the
polypeptide chain. Therefore, tRNAs are the molecules that actually “translate” the language
of RNA into the language of proteins.

Of the 64 possible mRNA codons—or triplet combinations of A, U, G, and C—three specify

the termination of protein synthesis and 61 specify the addition of amino acids to the
polypeptide chain. Of these 61, one codon (AUG) also encodes the initiation of translation.
Each tRNA anticodon can base pair with one or more of the mRNA codons for its amino
acid. For instance, if the sequence CUA occurred on an mRNA template in the proper reading
frame, it would bind a leucine tRNA expressing the complementary sequence, GAU. The
ability of some tRNAs to match more than one codon is what gives the genetic code its
blocky structure.

As the adaptor molecules of translation, it is surprising that tRNAs can fit so much specificity
into such a small package. Consider that tRNAs need to interact with three factors: 1) they
must be recognized by the correct aminoacyl synthetase (see below); 2) they must be
recognized by ribosomes; and 3) they must bind to the correct sequence in mRNA.
 EU Test Material

Figure .16 The ribosome and its function. The ribosome is responsible for translating the
mRNA into protein. A. The ribosome consists of a large and small ribosomal subunit.
Assembly of the subunits on the mRNA forms three tRNA binding sites. B. During
translation, charged tRNAs enter the Acceptor site, and the anticodon on the tRNA base pairs
with the codon in the mRNA. After the incoming amino acid forms a peptide bond with the
growing polypeptide chain, the ribosome will move three nucleotides toward the 3’ end of the
mRNA. This movement will transfer the tRNA with the growing polypeptide to the Peptidyl-
tRNA binding site and allow the empty tRNA to exit at the Exit site. Credit: Rao, A., Ryan,
K. and Fletcher, S. Department of Biology, Texas A&M University.

Aminoacyl tRNA Synthetases

The process of pre-tRNA synthesis by RNA polymerase III only creates the RNA portion of
the adaptor molecule. The corresponding amino acid must be added later, once the tRNA is
processed and exported to the cytoplasm. Through the process of tRNA “charging,” each
tRNA molecule is linked to its correct amino acid by one of a group of enzymes
 EU Test Material

called aminoacyl tRNA synthetases. At least one type of aminoacyl tRNA synthetase exists
for each of the 20 amino acids; the exact number of aminoacyl tRNA synthetases varies by
species. These enzymes first bind and hydrolyze ATP to catalyze a high-energy bond
between an amino acid and adenosine monophosphate (AMP); a pyrophosphate molecule is
expelled in this reaction. The activated amino acid is then transferred to the tRNA, and AMP
is released. The term "charging" is appropriate, since the high-energy bond that attaches an
amino acid to its tRNA is later used to drive the formation of the peptide bond. Each tRNA is
named for its amino acid.

Figure .17 Charging of tRNAs with correct amino acids. Aminoacyl-tRNA synthetases
catalyze covalent bond formation between the tRNA and the correct amino acid in
preparation for translation. Because there are multiple amino acids, there are multiple
different tRNA synthetases. All of the synthetases require energy, in the form of ATP, to
make sure the correct amino acid is attached to the tRNA with the correct anticodon
sequence. Credit: Rao, A., Ryan, K. and Tag, A. Department of Biology, Texas A&M
University.

The Mechanism of Protein Synthesis

As with mRNA synthesis, protein synthesis can be divided into three phases: initiation,
elongation, and termination. The process of translation is similar in prokaryotes and
eukaryotes. Here we’ll explore how translation occurs in E. coli, a representative prokaryote,
and specify any differences between prokaryotic and eukaryotic translation.

Initiation of Translation

Protein synthesis begins with the formation of an initiation complex. In E. coli, this complex
involves the small 30S ribosome, the mRNA template, three initiation factors (IFs; IF-1, IF-2,
and IF-3), and a special initiator tRNA, called tRNAMetf.
 EU Test Material

In E. coli mRNA, a sequence upstream of the first AUG codon, called the Shine-Dalgarno
sequence (AGGAGG), interacts with the rRNA molecules that compose the ribosome. This
interaction anchors the 30S ribosomal subunit at the correct location on the mRNA template.
Guanosine triphosphate (GTP), which is a purine nucleotide triphosphate, acts as an energy
source during translation—both at the start of elongation and during the ribosome’s
translocation. Binding of the mRNA to the 30S ribosome also requires IF-III.

The initiator tRNA then interacts with the start codon AUG (or rarely, GUG). This tRNA
carries the amino acid methionine, which is formylated after its attachment to the tRNA. The
formylation creates a "faux" peptide bond between the formyl carboxyl group and the amino
group of the methionine. Binding of the fMet-tRNAMetf is mediated by the initiation factor IF-
2. The fMet begins every polypeptide chain synthesized by E. coli, but it is usually removed
after translation is complete. When an in-frame AUG is encountered during translation
elongation, a non-formylated methionine is inserted by a regular Met-tRNAMet. After the
formation of the initiation complex, the 30S ribosomal subunit is joined by the 50S subunit to
form the translation complex. In eukaryotes, a similar initiation complex forms, comprising
mRNA, the 40S small ribosomal subunit, eukaryotic IFs, and nucleoside triphosphates (GTP
and ATP). The methionine on the charged initiator tRNA, called Met-tRNAi, is not
formylated. However, Met-tRNAi is distinct from other Met-tRNAs in that it can bind IFs.

Instead of depositing at the Shine-Dalgarno sequence, the eukaryotic initiation complex

recognizes the 7-methylguanosine cap at the 5' end of the mRNA. A cap-binding protein
(CBP) and several other IFs assist the movement of the ribosome to the 5' cap. Once at the
cap, the initiation complex tracks along the mRNA in the 5' to 3' direction, searching for the
AUG start codon. Many eukaryotic mRNAs are translated from the first AUG, but this is not
always the case. According to Kozak’s rules, the nucleotides around the AUG indicate
whether it is the correct start codon. Kozak’s rules state that the following consensus
sequence must appear around the AUG of vertebrate genes: 5'-gccRccAUGG-3'. The R (for
purine) indicates a site that can be either A or G, but cannot be C or U. Essentially, the closer
the sequence is to this consensus, the higher the efficiency of translation.

Once the appropriate AUG is identified, the other proteins and CBP dissociate, and the 60S
subunit binds to the complex of Met-tRNAi, mRNA, and the 40S subunit. This step
completes the initiation of translation in eukaryotes.

Translation, Elongation, and Termination

In prokaryotes and eukaryotes, the basics of elongation are the same, so we will review
elongation from the perspective of E. coli. When the translation complex is formed, the tRNA
binding region of the ribosome consists of three compartments. The A (aminoacyl) site binds
incoming charged aminoacyl tRNAs. The P (peptidyl) site binds charged tRNAs carrying
amino acids that have formed peptide bonds with the growing polypeptide chain but have not
yet dissociated from their corresponding tRNA. The E (exit) site releases dissociated tRNAs
so that they can be recharged with free amino acids. The initiating methionyl-tRNA,
however, occupies the P site at the beginning of the elongation phase of translation in both
prokaryotes and eukaryotes.

During translation elongation, the mRNA template provides tRNA binding specificity. As the
ribosome moves along the mRNA, each mRNA codon comes into register, and specific
binding with the corresponding charged tRNA anticodon is ensured. If mRNA were not
 EU Test Material

present in the elongation complex, the ribosome would bind tRNAs nonspecifically and
randomly.

Elongation proceeds with charged tRNAs sequentially entering and leaving the ribosome as
each new amino acid is added to the polypeptide chain. Movement of a tRNA from A to P to
E site is induced by conformational changes that advance the ribosome by three bases in the
3' direction. The energy for each step along the ribosome is donated by elongation factors that
hydrolyze GTP. GTP energy is required both for the binding of a new aminoacyl-tRNA to the
A site and for its translocation to the P site after formation of the peptide bond. Peptide bonds
form between the amino group of the amino acid attached to the A-site tRNA and the
carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide
bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the
50S ribosomal subunit. The energy for each peptide bond formation is derived from the high-
energy bond linking each amino acid to its tRNA. After peptide bond formation, the A-site
tRNA that now holds the growing peptide chain moves to the P site, and the P-site tRNA that
is now empty moves to the E site and is expelled from the ribosome (Figure 18). Amazingly,
the E. coli translation apparatus takes only 0.05 seconds to add each amino acid, meaning that
a 200-amino-acid protein can be translated in just 10 seconds.

VISUAL CONNECTION

Visual Connection

Figure .18 Translation begins when an initiator tRNA anticodon recognizes a start codon on
mRNA bound to a small ribosomal subunit. The large ribosomal subunit joins the small
subunit, and a second tRNA is recruited. As the mRNA moves relative to the ribosome,
successive tRNAs move through the ribosome and the polypeptide chain is formed. Entry of
a release factor into the A site terminates translation and the components dissociate.
 EU Test Material

Many antibiotics inhibit bacterial protein synthesis. For example, tetracycline blocks the A
site on the bacterial ribosome, and chloramphenicol blocks peptidyl transfer. What specific
effect would you expect each of these antibiotics to have on protein synthesis?

Tetracycline would directly affect:

a. tRNA binding to the ribosome

b. ribosome assembly
c. growth of the protein chain

Chloramphenicol would directly affect:

a. tRNA binding to the ribosome

b. ribosome assembly
c. growth of the protein chain

Termination of translation occurs when a nonsense codon (UAA, UAG, or UGA) is

encountered. Upon aligning with the A site, these nonsense codons are recognized by protein
release factors that resemble tRNAs. The releasing factors in both prokaryotes and
eukaryotes instruct peptidyl transferase to add a water molecule to the carboxyl end of the P-
site amino acid. This reaction forces the P-site amino acid to detach from its tRNA, and the
newly made protein is released. The small and large ribosomal subunits dissociate from the
mRNA and from each other; they are recruited almost immediately into another translation
initiation complex. After many ribosomes have completed translation, the mRNA is degraded
so the nucleotides can be reused in another transcription reaction.

Figure .19 Translation Termination is Active. Translation is terminated when a STOP codon
is in the A-site of the ribosome. Since there are no tRNAs corresponding to the STOP codons,
the Release Factor protein enters and catalyzes the hydrolysis between the last amino acid
and its tRNA. This hydrolysis releases the free carboxyl terminus (C-term) of the protein.
Additional factors use the energy in GTP hydrolysis to disassemble the large and small
 EU Test Material

ribosomal subunits and mRNA. Credit: Rao, A. and Ryan, K. Department of Biology, Texas
A&M University.

Protein Folding, Modification, and Targeting

During and after translation, individual amino acids may be chemically modified, signal
sequences appended, and the new protein “folded” into a distinct three-dimensional structure
as a result of intramolecular interactions. A signal sequence is a short sequence at the amino
end of a protein that directs it to a specific cellular compartment. These sequences can be
thought of as the protein’s “train ticket” to its ultimate destination, and are recognized by
signal-recognition proteins that act as conductors. For instance, a specific signal sequence
terminus will direct a protein to the mitochondria or chloroplasts (in plants). Once the protein
reaches its cellular destination, the signal sequence is usually clipped off.

Many proteins fold spontaneously, but some proteins require helper molecules,
called chaperones, to prevent them from aggregating during the complicated process of
folding. Even if a protein is properly specified by its corresponding mRNA, it could take on a
completely dysfunctional shape if abnormal temperature or pH conditions prevent it from
folding correctly.

Figure .20 Proteins are co-translationally targeted into the ER for secretion. Proteins that will
be secreted from the cell will contain a signal sequence at the N-terminus. The signal will be
recognized by SRP as soon as the amino acids emerge from the ribosome, and the ribosome
will be targeted to the translocation channel in the ER membrane. The rest of the protein will
go directly from the ribosome, across the ER membrane and into the ER lumen. From the ER,
proteins can be secreted from the cell via vesicle trafficking. Credit: Rao, A. and Ryan, K.
Department of Biology, Texas A&M University.
EU Test Material