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Tuesday, November 28, 2006
ALL ABOUT CHEMISTRY
Chemistry (from Greek χημεία khemeia[1] meaning "alchemy") is the science of matter at the atomic to molecular scale, dealing primarily with collections of atoms, such as molecules, crystals, and metals. Chemistry deals with the composition and statistical properties of such structures, as well as their transformations and interactions to become materials encountered in everyday life. Chemistry also deals with understanding the properties and interactions of individual atoms with the purpose of applying that knowledge at the macroscopic level. According to modern chemistry, the physical properties of materials are generally determined by their structure at the atomic scale, which is itself defined by interatomic forces.
Introduction
Chemistry is often called the "central science" because it connects other sciences, such as physics, material science, nanotechnology, biology, pharmacy, medicine, bioinformatics, and geology.[2] These connections are formed through various sub-disciplines that utilize concepts from multiple scientific disciplines. For example, physical chemistry involves applying the principles of physics to materials at the atomic and molecular level.
Chemistry pertains to the interactions of matter. These interactions may be between two material substances or between matter and energy, especially in conjunction with the First Law of Thermodynamics. Traditional chemistry involves interactions between substances in chemical reactions, where one or more substances become one or more other substances. Sometimes these reactions are driven by energetic (enthalpic) considerations, such as when two highly energetic substances such as elemental hydrogen and oxygen react to form the less energetic substance water. Chemical reactions may be facilitated by a catalyst, which is generally another chemical substance present within the reaction media but unconsumed (such as sulfuric acid catalyzing the electrolysis of water) or a non-material phenomenon (such as electromagnetic radiation in photochemical reactions). Traditional chemistry also deals with the analysis of chemicals both in and apart from a reaction, as in spectroscopy.
All ordinary matter consists of atoms or the subatomic components that make up atoms; protons, electrons and neutrons. Atoms may be combined to produce more complex forms of matter such as ions, molecules or crystals. The structure of the world we commonly experience and the properties of the matter we commonly interact with are determined by properties of chemical substances and their interactions. Steel is harder than iron because its atoms are bound together in a more rigid crystalline lattice. Wood burns or undergoes rapid oxidation because it can react spontaneously with oxygen in a chemical reaction above a certain temperature.
Substances tend to be classified in terms of their energy or phase as well as their chemical compositions. The three phases of matter at low energy are Solid, Liquid and Gas. Solids have fixed structures at room temperature which can resist gravity and other weak forces attempting to rearrange them, due to their tight bonds. Liquids have limited bonds, with no structure and flow with gravity. Gases have no bonds and act as free particles. Another way to view the three phases is by volume and shape: roughly speaking, solids have fixed volume and shape, liquids have fixed volume but no fixed shape, and gases have neither fixed volume nor fixed shape.
Water (H2O) is a liquid at room temperature because its molecules are bound by intermolecular forces called Hydrogen bonds. Hydrogen sulfide (H2S) on the other hand is a gas at room temperature and standard pressure, as its molecules are bound by weaker dipole-dipole interactions. The hydrogen bonds in water have enough energy to keep the water molecules from separating from each other but not from sliding around, making it a liquid at temperatures between 0 °C and 100 °C at sea level. Lowering the temperature or energy further, allows for a tighter organization to form, creating a solid, and releasing energy. Increasing the energy (see heat of fusion) will melt the ice although the temperature will not change until all the ice is melted. Increasing the temperature of the water will eventually cause boiling (see heat of vaporization) when there is enough energy to overcome the polar attractions between individual water molecules (100 °C at 1 atmosphere of pressure), allowing the H2O molecules to disperse enough to be a gas. Note that in each case there is energy required to overcome the intermolecular attractions and thus allow the molecules to move away from each other.
Scientists who study chemistry are known as chemists. Most chemists specialize in one or more sub-disciplines. The chemistry taught at the high school or early college level is often called "general chemistry" and is intended to be an introduction to a wide variety of fundamental concepts and to give the student the tools to continue on to more advanced subjects. Many concepts presented at this level are often incomplete and technically inaccurate, yet they are of extraordinary utility. Chemists regularly use these simple, elegant tools and explanations in their work because they have been proven to accurately model a very wide array of chemical reactivity, are generally sufficient, and more precise solutions may be prohibitively difficult to obtain.
The science of chemistry is historically a recent development but has its roots in alchemy which has been practiced for millennia throughout the world. The word chemistry is directly derived from the word alchemy; however, the etymology of alchemy is unclear (see alchemy).
History of chemistry
The roots of chemistry can be traced to the phenomenon of burning. Fire was a mystical force that transformed one substance into another and thus was of primary interest to mankind. It was fire that led to the discovery of iron and glass. After gold was discovered and became a precious metal, many people were interested to find a method that could convert other substances into gold. This led to the protoscience called Alchemy. Alchemy was practiced by many cultures throughout history and often contained a mixture of philosophy, mysticism, and protoscience (see Alchemy).
Alchemists discovered many chemical processes that led to the development of modern chemistry. As history progressed the more notable alchemists (esp. Geber and Paracelsus) evolved alchemy away from philosophy and mysticism and developed more systematic and scientific approaches. The first alchemist considered to apply the scientific method to alchemy and to distinguish chemistry from alchemy was Robert Boyle (1627–1691); however, chemistry as we know it today was invented by Antoine Lavoisier with his law of Conservation of mass in 1783. The discoveries of the chemical elements has a long history culminating in the creation of the periodic table of the chemical elements by Dmitri Mendeleyev.
The Nobel Prize in Chemistry created in 1901 gives an excellent overview of chemical discovery in the past 100 years. In the early part of the 20th century the subatomic nature of atoms were revealed and the science of quantum mechanics began to explain the physical nature of the chemical bond. By the mid 20th century chemistry had developed to the point of being able to understand and predict aspects of biology spawning the field of biochemistry.
The chemical industry represents an important economic activity. The global top 50 chemical producers in 2004 had sales of 587 billion US dollars with a profit margin of 8.1% and research and development spending of 2.1% of total chemical sales.[3]
Chemistry typically is divided into several major sub-disciplines. There are also several main cross-disciplinary and more specialized fields of chemistry.
• Analytical chemistry is the analysis of material samples to gain an understanding of their chemical composition and structure. Analytical chemistry incorporates standardized experimental methods in chemistry. These methods may be used in all subdisciplines of chemistry, excluding purely theoretical chemistry.
• Biochemistry is the study of the chemicals, chemical reactions and chemical interactions that take place in living organisms. Biochemistry and organic chemistry are closely related, as in medicinal chemistry or neurochemistry. Biochemistry is also associated with molecular biology and genetics.
• Inorganic chemistry is the study of the properties and reactions of inorganic compounds. The distinction between organic and inorganic disciplines is not absolute and there is much overlap, most importantly in the sub-discipline of organometallic chemistry.
• Organic chemistry is the study of the structure, properties, composition, mechanisms, and reactions of organic compounds. An organic compound is defined as any compound based on a carbon skeleton.
• Physical chemistry is the study of the physical and fundamental basis of chemical systems and processes. In particular, the energetics and dynamics of such systems and processes are of interest to physical chemists. Important areas of study include chemical thermodynamics, chemical kinetics, electrochemistry, statistical mechanics, and spectroscopy. Physical chemistry has large overlap with molecular physics. Physical chemistry involves the use of calculus in deriving equations. It is usually associated with quantum chemistry and theoretical chemistry.
• Theoretical chemistry is the study of chemistry via fundamental theoretical reasoning (usually within mathematics or physics). In particular the application of quantum mechanics to chemistry is called quantum chemistry. Since the end of the Second World War, the development of computers has allowed a systematic development of computational chemistry, which is the art of developing and applying computer programs for solving chemical problems. Theoretical chemistry has large overlap with (theoretical and experimental) condensed matter physics and molecular physics. Essentially from reductionism theoretical chemistry is just physics, just like fundamental biology is just chemistry and physics.
• Nuclear chemistry is the study of how subatomic particles come together and make nuclei. Modern Transmutation is a large component of nuclear chemistry, and the table of nuclides is an important result and tool for this field.
Other fields include Astrochemistry, Atmospheric chemistry, Chemical Engineering, Chemo-informatics, Electrochemistry, Environmental chemistry, Flow chemistry, Geochemistry, Green chemistry, History of chemistry, Materials science, Medicinal chemistry, Molecular Biology, Molecular genetics, Nanotechnology, Organometallic chemistry, Petrochemistry, Pharmacology, Photochemistry, Phytochemistry, Polymer chemistry, Solid-state chemistry, Sonochemistry, Supramolecular chemistry, Surface chemistry, and Thermochemistry.
Fundamental concepts
Nomenclature
Nomenclature refers to the system for naming chemical compounds. There are well-defined systems in place for naming chemical species. Organic compounds are named according to the organic nomenclature system. Inorganic compounds are named according to the inorganic nomenclature system.
Atoms
An atom is a collection of matter consisting of a positively charged core (the atomic nucleus) which contains protons and neutrons, and which maintains a number of electrons to balance the positive charge in the nucleus.
Elements
An element is a class of atoms which have the same number of protons in the nucleus. This number is known as the atomic number of the element. For example, all atoms with 6 protons in their nuclei are atoms of the chemical element carbon, and all atoms with 92 protons in their nuclei are atoms of the element uranium.
The most convenient presentation of the chemical elements is in the periodic table of the chemical elements, which groups elements by atomic number. Due to its ingenious arrangement, groups, or columns, and periods, or rows, of elements in the table either share several chemical properties, or follow a certain trend in characteristics such as atomic radius, electronegativity, electron affinity, and etc. Lists of the elements by name, by symbol, and by atomic number are also available. In addition, several isotopes of an element may exist.
Ions
An ion is a charged species, or an atom or a molecule that has lost or gained one or more electrons. Positively charged cations (e.g. sodium cation Na+) and negatively charged anions (e.g. chloride Cl−) can form neutral salts (e.g. sodium chloride NaCl). Examples of polyatomic ions that do not split up during acid-base reactions are hydroxide (OH−) and phosphate (PO43−).
Compounds
A compound is a substance with a fixed ratio of chemical elements which determines the composition, and a particular organization which determines chemical properties. For example, water is a compound containing hydrogen and oxygen in the ratio of two to one, with the oxygen between the hydrogens, and an angle of 104.5° between them. Compounds are formed and interconverted by chemical reactions.
Molecules
A molecule is the smallest indivisible portion of a pure compound or element that retains a set of unique chemical properties.
Substance
A chemical substance can be an element, compound or a mixture of compounds, elements or compounds and elements. Most of the matter we encounter in our daily life are one or another kind of mixtures, e.g. air, alloys, biomass etc.
Chemical bond
A chemical bond is the multipole balance between the positive charges in the nuclei and the negative charges oscillating about them. More than simple attraction and repulsion, the energies and distributions characterize the availability of an electron to bond to another atom. These potentials create the interactions which holds together atoms in molecules or crystals. In many simple compounds, Valence Bond Theory, the Valence Shell Electron Pair Repulsion model (VSEPR), and the concept of oxidation number can be used to predict molecular structure and composition. Similarly, theories from classical physics can be used to predict many ionic structures. With more complicated compounds, such as metal complexes, valence bond theory fails and alternative approaches, primarily based on principles of quantum chemistry such as the molecular orbital theory, are necessary. See diagram on electronic orbitals.
States of matter
phase is a set of states of a chemical system that have similar bulk structural properties, over a range of conditions, such as pressure or temperature. Physical properties, such as density and refractive index tend to fall within values characteristic of the phase. The phase of matter is defined by the phase transition, which is when energy put into or taken out of the system goes into rearranging the structure of the system, instead of changing the bulk conditions.
Sometimes the distinction between phases can be continuous instead of having a discrete boundary, in this case the matter is considered to be in a supercritical state. When three states meet based on the conditions, it is known as a triple point and since this is invariant, it is a convenient way to define a set of conditions.
The most familiar examples of phases are solids, liquids, and gases. Less familiar phases include plasmas, Bose-Einstein condensates and fermionic condensates and the paramagnetic and ferromagnetic phases of magnetic materials. Even the familiar ice has many different phases, depending on the pressure and temperature of the system. While most familiar phases deal with three-dimensional systems, it is also possible to define analogs in two-dimensional systems, which has received attention for its relevance to systems in biology.
Chemical reactions
A Chemical reaction is a process that results in the interconversion of chemical substances. Such reactions can result in molecules attaching to each other to form larger molecules, molecules breaking apart to form two or more smaller molecules, or rearrangement of atoms within or across molecules. Chemical reactions usually involve the making or breaking of chemical bonds. For example, substances that react with oxygen to produce other substances are said to undergo oxidation; similarly a group of substances called acids or alkalis can react with one another to neutralize each other's effect, a phenomenon known as neutralization. Substances can also be dissociated or synthesized from other substances by various different chemical processes.
Quantum chemistry
Quantum chemistry mathematically describes the fundamental behavior of matter at the molecular scale. It is, in principle, possible to describe all chemical systems using this theory. In practice, only the simplest chemical systems may realistically be investigated in purely quantum mechanical terms, and approximations must be made for most practical purposes (e.g., Hartree-Fock, post Hartree-Fock or Density functional theory, see computational chemistry for more details). Hence a detailed understanding of quantum mechanics is not necessary for most chemistry, as the important implications of the theory (principally the orbital approximation) can be understood and applied in simpler terms.
In quantum mechanics (several applications in computational chemistry and quantum chemistry), the Hamiltonian, or the physical state, of a particle can be expressed as the sum of two operators, one corresponding to kinetic energy and the other to potential energy. The Hamiltonian in the Schrödinger wave equation used in quantum chemistry does not contain terms for the spin of the electron.
Solutions of the Schrödinger equation for the hydrogen atom gives the form of the wave function for atomic orbitals, and the relative energy of say the 1s,2s,2p and 3s orbitals. The orbital approximation can be used to understand the other atoms e.g. helium, lithium and carbon.
Chemical Laws
The most fundamental concept in chemistry is the law of conservation of mass, which states that there is no detectable change in the quantity of matter during an ordinary chemical reaction. Modern physics shows that it is actually energy that is conserved, and that energy and mass are related; a concept which becomes important in nuclear chemistry. Conservation of energy leads to the important concepts of equilibrium, thermodynamics, and kinetics.
Further laws of chemistry elaborate on the law of conservation of mass. Joseph Proust's law of definite composition says that pure chemicals are composed of elements in a definite formulation; we now know that the structural arrangement of these elements is also important.
Dalton's law of multiple proportions says that these chemicals will present themselves in proportions that are small whole numbers (i.e. 1:2 O:H in water); although in many systems (notably biomacromolecules and minerals) the ratios tend to require large numbers, and are frequently represented as a fraction. Such compounds are known as non-stoichiometric compounds
Interpersonal chemistry
In the fields of sociology, behavioral psychology, and evolutionary psychology, with specific reference to intimate relationships or romantic relationships, interpersonal chemistry is a reaction between two people or the spontaneous reaction of two people to each other, especially a mutual sense of attraction or understanding.[4] In a colloquial sense, it is often intuited that people can have either good chemistry or bad chemistry together. Other related terms are team chemistry, a phrase often used in sports, and business chemistry, as between two companies.[5] Recent developments in neurochemistry have begun to shed light on the nature of the "chemistry of love", in terms of measurable changes neurotransmitters such as oxytocin, serotonin, and dopamine.
Etymology
The word chemistry comes from the earlier study of alchemy, which is basically the quest to make gold from earthen starting materials. As to the origin of the word “alchemy” the question is a debatable one; it certainly has Greek origins, and some, following E. Wallis Budge, have also asserted Egyptian origins. Alchemy, generally, derives from the old French alkemie; and the Arabic al-kimia: "the art of transformation." The Arabs borrowed the word “kimia” from the Greeks when they conquered Alexandria in the year 642 AD.
Saturday, November 25, 2006
FUNGAL DIVERSITY AND LIFE CYCLE
INTRODUCTION: The fungal functional type is comprised of sessile heterotrophs with cell walls. Rather than ingesting food as animals do, fungal organisms absorb food across the cell wall. The assemblage of organisms termed fungi are classified into two general categories. First, are the true fungi (Kingdom Fungi) which evolved from motile, aquatic protozoa that are also ancestors to the animal kingdom. True fungi first evolved as the chytrids (Phylum Chytridiomycota) who produce an enlarged globular cell from which numerous filaments grow into the food source. Chytrids produce motile spores and gametes, and the vegetative cells are ceonocytic (many nuclei float around in one big cell). Chytrids gave rise to the Zygomycetes (Phylum Zygomycota), which produce no motile cells but form ceonocytic hyphae. From the Zygomycetes, the advanced fungi arose and in time formed the Phylum Dikaryomycota. Organisms in the Dikaryomycota produce hyphae comprised of individual nucleated cells separated by walls (septate hyphae) with each cell having two haploid nuclei in what is called an N + N configuration. The two main groups of dikaryotic fungi are the ascomycetes (sac-forming fungi) and the basidiomycetes (club-forming fungi). The majority of fungi affecting humans are ascomycetes and basidiomycetes.
The second category of fungal organisms is the pseudofungi, made up of various unrelated protista groups. The pseudofungi were formerly classified into the catch-all kingdom Protista but have recently been reclassified into more-specific kingdoms that reflect genetic relationships. Important pseudofungi are the Oomycetes (egg fungi and water molds), and the slime molds. Oomycetes are closely related to the stramenopilous algae - the brown algae, golden-brown algae and diatoms of the Kingdom Stramenopila. The close relationship between the oomycetes and the brown algae is evident in that both have cellulose walls, and they share the same type of flagella. Oomycetes descend from algae that lost their chloroplasts, and hence have adopted a heterotrophic life form. Oomycetes also produce filamentous hyphae to better absorb nutrients from the food source. Because they have no relationship with the chytrids, it is clear that the oomycetes and chytridiomycetes independently evolved the mycelial life form.
Slime molds evolved from various ancient protozoa and have little genetic affinity to other fungi or algae groups. They are animal-like in that they ingest food early in the life-cycle, but are fungal-like in that they produce walled sporangia and spores.
In today’s lab, you will have a chance to examine the diversity of both false and true fungi. The lab will begin with the false fungi and then follow an evolutionary sequence through the true fungi. Many specimens illustrate important reproductive phases in the life-cycles of these organisms. You should examine these carefully because reproductive features are important to understanding and distinguishing the various groups of fungi.
Although not required, we urge you to draw and label the specimens you examine. Our experience has been that drawings with appropriate labels are the best way to learn the features emphasized in lab and lecture. Drawings are also an excellent study tool to refresh your memory just prior to the exam
PART I: THE PSEUDOFUNGI
A) Slime molds (Kingdoms Myxomycota and Dictyosteliomycota): Slime molds are largely saprophytic and are typically found on decaying wood in moist forests. During the vegetative phase of the life cycle (see figure 16-6 on page 353 of your text), they begin life as independent amoebae, ingesting microscopic bits of organic debris. The free-living amoebae eventually swarm together to produce a multicellular blob called a plasmodium. After a while, the plasmodium forms cellulose walls around the nuclei and produces sporangia (or fructifications). The sporangia release large numbers of air-borne spores which germinate in the presence of water to form free-living amoebae, thus completing the life cycle. Slime molds defy simple categorization. They are animal-like in that they ingest food during the amoeboid- and plasmodial-phase. They are plant-like in their formation of cell walls and sporangia during reproduction. Because of the cell-walls formed during the reproductive phase they are considered fungal in nature. There are two major types of slime molds:
A) the Myxomycota are the acellular, or true plasmodial slime modes: The plasmodium stage of this group is made up of large blobs of ceonocytic protoplasm with many nuclei inside.
B) the Dictyosteliomycota are the cellular clime molds, where the plasmodium is made of individual cells separated by membranes (but not cell walls).
Observations on Display: Fruiting bodies of slime molds are readily found in Ontario forests in the autumn. Some will be on display for you to examine, along with a diagram of the life cycle.
B) Oomycetes: the water molds, or egg fungi (Kingdom Stramenopila)
Oomycetes are characterized by the formation of large egg-bearing cells on the tips of specialized hyphae termed oogonia (see figure 17-4 on page 374 of your text). Large, non-motile eggs form inside the oogonia and are fertilized by male-like hyphae termed antheridia. The antheridia grow into the egg and deposit the male gametes, which then fuse with the egg to form a zygote, termed the oospore. The oospore undergoes mitosis and forms a sporangia. The spores that are produced disperse to infect leaves, seedlings, fish and dead organisms.
Oomycetes are important saprophytes in aquatic habitats. In terrestrial habitats, they are generally parasitic. Important diseases caused by water molds are downy mildews (Peronospora), potato late blight (Phytophthora infestans), and damping off disease (Pythium spp.). We will examine three species, Achlya, a saprophytic water mold; Albugo candida, the white rust of mustard plants and Phytophthora infestans.
Examine the following cultures with the dissecting microscope:
1. Achlya whole mounts: Achlya is a water mold that grows on organic debris in lakes and rivers. It forms floating mycelia mats arising from the food material, and in these mats sexual reproduction occurs. On your bench are prepared slides for you to examine with the light microscope. Focus on the blue-green material in the center of the slide. This is a hyphal mat with oogonia.
Observe the large conspicuous oogonia within the mycelium. Inside the oogonia you can see zygotes (oospores), that will later divide to form sporangia. Examine the oogonia closely to see if additional hyphae are attached to it. These would be antheridia, which present the sperm nucleus to the egg cells
within the oogonia. Note the properties of the vegetative hyphae. Do you see cross walls, or are the cells continous within a filament?
2. Phytophthora infestans: This organism causes potato late blight, one of the worst crop diseases in the history of humanity. In the 1840’s, Phytophthora infestans was introduced to Europe from Peru. It rapidly spread across the continent, destroying much of the potato crop. In Ireland, peasants were particularly dependent on the potato for survival, and the crop losses in 1845-1848 killed millions of Irish and forced the migration of millions more to North America. In continental Europe, the loss of the potato crop led to widespread economic failure and social revolt. Many of the radical worker movements that would later influence world history, such as Marxism-Leninism, arose in the wake of the food crisis caused by the Phytophthora outbreak.
Examine the Phytophthora slide prepared fresh from a culture stained with cotton blue. Identify the round oogonia in the slide and examine the hyphae for cross walls between the cells. Are there any oospores within the Oogonia? Can you see any antheridial hyphae attached to the oogonia? If you do, show other students in your group.
3. Albugo candida (White Rust of Mustards): Albugo infects plants of the mustard family, forming white pustules on the leaves. These pustules are comprised of many asexual conidiospores bursting through the plants epidermis. Inside the plant, fungal hyphae form oogonia and antheridia, which will mate and form oospores. The oospores develop sporangia which disperse genetically-distinct spores. The two prepared slides show the extent of the infection by the parasitic white rust fungus.
Examine the prepared cross section showing Albugo conidiaspores bursting through the epidermis of a mustard fruit. Note the strings of conidiospores forming under the epidermis. The bulge formed by the mass of conidia produce the rust pustule. Upon rupturing, the conidiospores are released on the wind to start a series of new infections.
Examine the prepared slide of the Albugo sexual organs (oogonia and antheridia). The oogonia are evident as dense, red-staining circles among the cells of the leaf tissue.
Notes and Drawings:
PART II: THE TRUE FUNGI (KINGDOM FUNGI)
We have assembled a range of specimens from the Kingdom Fungi, and they are arranged for you to examine beginning with the most primitive (the Chytrids) and then progressing through the Zygomycetes to the Ascomycetes and Basidiomycetes.
A. CHYTRIDIOMYCOTA (The Chytrids):
Chytrids are mainly aquatic or parasitic. They are important decomposers of pollen, dead insects and seeds that fall into ponds and rivers. Others are parasites of algae, higher fungi, mosquitoes, rotifers, and water molds. The general body plan is to form a large, central ceonocytic globule that either directly invades the body of the host, or produces diminutive hyphae termed rhizomycelium, which invade the surface cells of the host. The rhizomycelium grows into the food source and absorbs nutrients across the chitenous cell wall.
Chytrids are difficult to maintain in culture and have to be baited from natural sources. We have purchased slides showing a common parasitic chytrid, Synchytrium, invading a plant host, and have prepared a fresh culture of chytrids on snake skin. Chytrids are important decomposers of animal epithelial tissue in aquatic habitats. To capture them, we have placed strips of snake skin in pond water. Chytrid spores swim to the snake skin, and then germinate on the skin, forming a simple globular body with rhizomycelium.
Examine:
Synchytrium: Examine the prepared slide of Synchytrium that has infected either leaves or potato tubers. Note the simple globular body (the sorus) of the chytrid embedded in the tissue of the plant. Some of the globules may have matured into sporangia, and you may see spores inside.
Chytrids on snakeskin: In the dissecting scope, you can see the rhizomycelia extending from the snake skin, along with many protozoa. Note the pinhead-like cells within the mycelia. These are either the central globules from which multiple filaments of rhizomycelium arise, or they are sporangia.
In the light microscope, examine the globular cells and the hyphae growing away from them. The globular cell and rhizomycelium form the basic body plan of the saprophytic chytrids.
You will also see many tiny creatures swimming about the mycelium. Some of these may be zoospores released from sporangia. Examine the mycelium for any evidence of sporangia. Sporangia will be apparent as they will have only one hyphae attached to them.
B. ZYGOMYCOTA:
The Zygomycetes are important saprophytes, including species that are major decomposers of dung and food. Members of this group have a zygotic lifecycle (see figure 15-11 on page 316 of your text). The gametes are non-motile, and are born on the tips of specialized, fertile hyphae termed gametangia. The gametangia contain many haploid nuclei. In zygomycetes, the sexual act consists of two fertile hyphae growing towards each other. As they approach each other the ends of the hyphae form gametangia. The gametangia come in contact, after which the end walls disintegrate, releasing the haploid gamete nuclei into one common space. Pairs of haploid nuclei then fuse, creating many diploid nuclei.
The nucleate cell formed from the residuals of the two gametangia is termed a zygosporangium. Zygosporangia typically develop a thick wall that protects the diploid nuclei from harsh conditions, forming a many nucleate (ceonocytic) resting cell. Zygosporangia germinate when the diploid nuclei undergo meiosis to produce many new haploid nuclei. The haploid nuclei are walled off into distinct spores, which are released from a dispersal sporangium that grows out of the zygosporangium.
The most commonly encountered zygomycetes are the bread molds, which are important saprophytes that grow on carbohydrate-rich foods, including bread. The mycelium formed on the surface of the bread is a cottony mass that is initially white but soon darkens as the mycelium forms asexual sporanagia. Large numbers of mitospores are released, allowing the fungus to quickly spread.
Examine the following, first under the dissecting scope and then with the light microscope:
1. Zygorrhynchus moelleri: This mold growing on an agar plate shows the major stages of a typical Zygomycete life cycle. First examine the culture under the dissecting microscope, and then take a very small scraping of the agar with a dissecting needle or knife. Place this on a microscope slide, stain with the cotton blue stain on your bench, and examine at medium power with both phase contrast and normal visible light.
With the dissecting scope, examine the cottony matrix of the mycelium, and the sporangia rising above it, forming dark spheres on elongated stalks. These are mostly mitosporangia used in asexual reproduction. You may be able to see zygosporangia mixed in amongst the mycelium. They will be dark, barrel-shaped granules on the surface of the agar.
With the light microscope, observe the slide of agar, note the clear tubular nature of the hyphae and the absence of cross walls. Next observe any mitosporangia you may have opened while pressing down the cover slip.
Finally, observe the zygosporangia. These are dark, barrel-shaped structures with rough walls. Note the hyphae attached to the zygosprangia. These are the stalks of the gametangia, and are termed suspensors.
2. Rhizopus stolonifera: This is the common bread-mold, a regular feature of most pantries. We have provided you with a Petri plate of Rhizopus to examine. Note the following with the dissecting scope under high power:
Examine the mycelium and sporangia. You may be able to see elongated, horizontal hyphae connecting the sporangial stalks. Rhizopus spreads by these elongated hyphae, termed stolons (after the strawberry runners of the same name). Where stolons settle onto a food source, they produce anchoring hyphae that penetrate the food. Sporangia form above this contact point. This habit allows for rapid spread of Rhizopus over a loaf of bread. Typically, Zygomycetes reproduce asexually by mitospores when conditions are good, allowing for rapid spread over a new food source. They switch to sexual reproduction when the food is exhausted and conditions deteriorate.
3. Ungulate dung: We may display some moose or cow feces which may show a range of dung-zygomycetes, possibly including the hat-throwing fungus Pilobus. If anything of interest appears to be present, examine it with the dissecting scope and note the nature of the sporangia.
C. THE DIKARYOMYCOTA
The dikaryomycota were formerly classified as the phyla Ascomycota and Basiodiomycota (for example, see chapter 15 in your text), but recent advances in the systematic understanding have led to the merging of these two groups in a single phylum of higher fungi, the Dikaryomycota, with the ascomycetes and basiodiomycetes being separated into subphyla termed Ascomycotina and Basiodiomycotina. We will focus on these two groups.
The common feature of these groups is the formation of dikaryotic hyphae. The dikaryon arises when the protoplast of haploid hyphae fuse (they undergo plasmogamy). The nuclei do not initially fuse, and the resulting mycelium is made up of cells that are dikaryotic, or in the N + N state. Fusion of the two nuclei occurs in the fruiting body of the fungus, forming a diploid cell that immediately undergoes meiosis and mitosis to produce four to eight spores. The spores are released from sporangia formed in the fruiting body. In the ascomycetes, eight spores are released from sacs termed asci (singular is ascus, from the Latin word for sac). In the basiodiomycetes, the spores are formed on the end of a club-like sporangia termed a basidium (from the Latin word for club). The fruiting bodies of each fungus are termed an ascocarp (ascoma in your text), and the basidiocarp (basidioma in your text). The mushroom cap is a basidiocarp.
We have a number of specimens for you to examine today from each subphylum.
1. Subphyllum Ascomycotina
a. Unicellular forms: the yeasts: These are single-celled fungi that typically live within the food medium. Most are saprophytic, although some can become parasitic. The yeast fungus Candida albicans is an important pathogen in humans, forming diaper-rash, vaginal and urethral tract infections, and the potentially deadly sexually-transmitted disease candidiasis. The common yeast Saccharomyces cerviseae is the yeast of baking, brewing and enology (wine making). This yeast is preferred in fermentations as it rapidly grows, produces pleasant as opposed to noxious or toxic waste-products, and is tolerant of high (>10%) concentrations of ethanol.
Saccharomyces cereviseae: The common brewers yeast is growing on agar. Take a very small portion off the culture and smear into a drop of water on a microscope slide. Add a cover slip, and examine at low and then high power with the compound scope.
Look for budding cells amongst the large numbers of indistinct single yeast cells. These are apparent from the blob-like cellular extensions, termed buds that arise from mature yeast cells. Rather than simply dividing in two as most algae and plant cells do, yeasts divide by extruding protoplasm into a bud. This extrusion is then encapsulated in a wall and split off to form a new independent cell.
Occasionally, you may see some yeast forming asci: Yeasts live in both a haploid and diploid state. When conditions are harsh, two diploid yeast nuclei merge to form a zygote, which then undergoes meiosis to produce a four celled sac, the ascus. The ascus splits open to release the four cells, which then bud to start a new population of yeast cells. You may be able to see four-celled asci floating among the many cells in your slide. If so, show your classmates.
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b. Filamentous Ascomycetes: Multicellular ascomycetes produce hyphae and mycelium, and form ascocarps. Three types of ascocarps are produced by these fungi, cleistothecia (enclosed spheres), perithecia (vase-like) and apothecia (cup shaped). You should examine examples of each.
b.1 Cleistothecial species:
i. Powdery mildew (Uncinula spp.): Powdery mildews are common pathogenic fungi that infect leaves, forming a powdery mycelium on the surface. Powdery mildews reproduce asexually by forming chains of spores (conidiospores) on special hyphae termed conidophores. During sexual reproduction, they form a simple enclosed ascocarp, the cleistothecia. Cleistothecia are completely enclosed, with no opening for the developing spores to escape. When mature, the ascocarp wall ruptures, allowing enclosed asci with their ascospores to spill out and disperse. Often, cleistothecia have barbs and hooks, which can help disperse the entire ascocarp by clinging onto the fur of passing animals.
Examine the following:
Dissecting scope: Scan across the leaf infected with Uncinula to note the powdery mycelium, with chains of conidia rising above it. Periodically, you will see a pepper grain-like object with multiple elongated hooks attached. This is the cleistothecia.
Light microscope: Scrape some cleistothecia onto a microscope slide and cover gently with a cover slip. Examine at low power. Next press of the cover slip to rupture the ascocarp and release the spores inside.
ii. Powdery mildew on leaves: We also have specimens of unknown powdery mildews collected on leaves from around Toronto. Examine these under the dissecting scope for cleistothecia and conidia.
b.2 Perithecial species: The perithecium is a vase-shaped ascocarp with a narrow, open neck. Inside are multiple asci with spores. When mature, the asci protrude from the neck of the perithecia and forcibly eject the spores into the air. Sordaria is a dung saprophyte that is closely related to Neurospora, the fungus that has become one of the leading model organisms in genetic research.
Examine:
Dissecting scope: Sordaria is growing on agar plates, and the perithecia can be seen as dark pepper-like grains mixed in a mass of conidia-forming hyphae. Examine the perithecia closely and note the pear-like shape of the ascocarp. Are any asci protruding from the perithecia?
Light microscope: Scoop some perithecia onto a slide, cover and examine at low-to medium power. Gently push on the cover slip to squash open the perithecia. Note any football-shaped spores and asci that emerge.
b.3. Apothecial species: The apothecium is an ascocarp where the asci are directly exposed to the air in a cup, dome or invaginated surface. The fungi are commonly called the saucer, or cup fungi, and they include many beautiful, brightly colored forest species. The delectable morel is an apothecial ascocarp. To demonstrate the apothecium structure and form, we have a set of prepared slides and live specimens from a number of species.
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i. Bispora centrina (Yellow-fairy cups – live specimens): These are wood decomposing fungi that form small, brightly yellow cup shaped apothecium. Examine the stick with the fairly cups closely. You may take the stick to you bench to examine with a dissecting scope. The asci are formed on the inner surface of the cup. Return the stick to the display area when finished as we have only a few specimens.
ii. Peziza (prepared slides): Peziza is a cup fungus that grows on wood and is similar in form to Bispora. A life cycle of Peziza is shown in Fig. 15-14 on page 318 of your textbook.
Examine the prepared cross sections of the Peziza ascocarp under medium power with your light microscope. You will note the sac-like structure of the ascus with 8 haploid spores inside. These arose from meiosis and one subsequent round of mitosis. Note the zone of fertile tissue where the asci form. Below this are fattened vegetative cells that form the support structure of the ascocarp.
b.4 Ascomycetes of special note
i. Claviceps purpurea (Ergot of Rye)
We have display specimens of rye shoots infected with Ergot, caused by the perithecia-forming ascomycete Claviceps purpurea. Claviveps is an example of an endo-parasite, a fungus that grows within the stem and leaves of grasses. The fungus retards growth, but does not kill the host plant. In many instances, toxins produced by the fungus deter herbivory, and so the grass host can actually show superior performance relative to a non-infected plant that is eaten. In the case of ergot, the toxin produced is lysergic acid amide, from which the hallucinogenic drug lysergic acid diethyamide (LSD) was derived.
Examine the infected rye and note the grain heads with enlarged, dark-colored protrusions extending out from the grass stalk. These are sclerotia, which occur where the fungus has completely infected a developing grain and replaced the grain with a tight mat of interwoven mycelium. As the growing season ends, the sclerotia fall to the ground and overwinter. In the spring, they produce perithecia and in turn, large numbers of spores that infect the new rye crop.
Sclerotia break free and mix with the rye grain at harvest. People eating rye contaminated with ergot sclerotia experience severe poisoning, called ergotism. Symptoms include wild hallucinations coupled with extreme burning sensations in the extremities. Constriction of minor veins is common, leading to limbs dying and falling off. The pain is severe, and a typical victim would scream in agony while madly hallucinating. Before modern science explained the cause, people in the past would interpret the symptoms as an attack of demons, and in regions affected by ergot outbreaks, the citizens often turned to extreme religious practices to exorcise the devil. Throughout history, witch hunts, new religious movements, and mass hysteria have been attributed to ergot outbreaks.
Today, ergot poisoning is rare, and rye grain is routinely screened to filter out the larger sclerotia. Sclerotia are now intentionally grown as a source of drugs to control internal bleeding, migraine headaches, and to alter mental states in psychiatric patients.
ii. Peach Brown Rot (Monilinia fructicola): Many ascomycetes are severe pathogens of fruit crops. One of the worst is Peach Brown rot, which stunts trees and destroys mature peaches, apricots, cherries and related fruit. Infected trees form cankers on the twigs and leaves. Conidia erupting from the cankers are dispersed to infect other trees by asexual means. Fruits are infected as they near maturity.
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After infection, lesions develop on the fruit and it prematurely rots, falls to the ground and dries to a mummified carcass. Peach mummies are completely infected with the mycelium and in this form, the fungus will overwinter. In the spring, the fungus in the mummy form apothecia, from which spores will be released in huge numbers to infect new trees.
Examine the Monilinia cultures on agar with the dissecting microscope and note the lemon-shaped conidiospores arising from the mycelium. You can examine these more closely by taking a small piece of agar and preparing it on a slide for examination with the light microscope.
iii. Penecillium: Many ascomycetes are saprophytes that infect food and building materials in the home. Some are also sources of important drugs, while other produce powerful carcinogens. Penicillium is one of the most common molds in the household pantry, where it infects bread, fruits and milk products. Penicllium species are also important in making strong-flavored cheeses such rouquefort, gorgonzola, chamenbert, brie and Danish Blue. The blue-green color is actually the reproductive conidia of the Penicillium mold. Pennicilium is also the source of penicillin, the antibiotic that prevents wall synthesis in gram negative bacteria.
Examine the culture with the dissecting scope and note the green-blue broom-like conidiospore masses rising above the mycelium. These masses give Penicillium molds their characteristic color. Take a sample and prepare a microscope slide of it. Examine the conidiophores with conidia under the compound microscope. Note the broom-like structure of the spore-bearing mass.
We also have some blue-cheese on display. Examine the Pennicilium colony through the dissecting scope and try to identify the sporangia.
iv. Aspergillus: Aspergillus species are common black-colored molds in the household environment. They are frequently found on bread, drywall, and grains. Many species produce aflotoxins, which are powerful carcinogens of the liver found in stored grains, peanuts and cereals, including corn flakes. It is unwise to eat foods contaminated with wild Aspergillus species as they likely contain aflotoxins. (For example, never eat wild peanuts, or musty old grain). Beneficial Aspergillus spp. are used to produce soy sauce, miso (fermented soy paste), and to ferment rice in an early step in sake production.
Examine with a dissecting scope the culture on the agar plate and note the fan-shaped mass of conidia arising above the mycelium. Next, examine a piece of the mycelium to see the bulbous conidiophore. The dark masses of conidia give this fungus its particular color and shape.
Take a small chunk of infected agar and prepare a slide of the sporangia for the light microscope. Examine the swollen top of the condiophores and the attached fan-shaped array of conidia.
2. The subphylum Basidiomycotina
The most familiar fungi are the basidiomycetes. The fruiting bodies of the basidiomycetes (the basidiocarp) are the recognizable features of species of mushrooms, toadstools, coral fungi, shelf fungi and tooth fungi. In each, the main body lives underground or in wood as a dispersed mycelium. Although all basidiomycetes reproduce by forming spores on club-shaped basidia, there are actually two main groups: the homobasidiomycetes and the heterbasidiomycetes. The homobasidiomycetes produce one type of spore, the basidiospore. The heterobasidiomycetes produce two types of spores
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during the sexual life cycle. We will focus on the homobasidiomycete life cycle as exemplified by the common food mushroom, Agaricus campestris. Heterobasidiomycetes are the pathogenic rusts and smuts.
a. Basidiomycete yeast (Rhodotorula ruba): Some basidiomycetes also have evolved the unicellular life form. A common basidiomycete yeast is the red yeast, Rhodotorula ruba, a contaminant of bathroom curtains, tile and grout. The pink scum in filthy bathtubs and showers is caused by Rhodotorula. (You may remember the battle between the Cat-in-the-Hat and pink bathroom scum).
Examine the red yeast culture on display. If time permits, you may prepare a microscope slide of the cells from the agar culture. Examine them for budding and basidia, which are distinguished by an elongated shape and horn-like points on one end of the cell.
b. Heterobasidiomycetes: The heterobasidiomycetes include the rust diseases of grasses, and smut diseases of maize. Other members of this group are wood decomposers such as the jelly fungi. We may have a jelly fungus in the wild mushroom display.
Examine the specimens of grasses infected with wheat rust (Puccinia graminis). Note the rust-colored pustules forming on the blades of the grass. These are where asexual spores are formed to allow for continued infection of healthy plants during the summer. Black pustules appear in the late-summer. These are where teliospores are formed. Teliospores are overwintering spores that form basidiospores in the spring.
If available, examine any corn smut (Ustilago maydis) that may be on display. Corn smuts attack developing corn kernels and produce large, grey-colored smutballs that are filled with dark spores. Immature smutballs are served as a delicacy in Latin American cuisines.
Rusts and smuts are virulent parasites of grain crops, with the potential to wipe out the production of an entire region in any given year. The primary means of preventing infestation is to breed crop cultivars that are resistant to rust infections. The rusts eventually evolve new ways to infect the cultivar, so government agencies are continuously breeding resistance into varieties to stay ahead of the rust capacity to re-evolve virulence. Should breeding efforts fall behind (for example, via cost-cutting measures by governments and agribusiness), major rust outbreaks could result, ruining grain crops and causing food prices to sky-rocket.
Notes:
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c. Homobasidiomycetes: the Mushrooms
We have fresh specimens of the common store-bought mushroom for examination, along with cultures of the inky cap mushroom, and a range of wild mushrooms from southern Ontario. To aid in examining fine detail, we have prepared slides showing cross sections of mushroom caps for you to examine under the light microscope. A detailed diagram of the life-cycle of the mushroom is presented in Figure 15-19 of your textbook (page 321).
c.1. The common food mushroom Agricus bisporus:
Examine a) the mycelium of the spawn blocks on display, b) the young button mushrooms and c) mature-spore producing mushrooms from the collection of fresh mushrooms provided.
i. Mycelial stage: sample mycelia of the mushroom spawn that is available. Stain with cotton blue and view with the light microscope under both normal light and phase contrast. Find some isolated hyphae and examine this under high power. Note the septate nature of the hyphae. This is one of the diagnostic features of the basidiomycetes.
Each cell contains two haploid nuclei in the N + N configuration. A key feature of Basidiomycetes is the presence of clamp connections, which form after cell division in order to keep the N + N dikaryotic configuration intact (see figure 15-21 in your text). Clamp connections may be visible along the end walls of the hyphal cells, forming bulges or loops around the septate wall.
ii. The mushroom button stage: The basidiocarp forms from tightly woven mycelia. Initially, the basidiocarp form a button, or egg stage. Cut open a button and examine a) the immature stalk (or stipe), b) the young, white to pink gills, and c) the developing cap which extends down over the stipe. With a razor blade, cut a thin slice of a gill, and look at the slice with the light microscope. Stain the gill with Melzer’s blue. You may be able to see developing basidia with miniature spores.
iii. The mature mushroom: Note the features of the basidiocarp structure. The main parts are a) a well developed stalk, termed the stipe. The cap, termed the pileus, and the gills, where the spore bearing tissues occur. The gills have turn chocolate brown as millions of spores mature. Cut a thin slide of the mature gill and examine for spores and horned basidia. Stain with the Melzers stain placed on your bench. You should be able to isolate one or two good basidia from the mass of tissue.
iv. Prepared slide of the mushroom cap: Examine under the light microscope the cross sections prepared of an Agaricus pileaus. The cross sections of gills clearly demonstrate the club shaped basidia arising from the zone of fertile tissue (the hymenium). Examine the basidia under high power and note how the spores are attached to the horn-like basidial tips (the sterigmata). The spores appear as party balloons taped to a club.
v. Spore prints: As the spores mature, they are released and drift into the air below the cap. If the cap is place over paper, the spores cannot disperse and settle onto the paper to form a print of the mushroom gills. Spore prints have been prepared for you to examine. Spore prints are often used to identify particular mushrooms, and they are an easy way to verify the color of the spores.
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c.2. The Inky-cap mushroom, Coprinus cinereus. We have displays of the Inky-cap mushroom for you to examine. The cultures show how the mushroom arises from the mycelium. Prepared slides are also available if you wish to examine the gill structure and the basidia of Corpinus.
The cap self-digests upon maturity forming a purple-ink colored mass of goo.
c.3. The oyster mushroom Pleurotus ostreatus: Oyster mushrooms are delightful edible mushrooms that grow on logs and decaying stumps. They produce white spores on short gills, and exhibit a large, fan-shaped pileus with a short stipe. They are now commonly cultivated and are available in many supermarkets.
Examine the oyster mushrooms for morphological structure, then thin slice a gill and examine for the basidia and white spores under the microscope.
c.4. The chanterelle (Chanterellus cineareus): Chanterelles are a wonderful delicacy that is prized for its gentle, buttery flavor. Unlike the gilled mushrooms, chanterelles form their basidia on gently folded tissue underneath a wavy pileus. Assuming we have these available, take a small piece of the pileaus of a chanterelle and examine for basidia and spores. What color are the spores?
c.6. The wild mushroom display: We have collected a variety of wild mushrooms for you to examine for variation in form. Examine each carefully, paying attention to the pileus, the stipe (if present) and where the spore bearing tissues are located. In many of the samples, gills are not present. Instead, the spores are produced on elongated tubes that form pores on the underside of the pileus, on teeth-like protrusions that hang below the pileus, on coral-like prongs, or on rumpled folds of tissue that resemble elbow skin. These traits distinguish the major families of mushroom-forming species. You may take some of the specimens and examine them under the dissecting scope in order to better see the pores, teeth or prongs of the spore-bearing tissue.
PART IV: LICHENS
Lichens are structures formed by close symbiotic relationships between an algae and a fungus. Both green and blue-green algae can serve as the algal symbiont, while the fungus is typically an ascomycete. Because the sexual stage of the lichen that is visible is that of the fungal partner (the mycobiont), the lichen is typically named after the fungus.
In the symbiosis, the algae provide carbohydrates from photosynthesis while the fungus shelters the algae and gathers water and nutrients. Lichens can completely desiccate with no harm to the organisms inside. Upon wetting they rapidly rehydrate and resume activity. This ability allows them to live in extremely harsh surfaces, such as the branches and trunks of trees, the sides of rocks, and bare ground in deserts. In the boreal zone, lichens are important ground covers on bare soil, fallen branches and the surface of rocks. They also are common as epiphytes on trees. In general, they grow extremely slow, reflecting the harsh conditions of the habitats they live in.
Lichens come in three general categories based on morphology. Examine the specimens displayed in the lab room.
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A. Foliose lichens: Lichens that exhibit a leafy shape are termed the foliose lichens. These are common in wetter habitats, for example, forest interiors in eastern Canada. Foliose lichens are important epiphytes growing on branches of standing trees.
B. Fructicose lichens: These lichens are shrubby in appearance, with many narrow, highly branched stem-like structures. Fructicose lichens are common in the boreal forest, forming dense ground covers in spruce forests. When dry, they are extremely flammable and can be used as a fire starter. They also help wildfires spread and thus contribute to some of the severe forest fires that occur every summer in Canada. Fructicose lichens are common on the sides of trees, and often hang from the bra nches in dense growths termed Witch’s hair, or Old Man’s Beard.
C. Crustose lichens: Crustose lichens occur in the most extreme terrestrial environments where life is possible. They grown on the sides of rocks, buildings and on bare soil, and are common in arid and polar deserts, including the dry valleys of Antarctica. Crustose lichens form brilliant yellow, orange, red and yellow-green colors on the rock, and are some of the most beautiful features in what are otherwise barren landscapes.
Study Guide: You should be familiar with the major categories of fungi and the names of the common species of yeast, store mushrooms, and major disease organisms displayed in the lab. You need to know the terms presented in bold font, and should recognize an organism well enough to classify it to phylum or where relevant, to subphylum. Know and understand the distinguishing characteristics of the major phyla presented in lab, as well as that of the ascomycetes and basidiomycetes.
The second category of fungal organisms is the pseudofungi, made up of various unrelated protista groups. The pseudofungi were formerly classified into the catch-all kingdom Protista but have recently been reclassified into more-specific kingdoms that reflect genetic relationships. Important pseudofungi are the Oomycetes (egg fungi and water molds), and the slime molds. Oomycetes are closely related to the stramenopilous algae - the brown algae, golden-brown algae and diatoms of the Kingdom Stramenopila. The close relationship between the oomycetes and the brown algae is evident in that both have cellulose walls, and they share the same type of flagella. Oomycetes descend from algae that lost their chloroplasts, and hence have adopted a heterotrophic life form. Oomycetes also produce filamentous hyphae to better absorb nutrients from the food source. Because they have no relationship with the chytrids, it is clear that the oomycetes and chytridiomycetes independently evolved the mycelial life form.
Slime molds evolved from various ancient protozoa and have little genetic affinity to other fungi or algae groups. They are animal-like in that they ingest food early in the life-cycle, but are fungal-like in that they produce walled sporangia and spores.
In today’s lab, you will have a chance to examine the diversity of both false and true fungi. The lab will begin with the false fungi and then follow an evolutionary sequence through the true fungi. Many specimens illustrate important reproductive phases in the life-cycles of these organisms. You should examine these carefully because reproductive features are important to understanding and distinguishing the various groups of fungi.
Although not required, we urge you to draw and label the specimens you examine. Our experience has been that drawings with appropriate labels are the best way to learn the features emphasized in lab and lecture. Drawings are also an excellent study tool to refresh your memory just prior to the exam
PART I: THE PSEUDOFUNGI
A) Slime molds (Kingdoms Myxomycota and Dictyosteliomycota): Slime molds are largely saprophytic and are typically found on decaying wood in moist forests. During the vegetative phase of the life cycle (see figure 16-6 on page 353 of your text), they begin life as independent amoebae, ingesting microscopic bits of organic debris. The free-living amoebae eventually swarm together to produce a multicellular blob called a plasmodium. After a while, the plasmodium forms cellulose walls around the nuclei and produces sporangia (or fructifications). The sporangia release large numbers of air-borne spores which germinate in the presence of water to form free-living amoebae, thus completing the life cycle. Slime molds defy simple categorization. They are animal-like in that they ingest food during the amoeboid- and plasmodial-phase. They are plant-like in their formation of cell walls and sporangia during reproduction. Because of the cell-walls formed during the reproductive phase they are considered fungal in nature. There are two major types of slime molds:
A) the Myxomycota are the acellular, or true plasmodial slime modes: The plasmodium stage of this group is made up of large blobs of ceonocytic protoplasm with many nuclei inside.
B) the Dictyosteliomycota are the cellular clime molds, where the plasmodium is made of individual cells separated by membranes (but not cell walls).
Observations on Display: Fruiting bodies of slime molds are readily found in Ontario forests in the autumn. Some will be on display for you to examine, along with a diagram of the life cycle.
B) Oomycetes: the water molds, or egg fungi (Kingdom Stramenopila)
Oomycetes are characterized by the formation of large egg-bearing cells on the tips of specialized hyphae termed oogonia (see figure 17-4 on page 374 of your text). Large, non-motile eggs form inside the oogonia and are fertilized by male-like hyphae termed antheridia. The antheridia grow into the egg and deposit the male gametes, which then fuse with the egg to form a zygote, termed the oospore. The oospore undergoes mitosis and forms a sporangia. The spores that are produced disperse to infect leaves, seedlings, fish and dead organisms.
Oomycetes are important saprophytes in aquatic habitats. In terrestrial habitats, they are generally parasitic. Important diseases caused by water molds are downy mildews (Peronospora), potato late blight (Phytophthora infestans), and damping off disease (Pythium spp.). We will examine three species, Achlya, a saprophytic water mold; Albugo candida, the white rust of mustard plants and Phytophthora infestans.
Examine the following cultures with the dissecting microscope:
1. Achlya whole mounts: Achlya is a water mold that grows on organic debris in lakes and rivers. It forms floating mycelia mats arising from the food material, and in these mats sexual reproduction occurs. On your bench are prepared slides for you to examine with the light microscope. Focus on the blue-green material in the center of the slide. This is a hyphal mat with oogonia.
Observe the large conspicuous oogonia within the mycelium. Inside the oogonia you can see zygotes (oospores), that will later divide to form sporangia. Examine the oogonia closely to see if additional hyphae are attached to it. These would be antheridia, which present the sperm nucleus to the egg cells
within the oogonia. Note the properties of the vegetative hyphae. Do you see cross walls, or are the cells continous within a filament?
2. Phytophthora infestans: This organism causes potato late blight, one of the worst crop diseases in the history of humanity. In the 1840’s, Phytophthora infestans was introduced to Europe from Peru. It rapidly spread across the continent, destroying much of the potato crop. In Ireland, peasants were particularly dependent on the potato for survival, and the crop losses in 1845-1848 killed millions of Irish and forced the migration of millions more to North America. In continental Europe, the loss of the potato crop led to widespread economic failure and social revolt. Many of the radical worker movements that would later influence world history, such as Marxism-Leninism, arose in the wake of the food crisis caused by the Phytophthora outbreak.
Examine the Phytophthora slide prepared fresh from a culture stained with cotton blue. Identify the round oogonia in the slide and examine the hyphae for cross walls between the cells. Are there any oospores within the Oogonia? Can you see any antheridial hyphae attached to the oogonia? If you do, show other students in your group.
3. Albugo candida (White Rust of Mustards): Albugo infects plants of the mustard family, forming white pustules on the leaves. These pustules are comprised of many asexual conidiospores bursting through the plants epidermis. Inside the plant, fungal hyphae form oogonia and antheridia, which will mate and form oospores. The oospores develop sporangia which disperse genetically-distinct spores. The two prepared slides show the extent of the infection by the parasitic white rust fungus.
Examine the prepared cross section showing Albugo conidiaspores bursting through the epidermis of a mustard fruit. Note the strings of conidiospores forming under the epidermis. The bulge formed by the mass of conidia produce the rust pustule. Upon rupturing, the conidiospores are released on the wind to start a series of new infections.
Examine the prepared slide of the Albugo sexual organs (oogonia and antheridia). The oogonia are evident as dense, red-staining circles among the cells of the leaf tissue.
Notes and Drawings:
PART II: THE TRUE FUNGI (KINGDOM FUNGI)
We have assembled a range of specimens from the Kingdom Fungi, and they are arranged for you to examine beginning with the most primitive (the Chytrids) and then progressing through the Zygomycetes to the Ascomycetes and Basidiomycetes.
A. CHYTRIDIOMYCOTA (The Chytrids):
Chytrids are mainly aquatic or parasitic. They are important decomposers of pollen, dead insects and seeds that fall into ponds and rivers. Others are parasites of algae, higher fungi, mosquitoes, rotifers, and water molds. The general body plan is to form a large, central ceonocytic globule that either directly invades the body of the host, or produces diminutive hyphae termed rhizomycelium, which invade the surface cells of the host. The rhizomycelium grows into the food source and absorbs nutrients across the chitenous cell wall.
Chytrids are difficult to maintain in culture and have to be baited from natural sources. We have purchased slides showing a common parasitic chytrid, Synchytrium, invading a plant host, and have prepared a fresh culture of chytrids on snake skin. Chytrids are important decomposers of animal epithelial tissue in aquatic habitats. To capture them, we have placed strips of snake skin in pond water. Chytrid spores swim to the snake skin, and then germinate on the skin, forming a simple globular body with rhizomycelium.
Examine:
Synchytrium: Examine the prepared slide of Synchytrium that has infected either leaves or potato tubers. Note the simple globular body (the sorus) of the chytrid embedded in the tissue of the plant. Some of the globules may have matured into sporangia, and you may see spores inside.
Chytrids on snakeskin: In the dissecting scope, you can see the rhizomycelia extending from the snake skin, along with many protozoa. Note the pinhead-like cells within the mycelia. These are either the central globules from which multiple filaments of rhizomycelium arise, or they are sporangia.
In the light microscope, examine the globular cells and the hyphae growing away from them. The globular cell and rhizomycelium form the basic body plan of the saprophytic chytrids.
You will also see many tiny creatures swimming about the mycelium. Some of these may be zoospores released from sporangia. Examine the mycelium for any evidence of sporangia. Sporangia will be apparent as they will have only one hyphae attached to them.
B. ZYGOMYCOTA:
The Zygomycetes are important saprophytes, including species that are major decomposers of dung and food. Members of this group have a zygotic lifecycle (see figure 15-11 on page 316 of your text). The gametes are non-motile, and are born on the tips of specialized, fertile hyphae termed gametangia. The gametangia contain many haploid nuclei. In zygomycetes, the sexual act consists of two fertile hyphae growing towards each other. As they approach each other the ends of the hyphae form gametangia. The gametangia come in contact, after which the end walls disintegrate, releasing the haploid gamete nuclei into one common space. Pairs of haploid nuclei then fuse, creating many diploid nuclei.
The nucleate cell formed from the residuals of the two gametangia is termed a zygosporangium. Zygosporangia typically develop a thick wall that protects the diploid nuclei from harsh conditions, forming a many nucleate (ceonocytic) resting cell. Zygosporangia germinate when the diploid nuclei undergo meiosis to produce many new haploid nuclei. The haploid nuclei are walled off into distinct spores, which are released from a dispersal sporangium that grows out of the zygosporangium.
The most commonly encountered zygomycetes are the bread molds, which are important saprophytes that grow on carbohydrate-rich foods, including bread. The mycelium formed on the surface of the bread is a cottony mass that is initially white but soon darkens as the mycelium forms asexual sporanagia. Large numbers of mitospores are released, allowing the fungus to quickly spread.
Examine the following, first under the dissecting scope and then with the light microscope:
1. Zygorrhynchus moelleri: This mold growing on an agar plate shows the major stages of a typical Zygomycete life cycle. First examine the culture under the dissecting microscope, and then take a very small scraping of the agar with a dissecting needle or knife. Place this on a microscope slide, stain with the cotton blue stain on your bench, and examine at medium power with both phase contrast and normal visible light.
With the dissecting scope, examine the cottony matrix of the mycelium, and the sporangia rising above it, forming dark spheres on elongated stalks. These are mostly mitosporangia used in asexual reproduction. You may be able to see zygosporangia mixed in amongst the mycelium. They will be dark, barrel-shaped granules on the surface of the agar.
With the light microscope, observe the slide of agar, note the clear tubular nature of the hyphae and the absence of cross walls. Next observe any mitosporangia you may have opened while pressing down the cover slip.
Finally, observe the zygosporangia. These are dark, barrel-shaped structures with rough walls. Note the hyphae attached to the zygosprangia. These are the stalks of the gametangia, and are termed suspensors.
2. Rhizopus stolonifera: This is the common bread-mold, a regular feature of most pantries. We have provided you with a Petri plate of Rhizopus to examine. Note the following with the dissecting scope under high power:
Examine the mycelium and sporangia. You may be able to see elongated, horizontal hyphae connecting the sporangial stalks. Rhizopus spreads by these elongated hyphae, termed stolons (after the strawberry runners of the same name). Where stolons settle onto a food source, they produce anchoring hyphae that penetrate the food. Sporangia form above this contact point. This habit allows for rapid spread of Rhizopus over a loaf of bread. Typically, Zygomycetes reproduce asexually by mitospores when conditions are good, allowing for rapid spread over a new food source. They switch to sexual reproduction when the food is exhausted and conditions deteriorate.
3. Ungulate dung: We may display some moose or cow feces which may show a range of dung-zygomycetes, possibly including the hat-throwing fungus Pilobus. If anything of interest appears to be present, examine it with the dissecting scope and note the nature of the sporangia.
C. THE DIKARYOMYCOTA
The dikaryomycota were formerly classified as the phyla Ascomycota and Basiodiomycota (for example, see chapter 15 in your text), but recent advances in the systematic understanding have led to the merging of these two groups in a single phylum of higher fungi, the Dikaryomycota, with the ascomycetes and basiodiomycetes being separated into subphyla termed Ascomycotina and Basiodiomycotina. We will focus on these two groups.
The common feature of these groups is the formation of dikaryotic hyphae. The dikaryon arises when the protoplast of haploid hyphae fuse (they undergo plasmogamy). The nuclei do not initially fuse, and the resulting mycelium is made up of cells that are dikaryotic, or in the N + N state. Fusion of the two nuclei occurs in the fruiting body of the fungus, forming a diploid cell that immediately undergoes meiosis and mitosis to produce four to eight spores. The spores are released from sporangia formed in the fruiting body. In the ascomycetes, eight spores are released from sacs termed asci (singular is ascus, from the Latin word for sac). In the basiodiomycetes, the spores are formed on the end of a club-like sporangia termed a basidium (from the Latin word for club). The fruiting bodies of each fungus are termed an ascocarp (ascoma in your text), and the basidiocarp (basidioma in your text). The mushroom cap is a basidiocarp.
We have a number of specimens for you to examine today from each subphylum.
1. Subphyllum Ascomycotina
a. Unicellular forms: the yeasts: These are single-celled fungi that typically live within the food medium. Most are saprophytic, although some can become parasitic. The yeast fungus Candida albicans is an important pathogen in humans, forming diaper-rash, vaginal and urethral tract infections, and the potentially deadly sexually-transmitted disease candidiasis. The common yeast Saccharomyces cerviseae is the yeast of baking, brewing and enology (wine making). This yeast is preferred in fermentations as it rapidly grows, produces pleasant as opposed to noxious or toxic waste-products, and is tolerant of high (>10%) concentrations of ethanol.
Saccharomyces cereviseae: The common brewers yeast is growing on agar. Take a very small portion off the culture and smear into a drop of water on a microscope slide. Add a cover slip, and examine at low and then high power with the compound scope.
Look for budding cells amongst the large numbers of indistinct single yeast cells. These are apparent from the blob-like cellular extensions, termed buds that arise from mature yeast cells. Rather than simply dividing in two as most algae and plant cells do, yeasts divide by extruding protoplasm into a bud. This extrusion is then encapsulated in a wall and split off to form a new independent cell.
Occasionally, you may see some yeast forming asci: Yeasts live in both a haploid and diploid state. When conditions are harsh, two diploid yeast nuclei merge to form a zygote, which then undergoes meiosis to produce a four celled sac, the ascus. The ascus splits open to release the four cells, which then bud to start a new population of yeast cells. You may be able to see four-celled asci floating among the many cells in your slide. If so, show your classmates.
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b. Filamentous Ascomycetes: Multicellular ascomycetes produce hyphae and mycelium, and form ascocarps. Three types of ascocarps are produced by these fungi, cleistothecia (enclosed spheres), perithecia (vase-like) and apothecia (cup shaped). You should examine examples of each.
b.1 Cleistothecial species:
i. Powdery mildew (Uncinula spp.): Powdery mildews are common pathogenic fungi that infect leaves, forming a powdery mycelium on the surface. Powdery mildews reproduce asexually by forming chains of spores (conidiospores) on special hyphae termed conidophores. During sexual reproduction, they form a simple enclosed ascocarp, the cleistothecia. Cleistothecia are completely enclosed, with no opening for the developing spores to escape. When mature, the ascocarp wall ruptures, allowing enclosed asci with their ascospores to spill out and disperse. Often, cleistothecia have barbs and hooks, which can help disperse the entire ascocarp by clinging onto the fur of passing animals.
Examine the following:
Dissecting scope: Scan across the leaf infected with Uncinula to note the powdery mycelium, with chains of conidia rising above it. Periodically, you will see a pepper grain-like object with multiple elongated hooks attached. This is the cleistothecia.
Light microscope: Scrape some cleistothecia onto a microscope slide and cover gently with a cover slip. Examine at low power. Next press of the cover slip to rupture the ascocarp and release the spores inside.
ii. Powdery mildew on leaves: We also have specimens of unknown powdery mildews collected on leaves from around Toronto. Examine these under the dissecting scope for cleistothecia and conidia.
b.2 Perithecial species: The perithecium is a vase-shaped ascocarp with a narrow, open neck. Inside are multiple asci with spores. When mature, the asci protrude from the neck of the perithecia and forcibly eject the spores into the air. Sordaria is a dung saprophyte that is closely related to Neurospora, the fungus that has become one of the leading model organisms in genetic research.
Examine:
Dissecting scope: Sordaria is growing on agar plates, and the perithecia can be seen as dark pepper-like grains mixed in a mass of conidia-forming hyphae. Examine the perithecia closely and note the pear-like shape of the ascocarp. Are any asci protruding from the perithecia?
Light microscope: Scoop some perithecia onto a slide, cover and examine at low-to medium power. Gently push on the cover slip to squash open the perithecia. Note any football-shaped spores and asci that emerge.
b.3. Apothecial species: The apothecium is an ascocarp where the asci are directly exposed to the air in a cup, dome or invaginated surface. The fungi are commonly called the saucer, or cup fungi, and they include many beautiful, brightly colored forest species. The delectable morel is an apothecial ascocarp. To demonstrate the apothecium structure and form, we have a set of prepared slides and live specimens from a number of species.
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i. Bispora centrina (Yellow-fairy cups – live specimens): These are wood decomposing fungi that form small, brightly yellow cup shaped apothecium. Examine the stick with the fairly cups closely. You may take the stick to you bench to examine with a dissecting scope. The asci are formed on the inner surface of the cup. Return the stick to the display area when finished as we have only a few specimens.
ii. Peziza (prepared slides): Peziza is a cup fungus that grows on wood and is similar in form to Bispora. A life cycle of Peziza is shown in Fig. 15-14 on page 318 of your textbook.
Examine the prepared cross sections of the Peziza ascocarp under medium power with your light microscope. You will note the sac-like structure of the ascus with 8 haploid spores inside. These arose from meiosis and one subsequent round of mitosis. Note the zone of fertile tissue where the asci form. Below this are fattened vegetative cells that form the support structure of the ascocarp.
b.4 Ascomycetes of special note
i. Claviceps purpurea (Ergot of Rye)
We have display specimens of rye shoots infected with Ergot, caused by the perithecia-forming ascomycete Claviceps purpurea. Claviveps is an example of an endo-parasite, a fungus that grows within the stem and leaves of grasses. The fungus retards growth, but does not kill the host plant. In many instances, toxins produced by the fungus deter herbivory, and so the grass host can actually show superior performance relative to a non-infected plant that is eaten. In the case of ergot, the toxin produced is lysergic acid amide, from which the hallucinogenic drug lysergic acid diethyamide (LSD) was derived.
Examine the infected rye and note the grain heads with enlarged, dark-colored protrusions extending out from the grass stalk. These are sclerotia, which occur where the fungus has completely infected a developing grain and replaced the grain with a tight mat of interwoven mycelium. As the growing season ends, the sclerotia fall to the ground and overwinter. In the spring, they produce perithecia and in turn, large numbers of spores that infect the new rye crop.
Sclerotia break free and mix with the rye grain at harvest. People eating rye contaminated with ergot sclerotia experience severe poisoning, called ergotism. Symptoms include wild hallucinations coupled with extreme burning sensations in the extremities. Constriction of minor veins is common, leading to limbs dying and falling off. The pain is severe, and a typical victim would scream in agony while madly hallucinating. Before modern science explained the cause, people in the past would interpret the symptoms as an attack of demons, and in regions affected by ergot outbreaks, the citizens often turned to extreme religious practices to exorcise the devil. Throughout history, witch hunts, new religious movements, and mass hysteria have been attributed to ergot outbreaks.
Today, ergot poisoning is rare, and rye grain is routinely screened to filter out the larger sclerotia. Sclerotia are now intentionally grown as a source of drugs to control internal bleeding, migraine headaches, and to alter mental states in psychiatric patients.
ii. Peach Brown Rot (Monilinia fructicola): Many ascomycetes are severe pathogens of fruit crops. One of the worst is Peach Brown rot, which stunts trees and destroys mature peaches, apricots, cherries and related fruit. Infected trees form cankers on the twigs and leaves. Conidia erupting from the cankers are dispersed to infect other trees by asexual means. Fruits are infected as they near maturity.
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After infection, lesions develop on the fruit and it prematurely rots, falls to the ground and dries to a mummified carcass. Peach mummies are completely infected with the mycelium and in this form, the fungus will overwinter. In the spring, the fungus in the mummy form apothecia, from which spores will be released in huge numbers to infect new trees.
Examine the Monilinia cultures on agar with the dissecting microscope and note the lemon-shaped conidiospores arising from the mycelium. You can examine these more closely by taking a small piece of agar and preparing it on a slide for examination with the light microscope.
iii. Penecillium: Many ascomycetes are saprophytes that infect food and building materials in the home. Some are also sources of important drugs, while other produce powerful carcinogens. Penicillium is one of the most common molds in the household pantry, where it infects bread, fruits and milk products. Penicllium species are also important in making strong-flavored cheeses such rouquefort, gorgonzola, chamenbert, brie and Danish Blue. The blue-green color is actually the reproductive conidia of the Penicillium mold. Pennicilium is also the source of penicillin, the antibiotic that prevents wall synthesis in gram negative bacteria.
Examine the culture with the dissecting scope and note the green-blue broom-like conidiospore masses rising above the mycelium. These masses give Penicillium molds their characteristic color. Take a sample and prepare a microscope slide of it. Examine the conidiophores with conidia under the compound microscope. Note the broom-like structure of the spore-bearing mass.
We also have some blue-cheese on display. Examine the Pennicilium colony through the dissecting scope and try to identify the sporangia.
iv. Aspergillus: Aspergillus species are common black-colored molds in the household environment. They are frequently found on bread, drywall, and grains. Many species produce aflotoxins, which are powerful carcinogens of the liver found in stored grains, peanuts and cereals, including corn flakes. It is unwise to eat foods contaminated with wild Aspergillus species as they likely contain aflotoxins. (For example, never eat wild peanuts, or musty old grain). Beneficial Aspergillus spp. are used to produce soy sauce, miso (fermented soy paste), and to ferment rice in an early step in sake production.
Examine with a dissecting scope the culture on the agar plate and note the fan-shaped mass of conidia arising above the mycelium. Next, examine a piece of the mycelium to see the bulbous conidiophore. The dark masses of conidia give this fungus its particular color and shape.
Take a small chunk of infected agar and prepare a slide of the sporangia for the light microscope. Examine the swollen top of the condiophores and the attached fan-shaped array of conidia.
2. The subphylum Basidiomycotina
The most familiar fungi are the basidiomycetes. The fruiting bodies of the basidiomycetes (the basidiocarp) are the recognizable features of species of mushrooms, toadstools, coral fungi, shelf fungi and tooth fungi. In each, the main body lives underground or in wood as a dispersed mycelium. Although all basidiomycetes reproduce by forming spores on club-shaped basidia, there are actually two main groups: the homobasidiomycetes and the heterbasidiomycetes. The homobasidiomycetes produce one type of spore, the basidiospore. The heterobasidiomycetes produce two types of spores
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during the sexual life cycle. We will focus on the homobasidiomycete life cycle as exemplified by the common food mushroom, Agaricus campestris. Heterobasidiomycetes are the pathogenic rusts and smuts.
a. Basidiomycete yeast (Rhodotorula ruba): Some basidiomycetes also have evolved the unicellular life form. A common basidiomycete yeast is the red yeast, Rhodotorula ruba, a contaminant of bathroom curtains, tile and grout. The pink scum in filthy bathtubs and showers is caused by Rhodotorula. (You may remember the battle between the Cat-in-the-Hat and pink bathroom scum).
Examine the red yeast culture on display. If time permits, you may prepare a microscope slide of the cells from the agar culture. Examine them for budding and basidia, which are distinguished by an elongated shape and horn-like points on one end of the cell.
b. Heterobasidiomycetes: The heterobasidiomycetes include the rust diseases of grasses, and smut diseases of maize. Other members of this group are wood decomposers such as the jelly fungi. We may have a jelly fungus in the wild mushroom display.
Examine the specimens of grasses infected with wheat rust (Puccinia graminis). Note the rust-colored pustules forming on the blades of the grass. These are where asexual spores are formed to allow for continued infection of healthy plants during the summer. Black pustules appear in the late-summer. These are where teliospores are formed. Teliospores are overwintering spores that form basidiospores in the spring.
If available, examine any corn smut (Ustilago maydis) that may be on display. Corn smuts attack developing corn kernels and produce large, grey-colored smutballs that are filled with dark spores. Immature smutballs are served as a delicacy in Latin American cuisines.
Rusts and smuts are virulent parasites of grain crops, with the potential to wipe out the production of an entire region in any given year. The primary means of preventing infestation is to breed crop cultivars that are resistant to rust infections. The rusts eventually evolve new ways to infect the cultivar, so government agencies are continuously breeding resistance into varieties to stay ahead of the rust capacity to re-evolve virulence. Should breeding efforts fall behind (for example, via cost-cutting measures by governments and agribusiness), major rust outbreaks could result, ruining grain crops and causing food prices to sky-rocket.
Notes:
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c. Homobasidiomycetes: the Mushrooms
We have fresh specimens of the common store-bought mushroom for examination, along with cultures of the inky cap mushroom, and a range of wild mushrooms from southern Ontario. To aid in examining fine detail, we have prepared slides showing cross sections of mushroom caps for you to examine under the light microscope. A detailed diagram of the life-cycle of the mushroom is presented in Figure 15-19 of your textbook (page 321).
c.1. The common food mushroom Agricus bisporus:
Examine a) the mycelium of the spawn blocks on display, b) the young button mushrooms and c) mature-spore producing mushrooms from the collection of fresh mushrooms provided.
i. Mycelial stage: sample mycelia of the mushroom spawn that is available. Stain with cotton blue and view with the light microscope under both normal light and phase contrast. Find some isolated hyphae and examine this under high power. Note the septate nature of the hyphae. This is one of the diagnostic features of the basidiomycetes.
Each cell contains two haploid nuclei in the N + N configuration. A key feature of Basidiomycetes is the presence of clamp connections, which form after cell division in order to keep the N + N dikaryotic configuration intact (see figure 15-21 in your text). Clamp connections may be visible along the end walls of the hyphal cells, forming bulges or loops around the septate wall.
ii. The mushroom button stage: The basidiocarp forms from tightly woven mycelia. Initially, the basidiocarp form a button, or egg stage. Cut open a button and examine a) the immature stalk (or stipe), b) the young, white to pink gills, and c) the developing cap which extends down over the stipe. With a razor blade, cut a thin slice of a gill, and look at the slice with the light microscope. Stain the gill with Melzer’s blue. You may be able to see developing basidia with miniature spores.
iii. The mature mushroom: Note the features of the basidiocarp structure. The main parts are a) a well developed stalk, termed the stipe. The cap, termed the pileus, and the gills, where the spore bearing tissues occur. The gills have turn chocolate brown as millions of spores mature. Cut a thin slide of the mature gill and examine for spores and horned basidia. Stain with the Melzers stain placed on your bench. You should be able to isolate one or two good basidia from the mass of tissue.
iv. Prepared slide of the mushroom cap: Examine under the light microscope the cross sections prepared of an Agaricus pileaus. The cross sections of gills clearly demonstrate the club shaped basidia arising from the zone of fertile tissue (the hymenium). Examine the basidia under high power and note how the spores are attached to the horn-like basidial tips (the sterigmata). The spores appear as party balloons taped to a club.
v. Spore prints: As the spores mature, they are released and drift into the air below the cap. If the cap is place over paper, the spores cannot disperse and settle onto the paper to form a print of the mushroom gills. Spore prints have been prepared for you to examine. Spore prints are often used to identify particular mushrooms, and they are an easy way to verify the color of the spores.
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c.2. The Inky-cap mushroom, Coprinus cinereus. We have displays of the Inky-cap mushroom for you to examine. The cultures show how the mushroom arises from the mycelium. Prepared slides are also available if you wish to examine the gill structure and the basidia of Corpinus.
The cap self-digests upon maturity forming a purple-ink colored mass of goo.
c.3. The oyster mushroom Pleurotus ostreatus: Oyster mushrooms are delightful edible mushrooms that grow on logs and decaying stumps. They produce white spores on short gills, and exhibit a large, fan-shaped pileus with a short stipe. They are now commonly cultivated and are available in many supermarkets.
Examine the oyster mushrooms for morphological structure, then thin slice a gill and examine for the basidia and white spores under the microscope.
c.4. The chanterelle (Chanterellus cineareus): Chanterelles are a wonderful delicacy that is prized for its gentle, buttery flavor. Unlike the gilled mushrooms, chanterelles form their basidia on gently folded tissue underneath a wavy pileus. Assuming we have these available, take a small piece of the pileaus of a chanterelle and examine for basidia and spores. What color are the spores?
c.6. The wild mushroom display: We have collected a variety of wild mushrooms for you to examine for variation in form. Examine each carefully, paying attention to the pileus, the stipe (if present) and where the spore bearing tissues are located. In many of the samples, gills are not present. Instead, the spores are produced on elongated tubes that form pores on the underside of the pileus, on teeth-like protrusions that hang below the pileus, on coral-like prongs, or on rumpled folds of tissue that resemble elbow skin. These traits distinguish the major families of mushroom-forming species. You may take some of the specimens and examine them under the dissecting scope in order to better see the pores, teeth or prongs of the spore-bearing tissue.
PART IV: LICHENS
Lichens are structures formed by close symbiotic relationships between an algae and a fungus. Both green and blue-green algae can serve as the algal symbiont, while the fungus is typically an ascomycete. Because the sexual stage of the lichen that is visible is that of the fungal partner (the mycobiont), the lichen is typically named after the fungus.
In the symbiosis, the algae provide carbohydrates from photosynthesis while the fungus shelters the algae and gathers water and nutrients. Lichens can completely desiccate with no harm to the organisms inside. Upon wetting they rapidly rehydrate and resume activity. This ability allows them to live in extremely harsh surfaces, such as the branches and trunks of trees, the sides of rocks, and bare ground in deserts. In the boreal zone, lichens are important ground covers on bare soil, fallen branches and the surface of rocks. They also are common as epiphytes on trees. In general, they grow extremely slow, reflecting the harsh conditions of the habitats they live in.
Lichens come in three general categories based on morphology. Examine the specimens displayed in the lab room.
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A. Foliose lichens: Lichens that exhibit a leafy shape are termed the foliose lichens. These are common in wetter habitats, for example, forest interiors in eastern Canada. Foliose lichens are important epiphytes growing on branches of standing trees.
B. Fructicose lichens: These lichens are shrubby in appearance, with many narrow, highly branched stem-like structures. Fructicose lichens are common in the boreal forest, forming dense ground covers in spruce forests. When dry, they are extremely flammable and can be used as a fire starter. They also help wildfires spread and thus contribute to some of the severe forest fires that occur every summer in Canada. Fructicose lichens are common on the sides of trees, and often hang from the bra nches in dense growths termed Witch’s hair, or Old Man’s Beard.
C. Crustose lichens: Crustose lichens occur in the most extreme terrestrial environments where life is possible. They grown on the sides of rocks, buildings and on bare soil, and are common in arid and polar deserts, including the dry valleys of Antarctica. Crustose lichens form brilliant yellow, orange, red and yellow-green colors on the rock, and are some of the most beautiful features in what are otherwise barren landscapes.
Study Guide: You should be familiar with the major categories of fungi and the names of the common species of yeast, store mushrooms, and major disease organisms displayed in the lab. You need to know the terms presented in bold font, and should recognize an organism well enough to classify it to phylum or where relevant, to subphylum. Know and understand the distinguishing characteristics of the major phyla presented in lab, as well as that of the ascomycetes and basidiomycetes.
BRYOPHYTES
Overview
Hornworts, liverworts, and mosses - commonly referred to as bryophytes - are considered to be a pivotal group in our understanding of the origin of land plants because they are believed to be among the earliest diverging lineages of land plants. Mosses, liverworts and hornworts are found throughout the world in a variety of habitats, from the harsh environs of Antarctica to the lush conditions of the tropical rainforests. Bryophytes are unique among land plants in that they possess an alternation of generations, which involves a dominant, free-living, haploid gametophyte alternating with a reduced, generally dependent, diploid sporophyte. Bryophytes are small, herbaceous plants that grow closely packed together in mats or cushions on rocks, soil, or as epiphytes on the trunks and leaves of forest trees. Bryophytes are remarkably diverse for their small size and are well-adapted to moist habitats and flourish particularly well in moist, humid forests like the fog forests of the Pacific northwest or the montane rain forests of the southern hemisphere.
Significance of bryophytes
Bryophytes have a significant role in contributing to nutrient cycles, providing seed-beds for the larger plants of the community, and form microhabitats for insects and an entire array of microorganisms. Bryophytes are also very effective rainfall interceptors, and the overwhelming abundance of epiphytic liverworts in "cloud" or "mossy" forest zones is considered an important factor in eliminating the deteriorating effect of heavy rains, including adding to hill stability and helping to prevent soil erosion. The chemical compounds of some liverworts are also particularly interesting because they have important biological activities, for example, against certain cancer cell lines, anti-bacterial properties, anti-microbial, anti-fungal, and muscle relaxing activity.
Classification
Over the last decade, recent advances in DNA sequencing technology and analytical approaches to phylogenetic reconstruction, including the use of ultra-structural, morphological and anatomical data, have enabled unprecedented progress toward our understanding of plant evolution. A growing consensus suggests that the bryophytes possibly represent three separate evolutionary lineages, which are today recognized as mosses (phylum Bryophyta), liverworts (phylum Marchantiophyta) and hornworts (phylum Anthocerotophyta).
• Mosses (Bryophyta)
The greatest species diversity in bryophytes is found in the mosses, with estimates of the number of species ranging from 10,000 to 15,000. Higher-level classification of the mosses remains unresolved with considerable difference of opinion on the names of the major groups. However, generally four major groups or classes are recognised. These include: Sphagnopsida (peat or Sphagnum mosses), Andreaeopsida (rock or lantern mosses), Polytrichopsida (nematodontous mosses), and the Bryopsida (arthrodontous mosses). The Sphagnum mosses are one of the most ecologically and economically important groups of bryophytes. The class Bryopsida accounts for the largest and most diverse groups within the mosses with over 100 families.
• Liverworts (Marchantiophyta)
The estimated number of liverwort species range from 6000 to 8000. Traditionally, liverworts have been subdivided into two major groups or classes based, partially, on growth form. The class Marchantiopsida, includes the well-known genera Marchantia, Monoclea, Lunularia, and Riccia, and has a complex thalloid organisation. The class Jungermanniopsida represents an estimated 85% of liverwort species and shows an enormous amount of morphological, anatomical and ecological diversity; plants with leafy shoot systems are the most common growth form in this class, e.g., Frullania, Jubulopsis, Cololejeunea, and Radula.
• Hornworts (Anthocerotophyta)
Hornworts get their name from their long, horn-shaped sporophytes and are the smallest group of bryophytes with only approximately 100 species. Hornworts resemble some liverworts in having simple, unspecialized thalloid gametophytes, but they differ in many other characters. Hornworts differ from all other land plants in having only one large, algal-like chloroplast in each thallus cell.
Phylogeny
Scientific Name, Common Name
Plantae, Land Plants
Embryophytes, Green plants
The Bryophyta or mosses, unlike the liverworts, are present in most terrestrial habitats (even deserts) and may sometimes be the dominant plant life.
As with the liverworts the plant that we commonly see is the gametophyte. It shows the beginnings of differentiation of stem and leaves - but no roots. Mosses may have rhizoids and these may be multicellular but they do little more than hold the plant down.
The stem shows some internal differentiation into hydroids and leptoids which are like xylem and phloem of higher plants but very simply organized with no connection to leaves or branching stems.
The leaves are mostly one cell thick; sometimes the midrib is several cells thick but this does not contain conducting tissue so it is not equivalent to the vein of a leaf.
Male and female gametophytes look identical except when they produce reproductive structures.
The male plant produces clusters of antheridia which contain thousands of ciliate sperm.
The female produces archegonia, each containing a single egg.
Fertilization is dependent on water - sperm are splashed or swim to the archegonia. The zygote grows into the diploid sporophyte which remains attached to the female gametophyte It is a leafless stem with a seta or foot at one end, drawing nutrients from the gametophyte. At the other end is a capsule in which meiosis occurs to form spores.
The archegonium grows around the developing sporophyte for a while but becomes separated from the gametophyte and is carried up to form a cap or calyptra over the sporangium. Curiously, the sporangia of some mosses have stomata much like those on the leaves of vascular plants.
Immature moss capsules with calyptra
The calyptra is lost when the sporangium is mature as is the operculum or lid on the end of the capsule.
Underneath the operculum there are often peristome teeth which open under dry conditions and control spore release A spore germinates to produce a filamentous protonema which sooner or later produces buds that grow into new gametophytes.
Ecology of mosses
Mosses require abundant water for growth and reproduction. They can tolerate dry spells by drying out or,in the case of mosses like Sphagnum , by holding huge amounts of water in dead cells in the leaves.
They look pretty lowly and insignificant, but have become dominant in particular habitats and Sphagnum itself is said to occupy 1% of the earth's surface (half the area of the USA). Because of its ability to soak up blood and its relative freedom from bacterial contamination Sphagnum was used in dressings. The moss itself is used in some horticultural media and it is an important source of peat.
Polytrichum commune one of the larger mosses with mature sporophytes
If you have tried to grow a lawn in a shady location you have probably been troubled by mosses as weeds. Like many lower organisms they are very sensitive to copper salts and can be controlled in this way. On the other hand mosses are green and better adapted to shade than most grasses, so maybe we should accept them in this situation.
Natural Perspective
The Plant Kingdom : Mosses and Allies
Mosses and their allies are small green plants that are simlutaneously overlooked and deeply appreciated by the typical nature lover. On the one hand, very few people pay attention to individual moss plants and species. On the other hand, it is the mosses that imbues our forests with that wonderful lush "Rainforest" quality which soothes the soul and softens the contours of the earth.
These wonderfully soft carpets of green are, in fact, Nature's second line of attack in its war against rocks. After lichens have created a foothold in rocks the mosses move in, ultimately becoming a layer of topsoil for higher plants to take root. The mosses also hold loose dirt in place, thus preventing landslides.
Ecologically and structurally, mosses are closer to lichens than they are to other members of the plant kingdom. Both mosses and lichens depend upon external moisture to transport nutrients. Because of this they prefer damp places and have evolved special methods of dealing with long dry periods. Higher plants, on the other hand, have specialized organs for transporting fluid, allowing them to adapt to a wider variety of habitats.
Bryophytes used to be classified as three classes of a single phylum, Bryophyta . Modern texts, however, now assign each class to its own phylum: Mosses ( Bryophyta ), Liverworts ( Hepatophyta ), and Hornworts ( Anthoceraphyta ). This reflects the current taxonomic wisdom that the Liverworts and Hornworts are more primitive and only distantly related to Mosses and other plants.
Mosses (Phylum: Bryophyta )
All plants reproduce through alternating generations. Nowhere is this more apparent than in the mosses. The first generation, the gametophyte , forms the green leafy structure we ordinarily associate with moss. It produces a sperm and an egg (the gametes) which unite, when conditions are right, to grow into the next generation: the sporophyte or spore-bearing structure.
The moss sporophyte is typically a capsule growing on the end of a stalk called the seta . The sporophyte contains no clorophyl of its own: it grows parasitically on its gametophyte mother. As the sporophyte dries out, the capsule release spores which will grow into a new generation of gametophytes, if they germinate.
Mosses, the most common, diverse and advanced brypophytes, are categorized into three classes: Peat Mosses ( Sphagnopsida ), Granite Mosses ( Andreaopsida ), and "True" Mosses ( Bryopsida or Musci ) .
Shown: Class: Bryopsida ; Order: Hypnales ; Family: Brachythecia ; Homolathecium nutalli (probably)
Leafy Liverworts (Phylum: Hepatophyta , Class: Jungermanniidae )
While people typically know what a moss is, few have even heard of liverworts and hornworts.
These primitive plants function much like mosses and grow in the same places, often intertwined with each other. The liverworts take on one of two general forms, comprising the two classes of liverworts: Jungermanniidea are leafy, like moss; Marchantiopsida are leaf-like ( thalloid ) similar to foliose lichens .
The leafy liverworts look very much like mosses and, in fact, are difficult to tell apart when only gametophytes are present. The "leaves," however, are simpler than moss and dont have a midrib ( costa ). The stalk of the sporophyte is translucent to white; its capsule is typically black and egg-shaped. When it matures, the capsule splits open into four equal quarters, releasing the spores to the air.
The liverwort sporophyte shrivels up and disappears shortly after releasing its spores. Because of this one hardly ever sees liverwort sporophytes out of season. Moss sporophtyes, on the other hand, may persist much longer.
Shown: Class: Jungermanniidea ; Order: Jungermanniales ; Family: Scapaniaceae ; Scapania spp. (probably)
Leaf-like Liverworts (Phylum: Hepatophyta ; Class: Marchantiopsida )
The leaf-like ( thalloid ) liverworts are, on the whole, more substantial and easier to find than their leafy counterparts. The gametophyte is flat, green and more-or-less strap-shaped. The body may, however, branch out several times to round out the form.
When the gametophyte has become fertilized and is ready to produce its sporophyte generation it may grow a tall green umbrella-shaped structure called the carpocephalum . The sporophyte grows on the underside of this structure, often completely hidden from view.
During the dry season, leaf-like liverworts may shrivel up and completely disappear from view until the rains arrive again.
Thalloid liverworts are much easier to identify than their leafy counterparts due to the wider variety of gametophyte shapes.
Shown: Class: Marchnatiopsida ; Order: Marchantiales ; Family: Aytoniaceae ; Asterella californica
Hornworts (Phylum: Anthoceraphyta )
Hornworts are very similar to liverworts but differ in the shape of the sporophyte generation. Instead of generating spores in a capsule atop a stalk, the hornwort generates spores inside a green horn-like stalk. When the spores mature the stalk splits, releasing the spores.
Under the microscope, hornwort cells look quite distinct as well: they have a single, large chloroplast in each cell. Other plants typically have many small chloroplasts per cell. This structure imparts a particular quality of color and translucency to the body ( thallus ) of the plant.
Hornworts are all grouped into a single class, Anthocerotae , containing a single order, Anthocerotales .
Shown: Class: Anthocerotae ; Order: Anthocerotales ; Family: Anthocertaceae ; Phaeoceros spp.
Identification
Suggestions for the Use of Keys
1. Select appropriate keys for the materials to be identified. The keys may be in a flora, manual, guide' handbook, monograph, or revision (see Chapter 30). If the locality of an unknown plant is known, select a flora, guide, or manual treating the plants of that geographic area (see Guides to Floras in Chapter 30). If the family or genus is recognized, one may choose to use a monograph or revision. If locality is unknown. select a general work. If materials to be identified were cultivated, select one of the manuals treating such plants since most floras do not include cultivated plants unless naturalized.
2. Read the introductory comments on format details, abbreviations, etc. .before using the key.
3. Read both leads of a couplet before making a choice. Even though the first lead may seem to describe the unknown material, the second lead may be even more appropriate.
4. Use a glossary to check the meaning of terms you do not understand.
5. Measure several similar structures when measurements are used in the key, e.g. measure several leaves not a single leaf. Do not base your decisions on a single observation It is often desirable to examine several specimens.
6. Try both choices when dichotomies are not clear or when information is insufficient, and make a decision as to which of the two answers best fits the descriptions.
7. Verify your results by reading a description, comparing the specimen with an illustration or an authentically named herbarium specimen.
Suggestions for Construction of Keys
1. Identify all groups to be included in a key.
2. Prepare a description of each taxon (see Chapter 24 for details for description and descriptive format).
3. Select "key characters" with contrasting character states. Use macroscopic, morphological characters and constant character states when possible. Avoid characteristics that can only be seen in the field or on specially prepared specimens, i.e., use those characteristics that are generally available to the user.
4. Prepare a Comparison Chart (see Figure 25-3).
5. Construct strictly dichotomous keys.
6. Use parallel construction and comparative terminology in each lead of a couple.
7. Use at least two characters per lead when possible.
8. Follow key format (indented or bracketed see Figures 25-1 and 25-2).
9. Start both leads of a couple with the same word if at all possible and successive leads with different words.
10. Mention the name of the plant part before descriptive phrases, e.g., leaves or flowers blue not blue flowers, leaves alternate not alternate leaves.
11. Place those groups with numerous variable character states in a key several times when necessary.
12. Construct separate keys for dioecious plants, for flowering or fruiting materials and for vegetative materials when pertinent.
A DICHOTOMOUS KEY TO SELECTED GENERA OF SAXIFRAGACEAE
• Shrub or woody vine.
o Woody vine; petals 7 or more 3. Decumaria
o Shrub; petals 4 or 5.
Leaves alternate or on short spur branches.
Leaves pinnately veined; ovary superior; fruit a capsule 1. Itea
Leaves palmately veined; ovary inferior; fruit a berry 2. Ribes
Leaves opposite.
Petals usually 4;-stamens 20-40; fruit longitudinally dehiscent, not ribbed; 4. Philadelphus
Petals usually 5; stamens 8-10; fruit poricidally dehiscent, 10- to 15-ribbed 5. Hydrangea
• Herbs.
o Staminodia present; petals more than 10 mm long 6. Parnassia
o Staminodia absent; petals less than 10 mm long.
Leaves ternately decompound 7. Astilbe
Leaves simple.
Flowers solitary in leaf axils, or in short, leafy cimes.
Sepals 4; carpels 2 8. Chrysosplenium
Sepals 5; carpels 3 9. Lepuropetalon
Flowers in racemes or panicles.
Petals pinnatifid or fringed; stem leaves opposite 10. Mitella
Petals not pinnatifid or fringed; stem leaves alternate or absent.
Ovary 1-celled.
Inflorescence paniculate; stamens 5 11. Heuchera
Inflorescence racemose; stamens 10 12. Tiarella
Ovary 2-celled.
Stamens 5; leaves palmately lobed 13. Boykinia
Stamens 10; leaves not palmately lobed 14. Saxifraga
A DICHOTOMOUS KEY TO SELECTED GENERA OF SAXIFRAGACEAE
• 1. Shrub or woody vine 2.
• 1. Herbs 6.
o 2. Woody vine; petals 7 or more Decumaria.
o 2. Shrub; petals 4 or 5 3.
• 3. Leaves alternate or on short spur branches 4.
• 3. Leaves opposite 5.
o 4. Leaves pinnately veined; ovary superior; fruit a capsule Itea.
o 4. Leaves palmately veined; ovary inferior; fruit a berry Ribes.
• 5. Petals usually 4; stamens 20-40; fruit longitudinally dehiscent, not ribbed Philadelphus
• 5. Petals usually 5; stamens 8-10; fruit poricidally dehiscent, 10-15 ribbed Hydrangea.
o 6. Staminodia present; petals more than 10 mm long Parnassia.
o 6. Staminodia absent; petals less than 10 mm long 7.
• 7. Leaves ternately decompound Astilbe.
• 7. Leaves simple 8.
o 8. Flowers solitary in leaf axils, or in short, leafy cymes 9.
o 8. Flowers in racemes or panicles 10.
• 9. Sepals 4; carpels 2 Chrysosplenium.
• 9. Sepals 5; carpels 3 Lepuropetalon.
o 10. Petals pinnatifid or fringed; stem leaves opposite Mitella.
o 10. Petals not pinnatifid or fringed; stem leaves alternate or absent 11.
• 11. Ovary 1-celled 12.
• 11. Ovary 2-celled 13.
o 12. Inflorescence paniculate; stamens 5 Heuchera.
o 12. Inflorescence racemose; stamens 10 Tiarella.
• 13. Stamens 5; leaves palmately lobed Boykinia.
• 13. Stamens 10; leaves not palmately lobed Saxifraga.
Figure 25-2. Example of a bracketed key. (Modified from Radford, A. E., 11. E. Ahles, and C. R. Bell. 1968. Manual of the Vascular Flora of the Carolinas. University of North Carolina Press. Chapel Hill, North Carolina. Used with permission.)
PLANT IDENTIFICATION EXERCISE
1. Identification of an unknown. Select an unknown specimen and identify it by keying in an appropriate manual, flora, or monograph. Verify your results by reading a description, by comparing with an illustration or by checking with your instructor.
2. Preparation of a comparison chart. Select 5 or more specimens from the group provided by your instructor. Identify each by keying. Verify your results. Prepare a description of each similar to those in a flora or manual. Be sure characters and character states are in the same order. Select contrasting character states and prepare a comparison chart (see Figure 25-3).
3. Construction of keys. Construct a dichotomous key to these specimens using the information in the comparison chart.
COMPARISON CHART
Decumaria Itea Ribes Parnassia Heuchera Saxifraga
Habit Woody vine Shrub Shrub Herb Herb Herb
Leaf arrangement Opposite Alternate Alternate
or on spur roots Basal
(Rosulate) Basal
(Rosulate) Basal
(Rosulate)
Petal Number 7-10 5 5 5 5 5
Locule Number 7-10 2 1 1 1 2
Stamen Number 7+ 5 5 5
(stamonodia 5) 5 10
Fruit Type Capsule Capsule Berry Capsule Capsule Capsule
Figure 25-3. A comparison chart used in the construction of keys (for six of the genera in Figures 25-1 and 25-2).
Spirogyra
In almost every ditch in Holland with reasonably clean water we will in summer find slimy masses of filamentous algae, floating as scum on the surface. It looks rather distasteful, but a ditch like that is not polluted, only eutrophic (rich in nutrients). In spring these filamentous algae grow under water but when there is enough sunlight and the temperatures are not too low, they produce a lot of oxygen, sticking in little bubbles between the tangles of the algae. These come to the surface and become visible as slimy green masses. In these tangles we will find mainly three types of filamentous algae, Spirogyra, Mougeotia and Zygnema. In this article we will mainly write about Spirogyra.
From a distance these slimy tangles look perhaps a bit dirty, but under the microscope the filaments are very beautiful and moreover, they have a spectacular way of reproducing. Spirogyra owes its name to a chloroplast (the green part of the cell) that is wound into a spiral, a unique property of this genus which makes it easily to recognise. In The Netherlands up till now there are found more than 60 species of Spirogyra , in the whole world more than 400.
For the determination of a species it is necessary to look for reproducing specimens with spores. But a precise determination is not necessary for learning a lot of interesting facts from Spirogyra. It is easy to see that there are many species ; in a clean, eutrophic ditch with hard water in Holland we will find easily 20 different species. If we look at a filament of Spirogyra with the microscope, the first thing that attracts attention is the chloroplast, a narrow, banded spiral with serrated edges. The small round bodies in the chloroplast are pyrenoids, centres for the production of starch. In the middle of the cell we see the transparent nucleus, with fine strands linking it to the peripheral protoplasm. The filaments contain cells of different sizes and it is easy to find a new cell, just formed after a division.
The really interesting part comes as Spirogyra reproduces sexually. When two filaments are close together, the process starts. Cell outgrowths form connections between the filaments and a sort of ladder is formed. The contents of the cells in one filament will go through the connection tubes to the cells in the other filament. A zygospore is formed with a thick cell wall, round or oval and with a brownish colour. This conjugation process takes place especially between half May and half June. The spores are liberated, sink to the bottom and germinate in the next spring to form a new filament. It is very worthwhile to look in a sample of algae for the different stages of this conjugation process. It is always a nice surprise to find the conjugating filaments. Spirogyra can also exhibit, apart from the ladder like conjugation, another form of conjugation. Two neighbouring cells in the same filament can connect via a tube.
There are several other genera of related filamentous algae; Zygnema and Mougeotia, with respectively star like and plate like chloroplasts. These genera live in general in more acid, soft fresh water. The conjugation figures look different from those in Spirogyra, for instance X-like. Dune pools are a rich biotope for Spirogyra. In ditches the amount of species declines when the water becomes very eutrophic. Other filamentous algae then replace it, like Cladophora, Vaucheria and Enteromorpha. In the end we only will find duck weed. Then a ditch does not receive light, with disastrous consequences for the growths of plants and the production of oxygen.
The Filamentous Algae.
This gallery includes only the filamentous green algae. The group is a heterogeneous one in which the members, although superficially similar, show a wide diversity in their life cycle and modes of reproduction. Spirogyra, Oedogonium and Cladophora are amongst the varieties most frequently encountered.
All blue-green algae are now classified amongst the Bacteria, and will be found in the Cyanobacteria gallery.
Spirogyra.
Spirogyra is a filamentous green alga which is common in freshwater habitats. It has the appearance of very fine bright dark-green filaments moving gently with the currents in the water, and is slimy to the touch when attempts are made to collect it. The slime serves to deter creatures which otherwise attatch themselves to underwater plants, so Spirogyra under the microscope is usually spotless.
A field of Spirogyra filaments. Their appearance is not quite typical in that the nuclei are unusually prominent, and the characteristic spiral chloroplasts are so fine and tightly wound that close examination is needed to confirm the identification. In any case the possession of spiral chloroplasts is sufficient to positively identify Spirogyra to genus.
Darkfield, x120.
The central portion of a cell of Spirogyra showing the nucleus and giving an insight into the way the spiral chloroplast contacts with the wall of the cell. The filament in the background provides another view.
Brightfield. x1000.
Central portion of a Spirogyra cell showing nucleus and chloroplasts.
Brightfield, x1000.
This filament of Spirogyra is about to break into two filaments. The wall of each cell (centre of picture) has developed an inward indentation at the junction between the cells. Increase in pressure in each cell will cause the indentation to pop out, forcing separation of the filaments, and leaving them with highly convex ends.
Brightfield. x1000.
Two filaments of Spirogyra, the lower one clearly showing the nucleus. This picture also gives a good insight into the way the chloroplasts line the wall of the cell.
Brightfield. x1000.
Conjugation in Spirogyra.
In common with other members of its phylum (Gamophyta) Spirogyra lacks a motile variant at all stages of its life history; ie, no motile gametes (ova or sperm), no zoospores etc. Sexual reproduction is by a process called conjugation -- another of the famously remarkable sights available to the microscopist.
Although it is not possible to distinguish them visually, certain filaments in a loose parallel bundle of Spirogyra assume the female, and others the male, role in the process which follows. The cells of adjacent filaments develop bumps which grow towards one another and eventually fuse to form a continuous tube between the cells. Meanwhile the contents of each cell have detatched themselves from their respective cell walls and have formed a round ball. Over a relatively short space of time (minutes), the green spheres from the male filament squeeze their way down the connecting tubes to fuse with a similarly contracted female cell in the other filament. The result of this sexual union is the formation of a zygospore with a tough resistant outer covering within the chambers of the female filament. After a dormant period, these zygotes undergo meiosis and germinate, resulting in new filaments of Spirogyra.
Once seen never forgotten.
The central pair of cells are joined by a conjugation tube which has yet to fuse into form a continuous passage. The cell contents are at a similarly early stage of detatching themselves from the cell wall to form a ball.
By contrast, the two cells to the right contain newly formed zygospores as a result of consummated conjugation.
Male and female cells now occupy the same space, and are pictured before fusion to form a zygospore has taken place. The filament designated female is the one in which the zygospores have formed.
Two mature zygospores of Spirogyra from another part of the specimen which provided the above pictures. In this form, Spirogyra can survive winter or other adverse conditions and germinate in the spring to form new filaments. The hardened outer spore wall can be seen reflecting the light from the darkfield condenser.
Darkfield. x400.
A zygospore of Spirogyra against a background of decaying plant remains and other algal forms.
Darkfield, x400.
It's hard to say what is happening here. It looks like the stage in conjugation of Spirogyra in which the contraction of the cell contents to a ball is not quite complete, and the spiral nature of the chloroplast is still discernable.
Darkfield, x1000.
Cladophora and Microspora.
The filamentous alga Cladophora is a common inhabitatant of freshwater locations. It is called blanket weed in some places -- not an inappropriate name when in late summer dense floating rafts of Cladophora can be found both at the pond's edge and in the open water, buoyed up with the oxygen generated by its own photosynthesis.
Unlike Spirogyra, Cladophora is capable of branching, and seems to produce little or no mucilagineous secretion. This, and the fact that salts tend to crystallize on the filaments of older specimens, gives it a rougher, grittier feel than other filamentous algae. It is also more readily colonized by epiphytic diatoms and other algae, and provides a protected foraging environment for the smaller pond creatures such as protozoa, worms, small crustaceans and insect larvae.
Its springiness also makes it more difficult to prepare the thin, flat specimens required by the microscope.
Branching in this filament of Cladophora has begun with an outgrowth of the cell at the upper end near the cell wall junction. As the branch grows, differential growth of the main cell wall causes the branch to grow forwards rather than at right angles to the original cell.
An interesting feature of the picture is the distribution of plastids in the two cells shown. Since the plastids are the energy converters of the cell, large numbers have migrated into the growing branch, where the energy requirement is greatest.
The cell on the right shows a distribution of plastids normal to a resting cell.
Darkfield, x300.
Picture shows Cladophora at a branching point. The filaments are encrusted with diatoms (Gomphonema) and crystals of calcium carbonate which give the plant its rough gritty feel.
Darkfield, x400.
Microspora is common in ponds, especially in the winter months. It can be recognized by its reticulated chloroplast which covers the inside wall of the cell including the cell walls between one cell and the next.
Darkfield, x600.
Pteridium aquilinum
Bracken Fern
Bracken Fern
Photo © by Earl J.S. Rook
Flora, fauna, earth, and sky...
The natural history of the northwoods
Name: • Pteridium, from the Greek pteris (pteris), "fern"
• aquilinum, from the Latin, "eagle like"
• Bracken, an old English word for all large ferns, eventually applied to this species in particular.
• Other common names include: Brake, Brake Fern, Eagle Fern, Female Fern, Fiddlehead, Hog Brake, Pasture Brake, Western Brackenfern, Grande fougere, Fougere d'aigle, Warabi (Qué), Örnbräken, Bräken, Slokörnbräken, Taigaörnbräken, Vanlig Örnbräken (Swe), Einstape (Nor), Ørnebregne (Dan), Sananjalka (Fin), Adlerfarn (Ger), Kilpjalg, Kotkajalg, Põldsõnajalg, Seatinarohi, Sõnajalg (Estonia)
Taxonomy: • Kingdom Plantae, the Plants
o Division Polypodiophyta, the True Ferns
Class Filicopsida
Order Polypodiales
Family Dennstaedtiaceae
Genus Pteridium
• Taxonomic Serial Number: 17224
• Also known as Pteris aquilina, Asplenium aquilinum, Allosorus aquilinus, Ornithopteris aquilina, Filix aquilina, Filix-foemina aquilina, Pteris latiuscula
• Considered a single, worldwide species, although some disagree
Description: • A large, deciduous, rhizomatous fern
• Fronds 1'-3' w/leaf stalk up to 3'' but usually shorter than leaf blade. Blades of frond divided into pinnae, the bottom pair sometimes large enough to suggest a three part leaf. Pinna divided into pinnules. On fertile fronds the spores are borne in sori beneath the outer margins of the pinnules. Fronds are killed by frost each winter and new fronds grow in spring. Dead fronds form a mat of highly flammable litter that insulates the below-ground rhizomes from frost when there is no snow cover. This litter also delays the rise in soil temperature and emergence of frost-sensitive fronds in the spring.
• Rhizomes are the main carbohydrate and water storage organs (87% water). Rhizomes can be up to 1" diameter and branching is alternate. The rhizome system has two components. The long shoots form the main axis or stem of the plant. They elongate rapidly, have few lateral buds, do not produce fronds, and store carbohydrates. Short shoots, or leaf-bearing lateral branches, may be closer to the soil surface. They arise from the long shoots, are slow growing, and produce annual fronds and many dormant frond buds. Transition shoots start from both short and long shoots and may develop into either.
• Roots thin, black, brittle extending from the rhizome to over 20" inches into the soil.
• Brackenfern is a large, coarse, perennial fern that has almost horizontal leaves and can grow 1½ to 6½ feet tall (sometimes up to 10 feet). Unlike our more typical broadleaf perennials, this primitive perennial lacks true stems. Each leaf arises directly from a rhizome (horizontal underground stem), and is supported on a rigid leaf stalk. In addition, brackenfern does not produce flowers or seeds. Instead, it reproduces by spores and creeping rhizomes. This species often forms large colonies.
• Root system - The black, scaly, creeping rhizomes (horizontal underground stems) are ½ inch thick, and can grow as much as 20 feet long and 10 feet deep. Stout, black, wide-spreading roots grow sparsely along the rhizomes.
• Seedlings & Shoots - The curled leaves (fiddleheads) emerging from rhizomes in the spring are covered with silvery gray hair.
• Stems - The leaf stalk (not a true stem) is tall (about the same length as the leaf), smooth, rigid and grooved in front. It is green when young, but turns dark brown later in the season.
• Leaves - The leaf stalk supports a broad (3 feet long, 3 feet wide), triangular, dark green, leathery and coarse-textured leaf that often bends nearly horizontal. The leaf is divided into 3 parts, 1 terminal and 2 opposite. Each of the leaf parts is triangular and composed of numerous oblong, pointed leaflets, which are in turn composed of narrow, blunt-tipped subleaflets.
• Fruits & Seeds - A continuous line of spore cases (spore-producing structures) is formed along the underside edge of leaflets, but the spore cases are partially or completely covered by inrolled leaf margins and are difficult to see. Spore cases produce minute, brown spores.
• Biology: Spores of brackenfern are produced August through September. Brackenfern is one of the earliest ferns to appear in spring or after a fire. It sometimes forms large colonies of nearly solid stands. In the fall, it is one of the first plants to be killed by frost, resulting in large patches of crisp, brown foliage.
• Brackenfern is resistant to many herbicides and is tolerant of various forms of mechanical control. However, effective control has been obtained by repeated removal of aboveground growth, which eventually exhausts the food reserves in the rhizomes.
Identification: • Distinguished from other large North Country ferns by the large three part leaf atop a tall stalk.
• Field Marks
o broad triangular leaf held almost parallel to the ground
o smooth, grooved, rigid stalk about as long as the leaf
o narrowed tip to leaflets
Distribution: • Global; throughout the world with the exception of hot and cold deserts
Habitat: • Fossil evidence suggests that bracken fern has had at least 55 million years to evolve and perfect antidisease and antiherbivore chemicals. It produces bitter tasting sesquiterpenes and tannins, phytosterols that are closely related to the insect moulting-hormone, and cyanogenic glycosides that yield hydrogen cyanide (HCN) when crushed. It generates simple phenolic acids that reduce grazing, may act as fungicides, and are implicated in bracken fern's allelopathic activity. Severe disease outbreaks are very rare in bracken fern.
• Grows on a variety of soils with the exception of heavily waterlogged sites. Efficient stomatal control allows it to succeed on sites that would be too dry for most ferns, and its distribution does not normally seem limited by moisture. Grows best on deep, well-drained soils with good water-holding capacity, and may dominate other vegetation on such sites.
• Rhizomes are particularly effective at mobilizing phosphorus from inorganic sources into an available form for plant use. Bracken fern contributes to potassium cycling on sites and is associated with high levels of potassium.
• In northern climates bracken fern is frequently found on uplands and side slopes, since it is susceptible to spring frost damage. Fronds growing in the open or without litter cover are often killed as crosiers by spring frost damage, since the soil warms earlier and growth begins sooner. The result is that fronds appear earlier in shaded habitats.
• A shade intolerant pioneer and succession species that is sufficiently shade tolerant to survive in light spots in old growth forests.
• Light, windborne spores allow colonization of newly vacant areas.
• Despite production of bitter-tasting compounds, chemicals that interfere with insect growth, and toxic chemicals, bracken fern hosts a relatively large number and variety of herbivorous insects.
• Competition: Invades cultivated fields and disturbed areas, effectively competing for soil moisture and nutrients. Rhizomes grow under the roots of herbs and tree or shrub seedlings, and when the fronds emerge, they shade the smaller plants. In the winter dead fronds may bury other plants and press them to the ground. On some sites shading may protect tree seedlings and increase survival.
• Allelopathy: Bracken fern's production and release of allelopathic chemicals is an important factor in its ability to dominate other vegetation. Farther north no allelopathic chemicals are released from the green fronds but are readily leached from standing dead fronds. Herbs may be inhibited for a full growing season after bracken fern is removed, apparently because active plant toxins remain in the soil.
Fire: • A fire-adapted species throughout the world. Not merely well adapted to fire, it promotes fire by producing a highly flammable layer of dried fronds every fall. Repeated fires favor Bracken.
• Primary fire adaptation is deeply buried rhizomes which sprout vigorously following fires before most competing vegetation is established. Windborne spores may disperse over long distances.
• Fire removes competition and creates the alkaline soil conditions suitable for its establishment from spores
• Fuel loading in areas dominated by bracken fern can be quite high.
Associates: • Shrubs: Bunchberry (Cornus canadensis), Twinflower (Linnaea borealis)
• Herbs: Wild Sarsaparilla (Aralia nudicaulis), Large Leaf Aster (Aster macrophyllus), Blue Bead Lily (Clintonia borealis), Gold Thread (Coptis trifolia), Bedstraws (Galium ssp.), Oak Fern (Gymnocarpium dryopteris), Canada Mayflower (Maianthemum canadense), Bishop's Cap (Mitella nuda), One Flowered Pyrola (Moneses uniflora), One Sided Pyrola (Pyrola secunda), Rose Twisted Stalk (Streptopus rosea), Starflower (Trientalis borealis), Kidney Leaf Violet (Viola renifolia), Violets (Viola spp.)
• Mammals: Palatability is usually nil to poor
History: • Considered so valuable during the Middle Ages it was used to pay rents.
• Used as roofing thatch and as fuel when a quick hot fire was desired.
• The ash was used as a source of potash in the soap and glass industry until 1860 and for making soap and bleach. The rhizomes were used in tanning leathers and to dye wool yellow.
• Bracken still used for winter livestock bedding in parts of Wales since it is more absorbent, warmer, and easier to handle than straw.
• Also used as a green mulch and compost
Uses: • Most commonly used today as a food for humans. The newly emerging croziers or fiddleheads are picked in spring and may be consumed fresh or preserved by salting, pickling, or sun drying. Both fronds and rhizomes have been used in brewing beer, and rhizome starch has been used as a substitute for arrowroot. Bread can be made out of dried and powered rhizomes alone or with other flour. American Indians cooked the rhizomes, then peeled and ate them or pounded the starchy fiber into flour. In Japan starch from the rhizomes is used to make confections. Bracken fern is grown commercially for use as a food and herbal remedy in Canada, the United States, Siberia, China, Japan, and Brazil and is often listed as an edible wild plant. Powdered rhizome has been considered particularly effective against parasitic worms. American Indians ate raw rhizomes as a remedy for bronchitis
• Bracken fern has been found to be mutagenic and carcinogenic in rats and mice, usually causing stomach or intestinal cancer. It is implicated in some leukemias, bladder cancer, and cancer of the esophagus and stomach in humans. All parts of the plant, including the spores, are carcinogenic, and face masks are recommended for people working in dense bracken. The toxins in bracken fern pass into cow's milk. The growing tips of the fronds are more carcinogenic than the stalks. If young fronds are boiled under alkaline conditions, they will be safer to eat and less bitter.
• Bracken fern is a potential source of insecticides and it has potential as a biofuel. Bracken fern increases soil fertility by bringing larger amounts of phosphate, nitrogen, and potassium into circulation through litter leaching and stem flow; its rhizomes also mobilize mineral phosphate. Bracken fern fronds are particularly sensitive to acid rain which also reduces gamete fertilization. Both effects signal the amount of pollutants in rain water making bracken fern a useful indicator.
• Fronds may release hydrogen cyanide (HCN) when they are damaged (cyanogenesis), particularly the younger fronds. Herbivores, including sheep, selectively graze young fronds that are acyanogenic (without HCN) Lignin, tannin, and silicate levels tend to increase through the growing season making the plants less palatable. Cyanide (HCN) levels fall during the season as do the levels of a thiaminase which prevents utilization of B vitamins.
• Toxicity: Known to be poisonous to livestock throughout the US, Canada, and Europe. Simple stomach animals like horses, pigs, and rats develop a thiamine deficiency within a month. Acute bracken poisoning affects the bone marrow of both cattle and sheep, causing anemia and hemorrhaging which is often fatal. Blindness and tumors of the jaws, rumen, intestine, and liver are found in sheep feeding on bracken fern.
• Toxicity: All parts of brackenfern, including rootstocks, fresh or dry leaves, fiddleheads and spores, contain toxic compounds, and are poisonous to livestock and humans. Consumption of brackenfern causes vitamin B1 deficiency in horses, and toxins can pass into the milk of cattle. Young leaves of brackenfern have been used as a human food source, especially in Japan, and may be linked to increased incidence of stomach cancer. Humans working outdoors near abundant stands of the plant may be at risk from cancer-causing compounds in the spores.
• Facts and Folklore:
It was once thought that, if the spores of the brackenfern were gathered on St. John's Eve, it would make the possessor invisible.
In the 17th century, live brackenfern was set on fire in hopes of producing rain.
Brackenfern fiddleheads have been used as a food source; however, their consumption has been linked to various types of cancer in humans.
Reproduction: • Reproduces by spores and vegetatively by rhizomes
• Most regeneration is vegetative. Many have searched for young plants growing from spores, but few have found them. However, spores do germinate and grow readily in culture.
• Young plants produce spores by the end of the second growing season in cultivation but normally do not produce spores until the third or fourth growing season. A single, fertile frond can produce 300,000,000 spores annually. Spore production varies from year to year depending on plant age, frond development, weather, and light exposure. Production decreases with increasing shade. The wind-borne spores are extremely small. Dry spores are very resistant to extreme physical conditions, although the germination of bracken fern spores declines from 95-96% to around 30-35% after 3 years storage. The spores germinate without any dormancy requirement. Under favorable conditions, young plants could be found 6 to 7 weeks after the spores are shed. Under normal conditions the spores may not germinate until the spring after they are shed.
• Sufficient moisture and shelter from wind are important factors in fern spore germination. Bracken fern spore germination appears to require soil sterilized by fire. On unsterilized soils spores may germinate, but the new plants are quickly overwhelmed by other growth. Temperatures between 59º and 86º F are generally best for germination, although bracken fern is capable of germination at 33º-36ºF.
• A pH range of 5.5 to 7.5 is optimal for germination. Germination is indifferent to light quality; it is one of the few ferns that can germinate in the dark. Despite limitations on spore germination, genotype analysis in the Northeast indicates that many stands of bracken fern represent multiple establishment of individuals from spores.
• When spores germinate, they produce bisexual, gamete-bearing plants about ¼" in diameter and one cell thick. These tiny plants have no vascular system and require very moist conditions to survive. The young spore-bearing plant which develops from the fertilized egg is initially dependent on the gametopyte until it develops its first leaf and roots. The first fronds are simple and lobed. They develop into thin, delicate fronds divided into lobed pinnae. They do not look like adult plants and are frequently not recognized as bracken fern. Cultivated plants begin to resemble adult fern after 18 weeks. The rhizomes begin to develop after there are a number (up to 10) of fronds and a well-developed root system or in the fifteenth week of growth under optimal conditions.In the first year rhizomes may grow to 86 inches long. By the end of a second year the rhizome system may exceed 6' in diameter.
• Aggressive rhizome system gives it the ability to reproduce vegetatively and reduces dependence on water for reproduction. The rhizomatous clones can be up to 400' in diameter and hundreds of years old; some clones alive today may be over 1,000 years old.
• Rhizomes have a high proportion of dormant buds. When disturbed or broken off, all portions of the rhizome may sprout, and plants growing from small rhizome fragments revert temporarily to a juvenile morphology.
• Shaded plants produce fewerspores than plants in full sun
• Bracken fern is a survivor. The fronds are generally killed by fire, but some rhizomes survive. The rhizomes are sensitive to elevated temperatures. During fires the rhizome system is insulated by mineral soil. Depth of the main rhizome system is normally between 3½" and 12" short rhizomes may be within 1½" of the surface and some rhizomes may be as deep as 40".
• Well known postfire colonizer in eastern pine and oak forests. Fire benefits bracken by removing competition while it sprouts profusely from surviving rhizomes. New sprouts are more vigorous following fire, and bracken fern becomes more fertile, producing far more spores than it does in the shade
• Spores germinate well on alkaline soils, allowing them to establish in the basic conditions created by fire.
Propagation: • Division most successful method
Cultivation: • Hardy to USDA Zone 3 (average minimum annual temperature -40ºF)
• Characteristically found on soils with medium to very rich nutrients.
• Cultivated and shaded plants produce fewer, thinner but larger fronds than open-grown plants
Population Genetics and Evolution
Introduction:
In 1908, G.H.Hardy and W. Weinberg independently suggested a scheme whereby evolution could be viewed as changes in frequency of alleles in a population of organisms. In this scheme, if A and a are alleles for a particular gene locus and each diploid individual has two such loci, then p can be designated as the frequency of the A allele and q as the frequency of the a allele. For example, in a population of 100 individuals ( each with two loci ) in which 40% of the alleles are A, p would be 0.40. The rest of the alleles would be ( 60%) would be a and q would be equal to 0.60. p + q = 1 These are referred to as allele frequencies. The frequency of the possible diploid combinations of these alleles ( AA, Aa, aa ) is expressed as p2 +2pq +q2 = 1.0. Hardy and Weinberg also argued that if 5 conditions are met, the population's alleles and genotype frequencies will remain constant from generation to generation. These conditions are as follows:
• The breeding population is large. ( Reduces the problem of genetic drift.)
• Mating is random. ( Individual show no preference for a particular mating type.)
• There is no mutation of the alleles.
• No differential migration occurs. ( No immigration or emigration.)
• There is no selection. ( All genotypes have an equal chance of surviving and reproducing.)
The Hardy-Weinberg equation describes an existing situation. Of what value is such a rule? It provides a yardstick by which changes in allelic frequencies can be measured. If a population's allelic frequencies change, it is undergoing evolution.
Estimating Allele Frequencies for a Specific Trait within a Sample Population:
Using the class as a sample population, the allele frequency of a gene controlling the ability to taste the chemical PTC (phenylthiocarbamide) could be estimated. A bitter taste reaction is evidence of the presence of a dominant allele in either a homozygous (AA) or heterozygous (Aa) condition. The inability to taste the PTC is dependent on the presence of the two recessive alleles (aa). Instead of using the PTC paper the trait for tongue rolling may be substituted. To estimate the frequency of the PTC -tasting allele in the population, one must find p. To find p, one must first determine q ( the frequency of the non tasting allele).
Procedure:
1. Using the PTC taste test paper, tear off a short strip and press it to your tongue tip. PTC tasters will sense a bitter taste.
2. A decimal number representing the frequency of tasters (p2+2pq) should be calculated by dividing the number of tasters in the class by the total number of students in the class. A decimal number representing the frequency of the non tasters (q2) can be obtained by dividing the number of non tasters by the total number of students. You should then record these numbers in Table 8.1.
3. Use the Hardy-Weinberg equation to determine the frequencies (p and q ) of the two alleles. The frequency q can be calculated by taking the square root of q2. Once q has been determined, p can be determined because 1-q=p. Record these values in Table 8.1 for the class and also calculate and record values of p and q for the North American population.
Table 8.1 Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles
Phenotypes Allele Frequency Based on the H-W Equation
Tasters (p2+2pq) Non Tastes(q2) p q
Class Population #= %= #= %=
North American Population 0.55 0.45
Topics for Discussion:
1. What is the percentage of heterozygous tasters (2pq) in your class? ______________________.
2. What percentage of the North American population is heterozygous for the taster allele? _____________
Case Studies:
Case 1 ( Test of an Ideal Hardy-Weinberg Community)
The entire class will represent a breeding population, so find a large open space for its simulation. In order to ensure random mating, choose another student at random. In this simulation, we will assume that gender and genotype are irrelevant to mate selection.
The class will simulate a population of randomly mating heterozygous individuals with an initial gene frequency of 0.5 for the dominant allele A and the recessive allele a and genotype frequencies of 0.25AA, 0.50Aa, and 0.25aa. Record this on the Data page at the end of the lab. Each member of the class will receive four cards. Two cards will have A and two cards will have a. The four cars represent the products of meiosis. Each "parent" will contribute a haploid set of chromosomes to the next generation.
Procedure:
1. Turn the four cards over so the letters are not showing, shuffle them, and take the card on top to contribute to the production of the first offspring. Your partner should do the same. Put the cards together. The two cards represent the alleles of the first offspring. One of you should record the genotype of this offspring in the Case 1 section at the end of the lab. Each student pair must produce two offspring, so all four cards must be reshuffled and the process repeated to produce a second offspring.
2. The other partner should then record the genotype of the second offspring in the Case 1 section at the end of the lab. Using the genotypes produced from the matings, you and your partner will mate again using the genotypes of the two offspring. That is , student 1 assumes the genotype of the first offspring, and student 2 assumes the genotype of the second offspring.
3. Each student should obtain, if necessary, new cards representing their alleles in his or her respective gametes after the process of meiosis. For example, student 1 becomes the genotype Aa and obtains cards A,A,a,a; student 2 becomes aa and obtains cards,a,a,a,a. Each participant should randomly seek out another person with whom to mate in order to produce offspring of the next generation. You should follow the same mating procedure as for the first generation, being sure you record your new genotype after each generation in the Case 1 section. Class data should be collected after each generation for five generations. At the end of each generation, remember to record the genotype that you have assumed. Your teacher will collect class data after each generation by asking you to raise your hand to report your genotype.
Allele frequency: The allele frequencies, p and q, should be calculated for the population after five generations of simulated random mating.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA _____________ X 2= _______________ A alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ A alleles
p = Total number of A alleles =
Total number of alleles in the population
In this case, the total number of alleles in the population is equal to the number of students in the class X 2.
Number of a alleles present at the fifth generation
Number of offspring with genotype aa _____________ X 2= _______________ a alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ a alleles
q = Total number of a alleles =
Total number of alleles in the population
1. What does the Hardy-Weinberg equation predict for the new p and q?.
_____________________________________________________________________
_____________________________________________________________________
2. Do the results you obtained in this simulation agree? __________ If not, why not?
_____________________________________________________________________
_____________________________________________________________________
3. What major assumption(s) were not strictly followed in this simulation?
_____________________________________________________________________
_____________________________________________________________________
Case 2 ( Selection )
In this case you will modify the simulation to make it more realistic. in the natural environment , not all genotypes have the same rate of survival; that is, the environment might favor some genotypes while selecting against others. An example is the human condition sickle-celled anemia. It is a condition caused by a mutation on one allele, in which a homozygous recessive does not survive to reproduce. For this simulation you will assume that the homozygous recessive individuals never survive. Heterozygous and homozygous dominant individuals always survive.
The procedure is similar to that for Case 1. Start again with your initial genotype, and produce your "offspring" as in Case 1. This time, However, there is one important difference. Every time your offspring is aa it does not reproduce. Since we want to maintain a constant population size, the same two parents must try again until they produce two surviving offspring. You may need to get new allele cards from the pool.
Proceed through five generations, selecting against the homozygous offspring 100% of the time. Then add up the genotype frequencies that exist in the population and calculate the new p and q frequencies in the same way as it was done in Case 1.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA _____________ X 2= _______________ A alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ A alleles
p = Total number of A alleles =
Total number of alleles in the population
In this case, the total number of alleles in the population is equal to the number of students in the class X 2.
Number of a alleles present at the fifth generation
Number of offspring with genotype aa _____________ X 2= _______________ a alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ a alleles
q = Total number of a alleles =
Total number of alleles in the population
1. How do the new frequencies of p and q compare to the initial frequencies in Case 1?
_____________________________________________________________________
_____________________________________________________________________
2. How has the allelic frequency of the population changed?
_____________________________________________________________________
_____________________________________________________________________
3. Predict what would happen to the frequencies of p and q if you simulated another 5 generations.
_____________________________________________________________________
_____________________________________________________________________
4. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Hardy-Weinberg Problems
1. In Drosophila, the allele for normal length wings is dominant over the allele for vestigial wings. In a population of 1,000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
2. The allele for the ability to roll one's tongue is dominant over the allele for the lack of this ability. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
3. The allele for the hair pattern called "widow's peak" is dominant over the allele for no "widow's peak." In a population of 1,000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?
4. In a certain population, the dominant phenotype of a certain trait occurs 91 % of the time. What is the frequency of the dominant allele?
Data Page:
Case 1 ( Hardy-Weinberg Equilibrium )
Initial Class Frequencies:
AA ________ Aa________ aa_________
My initial genotype :_______________
F1 Genotype ______
F2 Genotype ______
F3 Genotype ______
F4 Genotype ______
F5 Genotype ______
Final Class Frequencies:
AA ________ Aa________ aa_________
p _________ q __________
Case 2 ( Selection )
Initial Class Frequencies:
AA ________ Aa________ aa_________
My initial genotype :_______________
F1 Genotype ______
F2 Genotype ______
F3 Genotype ______
F4 Genotype ______
F5 Genotype ______
Final Class Frequencies:
AA ________ Aa________ aa_________
p _________ q __________
Biology 198
PRINCIPLES OF BIOLOGY
Hardy-Weinberg practice questions
POPULATION GENETICS AND THE HARDY-WEINBERG LAW
The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:
1. No mutations must occur so that new alleles do not enter the population.
2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
3. Random mating must occur (i.e. individuals must pair by chance)
4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
5. No selection can occur so that certain alleles are not selected for, or against.
Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place:
ANSWERS TO THE QUESTIONS
Remember the basic formulas:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
1. PROBLEM #1.
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
A. The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself.
B. The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
C. The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
D. The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.
2. PROBLEM #2.
Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene? Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).
3. PROBLEM #3.
There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:
A. The frequency of the recessive allele. Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%).
B. The frequency of the dominant allele. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%).
C. The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).
4. PROBLEM #4.
Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:
A. The percentage of butterflies in the population that are heterozygous.
B. The frequency of homozygous dominant individuals.
Answers: The first thing you'll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency of homozygous dominant individuals? That would be p2 or (0.37)2 = 0.14.
5. PROBLEM #5.
A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:
A. The allele frequencies of each allele. Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don't have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355.
B. The expected genotype frequencies. Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above).
C. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0.458 x 953 = about 436.
D. The expected phenotype frequencies. Answer: Well, the "A" phenotype = 0.126 + 0.458 = 0.584 and the "a" phenotype = 0.416 (you already knew this from part A above).
E. Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? Answer: Simply put, The "A" phenotype = 0.584 x 1,245 = 727 tan-sided and the "a" phenotype = 0.416 x 1,245 = 518 red-sided ( or 1,245 - 727 = 518).
6. PROBLEM #6.
A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.
7. PROBLEM #7.
After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island? Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.
8. PROBLEM #8.
You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:
BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY
M MM 490 0.49
MN MN 420 0.42
N NN 90 0.09
Using the data provide above, calculate the following:
A. The frequency of each allele in the population. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3.
B. Supposing the matings are random, the frequencies of the matings. Answer: This is a little harder to figure out. Try setting up a "Punnett square" type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:
MM (0.49) MN (0.42) NN (0.09)
MM (0.49) 0.2401* 0.2058 0.0441
MN (0.42) 0.2058 0.1764* 0.0378
NN (0.09) 0.0441 0.0378 0.0081*
C. Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three "unique" combinations. Thus, three of the possibilities must be doubled.
D. MM x MM = 0.49 x 0.49 = 0.2401
MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
MN x MN = 0.42 x 0.42 = 0.1764
MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
NN x NN = 0.09 x 0.09 = 0.0081
E. The probability of each genotype resulting from each potential cross. Answer: You may wish to do a simple Punnett's square monohybrid cross and, if you do, you'll come out with the following result:
MM x MM = 1.0 MM
MM x MN = 0.5 MM 0.5 MN
MM x NN = 1.0 MN
MN x MN = 0.25 MM 0.5 MN 0.25 NN
MN x NN = 0.5 MN 0.5 NN
NN x NN = 1.0 NN
9. PROBLEM #9.
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
A. The frequency of the recessive allele in the population. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
B. The frequency of the dominant allele in the population. Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%).
C. The percentage of heterozygous individuals (carriers) in the population. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.
10. PROBLEM #10.
In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)? Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.
11. PROBLEM #11.
The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies. Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. You already know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 = 0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and 0.495 and these should equal 1.0 or very close to it. This type of problem may be on the exam.
12. PROBLEM #12. (You will not have this type of problem on the exam)
What allelic frequency will generate twice as many recessive homozygotes as heterozygotes? Answer: We need to solve for the following equation: q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or another way of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q, we can substitute 1 - q for p and, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 - 4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, we get -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is 0.8 =q. I cannot imagine you getting this type of problem in this general biology course although if you take algebra good luck
Hornworts, liverworts, and mosses - commonly referred to as bryophytes - are considered to be a pivotal group in our understanding of the origin of land plants because they are believed to be among the earliest diverging lineages of land plants. Mosses, liverworts and hornworts are found throughout the world in a variety of habitats, from the harsh environs of Antarctica to the lush conditions of the tropical rainforests. Bryophytes are unique among land plants in that they possess an alternation of generations, which involves a dominant, free-living, haploid gametophyte alternating with a reduced, generally dependent, diploid sporophyte. Bryophytes are small, herbaceous plants that grow closely packed together in mats or cushions on rocks, soil, or as epiphytes on the trunks and leaves of forest trees. Bryophytes are remarkably diverse for their small size and are well-adapted to moist habitats and flourish particularly well in moist, humid forests like the fog forests of the Pacific northwest or the montane rain forests of the southern hemisphere.
Significance of bryophytes
Bryophytes have a significant role in contributing to nutrient cycles, providing seed-beds for the larger plants of the community, and form microhabitats for insects and an entire array of microorganisms. Bryophytes are also very effective rainfall interceptors, and the overwhelming abundance of epiphytic liverworts in "cloud" or "mossy" forest zones is considered an important factor in eliminating the deteriorating effect of heavy rains, including adding to hill stability and helping to prevent soil erosion. The chemical compounds of some liverworts are also particularly interesting because they have important biological activities, for example, against certain cancer cell lines, anti-bacterial properties, anti-microbial, anti-fungal, and muscle relaxing activity.
Classification
Over the last decade, recent advances in DNA sequencing technology and analytical approaches to phylogenetic reconstruction, including the use of ultra-structural, morphological and anatomical data, have enabled unprecedented progress toward our understanding of plant evolution. A growing consensus suggests that the bryophytes possibly represent three separate evolutionary lineages, which are today recognized as mosses (phylum Bryophyta), liverworts (phylum Marchantiophyta) and hornworts (phylum Anthocerotophyta).
• Mosses (Bryophyta)
The greatest species diversity in bryophytes is found in the mosses, with estimates of the number of species ranging from 10,000 to 15,000. Higher-level classification of the mosses remains unresolved with considerable difference of opinion on the names of the major groups. However, generally four major groups or classes are recognised. These include: Sphagnopsida (peat or Sphagnum mosses), Andreaeopsida (rock or lantern mosses), Polytrichopsida (nematodontous mosses), and the Bryopsida (arthrodontous mosses). The Sphagnum mosses are one of the most ecologically and economically important groups of bryophytes. The class Bryopsida accounts for the largest and most diverse groups within the mosses with over 100 families.
• Liverworts (Marchantiophyta)
The estimated number of liverwort species range from 6000 to 8000. Traditionally, liverworts have been subdivided into two major groups or classes based, partially, on growth form. The class Marchantiopsida, includes the well-known genera Marchantia, Monoclea, Lunularia, and Riccia, and has a complex thalloid organisation. The class Jungermanniopsida represents an estimated 85% of liverwort species and shows an enormous amount of morphological, anatomical and ecological diversity; plants with leafy shoot systems are the most common growth form in this class, e.g., Frullania, Jubulopsis, Cololejeunea, and Radula.
• Hornworts (Anthocerotophyta)
Hornworts get their name from their long, horn-shaped sporophytes and are the smallest group of bryophytes with only approximately 100 species. Hornworts resemble some liverworts in having simple, unspecialized thalloid gametophytes, but they differ in many other characters. Hornworts differ from all other land plants in having only one large, algal-like chloroplast in each thallus cell.
Phylogeny
Scientific Name, Common Name
Plantae, Land Plants
Embryophytes, Green plants
The Bryophyta or mosses, unlike the liverworts, are present in most terrestrial habitats (even deserts) and may sometimes be the dominant plant life.
As with the liverworts the plant that we commonly see is the gametophyte. It shows the beginnings of differentiation of stem and leaves - but no roots. Mosses may have rhizoids and these may be multicellular but they do little more than hold the plant down.
The stem shows some internal differentiation into hydroids and leptoids which are like xylem and phloem of higher plants but very simply organized with no connection to leaves or branching stems.
The leaves are mostly one cell thick; sometimes the midrib is several cells thick but this does not contain conducting tissue so it is not equivalent to the vein of a leaf.
Male and female gametophytes look identical except when they produce reproductive structures.
The male plant produces clusters of antheridia which contain thousands of ciliate sperm.
The female produces archegonia, each containing a single egg.
Fertilization is dependent on water - sperm are splashed or swim to the archegonia. The zygote grows into the diploid sporophyte which remains attached to the female gametophyte It is a leafless stem with a seta or foot at one end, drawing nutrients from the gametophyte. At the other end is a capsule in which meiosis occurs to form spores.
The archegonium grows around the developing sporophyte for a while but becomes separated from the gametophyte and is carried up to form a cap or calyptra over the sporangium. Curiously, the sporangia of some mosses have stomata much like those on the leaves of vascular plants.
Immature moss capsules with calyptra
The calyptra is lost when the sporangium is mature as is the operculum or lid on the end of the capsule.
Underneath the operculum there are often peristome teeth which open under dry conditions and control spore release A spore germinates to produce a filamentous protonema which sooner or later produces buds that grow into new gametophytes.
Ecology of mosses
Mosses require abundant water for growth and reproduction. They can tolerate dry spells by drying out or,in the case of mosses like Sphagnum , by holding huge amounts of water in dead cells in the leaves.
They look pretty lowly and insignificant, but have become dominant in particular habitats and Sphagnum itself is said to occupy 1% of the earth's surface (half the area of the USA). Because of its ability to soak up blood and its relative freedom from bacterial contamination Sphagnum was used in dressings. The moss itself is used in some horticultural media and it is an important source of peat.
Polytrichum commune one of the larger mosses with mature sporophytes
If you have tried to grow a lawn in a shady location you have probably been troubled by mosses as weeds. Like many lower organisms they are very sensitive to copper salts and can be controlled in this way. On the other hand mosses are green and better adapted to shade than most grasses, so maybe we should accept them in this situation.
Natural Perspective
The Plant Kingdom : Mosses and Allies
Mosses and their allies are small green plants that are simlutaneously overlooked and deeply appreciated by the typical nature lover. On the one hand, very few people pay attention to individual moss plants and species. On the other hand, it is the mosses that imbues our forests with that wonderful lush "Rainforest" quality which soothes the soul and softens the contours of the earth.
These wonderfully soft carpets of green are, in fact, Nature's second line of attack in its war against rocks. After lichens have created a foothold in rocks the mosses move in, ultimately becoming a layer of topsoil for higher plants to take root. The mosses also hold loose dirt in place, thus preventing landslides.
Ecologically and structurally, mosses are closer to lichens than they are to other members of the plant kingdom. Both mosses and lichens depend upon external moisture to transport nutrients. Because of this they prefer damp places and have evolved special methods of dealing with long dry periods. Higher plants, on the other hand, have specialized organs for transporting fluid, allowing them to adapt to a wider variety of habitats.
Bryophytes used to be classified as three classes of a single phylum, Bryophyta . Modern texts, however, now assign each class to its own phylum: Mosses ( Bryophyta ), Liverworts ( Hepatophyta ), and Hornworts ( Anthoceraphyta ). This reflects the current taxonomic wisdom that the Liverworts and Hornworts are more primitive and only distantly related to Mosses and other plants.
Mosses (Phylum: Bryophyta )
All plants reproduce through alternating generations. Nowhere is this more apparent than in the mosses. The first generation, the gametophyte , forms the green leafy structure we ordinarily associate with moss. It produces a sperm and an egg (the gametes) which unite, when conditions are right, to grow into the next generation: the sporophyte or spore-bearing structure.
The moss sporophyte is typically a capsule growing on the end of a stalk called the seta . The sporophyte contains no clorophyl of its own: it grows parasitically on its gametophyte mother. As the sporophyte dries out, the capsule release spores which will grow into a new generation of gametophytes, if they germinate.
Mosses, the most common, diverse and advanced brypophytes, are categorized into three classes: Peat Mosses ( Sphagnopsida ), Granite Mosses ( Andreaopsida ), and "True" Mosses ( Bryopsida or Musci ) .
Shown: Class: Bryopsida ; Order: Hypnales ; Family: Brachythecia ; Homolathecium nutalli (probably)
Leafy Liverworts (Phylum: Hepatophyta , Class: Jungermanniidae )
While people typically know what a moss is, few have even heard of liverworts and hornworts.
These primitive plants function much like mosses and grow in the same places, often intertwined with each other. The liverworts take on one of two general forms, comprising the two classes of liverworts: Jungermanniidea are leafy, like moss; Marchantiopsida are leaf-like ( thalloid ) similar to foliose lichens .
The leafy liverworts look very much like mosses and, in fact, are difficult to tell apart when only gametophytes are present. The "leaves," however, are simpler than moss and dont have a midrib ( costa ). The stalk of the sporophyte is translucent to white; its capsule is typically black and egg-shaped. When it matures, the capsule splits open into four equal quarters, releasing the spores to the air.
The liverwort sporophyte shrivels up and disappears shortly after releasing its spores. Because of this one hardly ever sees liverwort sporophytes out of season. Moss sporophtyes, on the other hand, may persist much longer.
Shown: Class: Jungermanniidea ; Order: Jungermanniales ; Family: Scapaniaceae ; Scapania spp. (probably)
Leaf-like Liverworts (Phylum: Hepatophyta ; Class: Marchantiopsida )
The leaf-like ( thalloid ) liverworts are, on the whole, more substantial and easier to find than their leafy counterparts. The gametophyte is flat, green and more-or-less strap-shaped. The body may, however, branch out several times to round out the form.
When the gametophyte has become fertilized and is ready to produce its sporophyte generation it may grow a tall green umbrella-shaped structure called the carpocephalum . The sporophyte grows on the underside of this structure, often completely hidden from view.
During the dry season, leaf-like liverworts may shrivel up and completely disappear from view until the rains arrive again.
Thalloid liverworts are much easier to identify than their leafy counterparts due to the wider variety of gametophyte shapes.
Shown: Class: Marchnatiopsida ; Order: Marchantiales ; Family: Aytoniaceae ; Asterella californica
Hornworts (Phylum: Anthoceraphyta )
Hornworts are very similar to liverworts but differ in the shape of the sporophyte generation. Instead of generating spores in a capsule atop a stalk, the hornwort generates spores inside a green horn-like stalk. When the spores mature the stalk splits, releasing the spores.
Under the microscope, hornwort cells look quite distinct as well: they have a single, large chloroplast in each cell. Other plants typically have many small chloroplasts per cell. This structure imparts a particular quality of color and translucency to the body ( thallus ) of the plant.
Hornworts are all grouped into a single class, Anthocerotae , containing a single order, Anthocerotales .
Shown: Class: Anthocerotae ; Order: Anthocerotales ; Family: Anthocertaceae ; Phaeoceros spp.
Identification
Suggestions for the Use of Keys
1. Select appropriate keys for the materials to be identified. The keys may be in a flora, manual, guide' handbook, monograph, or revision (see Chapter 30). If the locality of an unknown plant is known, select a flora, guide, or manual treating the plants of that geographic area (see Guides to Floras in Chapter 30). If the family or genus is recognized, one may choose to use a monograph or revision. If locality is unknown. select a general work. If materials to be identified were cultivated, select one of the manuals treating such plants since most floras do not include cultivated plants unless naturalized.
2. Read the introductory comments on format details, abbreviations, etc. .before using the key.
3. Read both leads of a couplet before making a choice. Even though the first lead may seem to describe the unknown material, the second lead may be even more appropriate.
4. Use a glossary to check the meaning of terms you do not understand.
5. Measure several similar structures when measurements are used in the key, e.g. measure several leaves not a single leaf. Do not base your decisions on a single observation It is often desirable to examine several specimens.
6. Try both choices when dichotomies are not clear or when information is insufficient, and make a decision as to which of the two answers best fits the descriptions.
7. Verify your results by reading a description, comparing the specimen with an illustration or an authentically named herbarium specimen.
Suggestions for Construction of Keys
1. Identify all groups to be included in a key.
2. Prepare a description of each taxon (see Chapter 24 for details for description and descriptive format).
3. Select "key characters" with contrasting character states. Use macroscopic, morphological characters and constant character states when possible. Avoid characteristics that can only be seen in the field or on specially prepared specimens, i.e., use those characteristics that are generally available to the user.
4. Prepare a Comparison Chart (see Figure 25-3).
5. Construct strictly dichotomous keys.
6. Use parallel construction and comparative terminology in each lead of a couple.
7. Use at least two characters per lead when possible.
8. Follow key format (indented or bracketed see Figures 25-1 and 25-2).
9. Start both leads of a couple with the same word if at all possible and successive leads with different words.
10. Mention the name of the plant part before descriptive phrases, e.g., leaves or flowers blue not blue flowers, leaves alternate not alternate leaves.
11. Place those groups with numerous variable character states in a key several times when necessary.
12. Construct separate keys for dioecious plants, for flowering or fruiting materials and for vegetative materials when pertinent.
A DICHOTOMOUS KEY TO SELECTED GENERA OF SAXIFRAGACEAE
• Shrub or woody vine.
o Woody vine; petals 7 or more 3. Decumaria
o Shrub; petals 4 or 5.
Leaves alternate or on short spur branches.
Leaves pinnately veined; ovary superior; fruit a capsule 1. Itea
Leaves palmately veined; ovary inferior; fruit a berry 2. Ribes
Leaves opposite.
Petals usually 4;-stamens 20-40; fruit longitudinally dehiscent, not ribbed; 4. Philadelphus
Petals usually 5; stamens 8-10; fruit poricidally dehiscent, 10- to 15-ribbed 5. Hydrangea
• Herbs.
o Staminodia present; petals more than 10 mm long 6. Parnassia
o Staminodia absent; petals less than 10 mm long.
Leaves ternately decompound 7. Astilbe
Leaves simple.
Flowers solitary in leaf axils, or in short, leafy cimes.
Sepals 4; carpels 2 8. Chrysosplenium
Sepals 5; carpels 3 9. Lepuropetalon
Flowers in racemes or panicles.
Petals pinnatifid or fringed; stem leaves opposite 10. Mitella
Petals not pinnatifid or fringed; stem leaves alternate or absent.
Ovary 1-celled.
Inflorescence paniculate; stamens 5 11. Heuchera
Inflorescence racemose; stamens 10 12. Tiarella
Ovary 2-celled.
Stamens 5; leaves palmately lobed 13. Boykinia
Stamens 10; leaves not palmately lobed 14. Saxifraga
A DICHOTOMOUS KEY TO SELECTED GENERA OF SAXIFRAGACEAE
• 1. Shrub or woody vine 2.
• 1. Herbs 6.
o 2. Woody vine; petals 7 or more Decumaria.
o 2. Shrub; petals 4 or 5 3.
• 3. Leaves alternate or on short spur branches 4.
• 3. Leaves opposite 5.
o 4. Leaves pinnately veined; ovary superior; fruit a capsule Itea.
o 4. Leaves palmately veined; ovary inferior; fruit a berry Ribes.
• 5. Petals usually 4; stamens 20-40; fruit longitudinally dehiscent, not ribbed Philadelphus
• 5. Petals usually 5; stamens 8-10; fruit poricidally dehiscent, 10-15 ribbed Hydrangea.
o 6. Staminodia present; petals more than 10 mm long Parnassia.
o 6. Staminodia absent; petals less than 10 mm long 7.
• 7. Leaves ternately decompound Astilbe.
• 7. Leaves simple 8.
o 8. Flowers solitary in leaf axils, or in short, leafy cymes 9.
o 8. Flowers in racemes or panicles 10.
• 9. Sepals 4; carpels 2 Chrysosplenium.
• 9. Sepals 5; carpels 3 Lepuropetalon.
o 10. Petals pinnatifid or fringed; stem leaves opposite Mitella.
o 10. Petals not pinnatifid or fringed; stem leaves alternate or absent 11.
• 11. Ovary 1-celled 12.
• 11. Ovary 2-celled 13.
o 12. Inflorescence paniculate; stamens 5 Heuchera.
o 12. Inflorescence racemose; stamens 10 Tiarella.
• 13. Stamens 5; leaves palmately lobed Boykinia.
• 13. Stamens 10; leaves not palmately lobed Saxifraga.
Figure 25-2. Example of a bracketed key. (Modified from Radford, A. E., 11. E. Ahles, and C. R. Bell. 1968. Manual of the Vascular Flora of the Carolinas. University of North Carolina Press. Chapel Hill, North Carolina. Used with permission.)
PLANT IDENTIFICATION EXERCISE
1. Identification of an unknown. Select an unknown specimen and identify it by keying in an appropriate manual, flora, or monograph. Verify your results by reading a description, by comparing with an illustration or by checking with your instructor.
2. Preparation of a comparison chart. Select 5 or more specimens from the group provided by your instructor. Identify each by keying. Verify your results. Prepare a description of each similar to those in a flora or manual. Be sure characters and character states are in the same order. Select contrasting character states and prepare a comparison chart (see Figure 25-3).
3. Construction of keys. Construct a dichotomous key to these specimens using the information in the comparison chart.
COMPARISON CHART
Decumaria Itea Ribes Parnassia Heuchera Saxifraga
Habit Woody vine Shrub Shrub Herb Herb Herb
Leaf arrangement Opposite Alternate Alternate
or on spur roots Basal
(Rosulate) Basal
(Rosulate) Basal
(Rosulate)
Petal Number 7-10 5 5 5 5 5
Locule Number 7-10 2 1 1 1 2
Stamen Number 7+ 5 5 5
(stamonodia 5) 5 10
Fruit Type Capsule Capsule Berry Capsule Capsule Capsule
Figure 25-3. A comparison chart used in the construction of keys (for six of the genera in Figures 25-1 and 25-2).
Spirogyra
In almost every ditch in Holland with reasonably clean water we will in summer find slimy masses of filamentous algae, floating as scum on the surface. It looks rather distasteful, but a ditch like that is not polluted, only eutrophic (rich in nutrients). In spring these filamentous algae grow under water but when there is enough sunlight and the temperatures are not too low, they produce a lot of oxygen, sticking in little bubbles between the tangles of the algae. These come to the surface and become visible as slimy green masses. In these tangles we will find mainly three types of filamentous algae, Spirogyra, Mougeotia and Zygnema. In this article we will mainly write about Spirogyra.
From a distance these slimy tangles look perhaps a bit dirty, but under the microscope the filaments are very beautiful and moreover, they have a spectacular way of reproducing. Spirogyra owes its name to a chloroplast (the green part of the cell) that is wound into a spiral, a unique property of this genus which makes it easily to recognise. In The Netherlands up till now there are found more than 60 species of Spirogyra , in the whole world more than 400.
For the determination of a species it is necessary to look for reproducing specimens with spores. But a precise determination is not necessary for learning a lot of interesting facts from Spirogyra. It is easy to see that there are many species ; in a clean, eutrophic ditch with hard water in Holland we will find easily 20 different species. If we look at a filament of Spirogyra with the microscope, the first thing that attracts attention is the chloroplast, a narrow, banded spiral with serrated edges. The small round bodies in the chloroplast are pyrenoids, centres for the production of starch. In the middle of the cell we see the transparent nucleus, with fine strands linking it to the peripheral protoplasm. The filaments contain cells of different sizes and it is easy to find a new cell, just formed after a division.
The really interesting part comes as Spirogyra reproduces sexually. When two filaments are close together, the process starts. Cell outgrowths form connections between the filaments and a sort of ladder is formed. The contents of the cells in one filament will go through the connection tubes to the cells in the other filament. A zygospore is formed with a thick cell wall, round or oval and with a brownish colour. This conjugation process takes place especially between half May and half June. The spores are liberated, sink to the bottom and germinate in the next spring to form a new filament. It is very worthwhile to look in a sample of algae for the different stages of this conjugation process. It is always a nice surprise to find the conjugating filaments. Spirogyra can also exhibit, apart from the ladder like conjugation, another form of conjugation. Two neighbouring cells in the same filament can connect via a tube.
There are several other genera of related filamentous algae; Zygnema and Mougeotia, with respectively star like and plate like chloroplasts. These genera live in general in more acid, soft fresh water. The conjugation figures look different from those in Spirogyra, for instance X-like. Dune pools are a rich biotope for Spirogyra. In ditches the amount of species declines when the water becomes very eutrophic. Other filamentous algae then replace it, like Cladophora, Vaucheria and Enteromorpha. In the end we only will find duck weed. Then a ditch does not receive light, with disastrous consequences for the growths of plants and the production of oxygen.
The Filamentous Algae.
This gallery includes only the filamentous green algae. The group is a heterogeneous one in which the members, although superficially similar, show a wide diversity in their life cycle and modes of reproduction. Spirogyra, Oedogonium and Cladophora are amongst the varieties most frequently encountered.
All blue-green algae are now classified amongst the Bacteria, and will be found in the Cyanobacteria gallery.
Spirogyra.
Spirogyra is a filamentous green alga which is common in freshwater habitats. It has the appearance of very fine bright dark-green filaments moving gently with the currents in the water, and is slimy to the touch when attempts are made to collect it. The slime serves to deter creatures which otherwise attatch themselves to underwater plants, so Spirogyra under the microscope is usually spotless.
A field of Spirogyra filaments. Their appearance is not quite typical in that the nuclei are unusually prominent, and the characteristic spiral chloroplasts are so fine and tightly wound that close examination is needed to confirm the identification. In any case the possession of spiral chloroplasts is sufficient to positively identify Spirogyra to genus.
Darkfield, x120.
The central portion of a cell of Spirogyra showing the nucleus and giving an insight into the way the spiral chloroplast contacts with the wall of the cell. The filament in the background provides another view.
Brightfield. x1000.
Central portion of a Spirogyra cell showing nucleus and chloroplasts.
Brightfield, x1000.
This filament of Spirogyra is about to break into two filaments. The wall of each cell (centre of picture) has developed an inward indentation at the junction between the cells. Increase in pressure in each cell will cause the indentation to pop out, forcing separation of the filaments, and leaving them with highly convex ends.
Brightfield. x1000.
Two filaments of Spirogyra, the lower one clearly showing the nucleus. This picture also gives a good insight into the way the chloroplasts line the wall of the cell.
Brightfield. x1000.
Conjugation in Spirogyra.
In common with other members of its phylum (Gamophyta) Spirogyra lacks a motile variant at all stages of its life history; ie, no motile gametes (ova or sperm), no zoospores etc. Sexual reproduction is by a process called conjugation -- another of the famously remarkable sights available to the microscopist.
Although it is not possible to distinguish them visually, certain filaments in a loose parallel bundle of Spirogyra assume the female, and others the male, role in the process which follows. The cells of adjacent filaments develop bumps which grow towards one another and eventually fuse to form a continuous tube between the cells. Meanwhile the contents of each cell have detatched themselves from their respective cell walls and have formed a round ball. Over a relatively short space of time (minutes), the green spheres from the male filament squeeze their way down the connecting tubes to fuse with a similarly contracted female cell in the other filament. The result of this sexual union is the formation of a zygospore with a tough resistant outer covering within the chambers of the female filament. After a dormant period, these zygotes undergo meiosis and germinate, resulting in new filaments of Spirogyra.
Once seen never forgotten.
The central pair of cells are joined by a conjugation tube which has yet to fuse into form a continuous passage. The cell contents are at a similarly early stage of detatching themselves from the cell wall to form a ball.
By contrast, the two cells to the right contain newly formed zygospores as a result of consummated conjugation.
Male and female cells now occupy the same space, and are pictured before fusion to form a zygospore has taken place. The filament designated female is the one in which the zygospores have formed.
Two mature zygospores of Spirogyra from another part of the specimen which provided the above pictures. In this form, Spirogyra can survive winter or other adverse conditions and germinate in the spring to form new filaments. The hardened outer spore wall can be seen reflecting the light from the darkfield condenser.
Darkfield. x400.
A zygospore of Spirogyra against a background of decaying plant remains and other algal forms.
Darkfield, x400.
It's hard to say what is happening here. It looks like the stage in conjugation of Spirogyra in which the contraction of the cell contents to a ball is not quite complete, and the spiral nature of the chloroplast is still discernable.
Darkfield, x1000.
Cladophora and Microspora.
The filamentous alga Cladophora is a common inhabitatant of freshwater locations. It is called blanket weed in some places -- not an inappropriate name when in late summer dense floating rafts of Cladophora can be found both at the pond's edge and in the open water, buoyed up with the oxygen generated by its own photosynthesis.
Unlike Spirogyra, Cladophora is capable of branching, and seems to produce little or no mucilagineous secretion. This, and the fact that salts tend to crystallize on the filaments of older specimens, gives it a rougher, grittier feel than other filamentous algae. It is also more readily colonized by epiphytic diatoms and other algae, and provides a protected foraging environment for the smaller pond creatures such as protozoa, worms, small crustaceans and insect larvae.
Its springiness also makes it more difficult to prepare the thin, flat specimens required by the microscope.
Branching in this filament of Cladophora has begun with an outgrowth of the cell at the upper end near the cell wall junction. As the branch grows, differential growth of the main cell wall causes the branch to grow forwards rather than at right angles to the original cell.
An interesting feature of the picture is the distribution of plastids in the two cells shown. Since the plastids are the energy converters of the cell, large numbers have migrated into the growing branch, where the energy requirement is greatest.
The cell on the right shows a distribution of plastids normal to a resting cell.
Darkfield, x300.
Picture shows Cladophora at a branching point. The filaments are encrusted with diatoms (Gomphonema) and crystals of calcium carbonate which give the plant its rough gritty feel.
Darkfield, x400.
Microspora is common in ponds, especially in the winter months. It can be recognized by its reticulated chloroplast which covers the inside wall of the cell including the cell walls between one cell and the next.
Darkfield, x600.
Pteridium aquilinum
Bracken Fern
Bracken Fern
Photo © by Earl J.S. Rook
Flora, fauna, earth, and sky...
The natural history of the northwoods
Name: • Pteridium, from the Greek pteris (pteris), "fern"
• aquilinum, from the Latin, "eagle like"
• Bracken, an old English word for all large ferns, eventually applied to this species in particular.
• Other common names include: Brake, Brake Fern, Eagle Fern, Female Fern, Fiddlehead, Hog Brake, Pasture Brake, Western Brackenfern, Grande fougere, Fougere d'aigle, Warabi (Qué), Örnbräken, Bräken, Slokörnbräken, Taigaörnbräken, Vanlig Örnbräken (Swe), Einstape (Nor), Ørnebregne (Dan), Sananjalka (Fin), Adlerfarn (Ger), Kilpjalg, Kotkajalg, Põldsõnajalg, Seatinarohi, Sõnajalg (Estonia)
Taxonomy: • Kingdom Plantae, the Plants
o Division Polypodiophyta, the True Ferns
Class Filicopsida
Order Polypodiales
Family Dennstaedtiaceae
Genus Pteridium
• Taxonomic Serial Number: 17224
• Also known as Pteris aquilina, Asplenium aquilinum, Allosorus aquilinus, Ornithopteris aquilina, Filix aquilina, Filix-foemina aquilina, Pteris latiuscula
• Considered a single, worldwide species, although some disagree
Description: • A large, deciduous, rhizomatous fern
• Fronds 1'-3' w/leaf stalk up to 3'' but usually shorter than leaf blade. Blades of frond divided into pinnae, the bottom pair sometimes large enough to suggest a three part leaf. Pinna divided into pinnules. On fertile fronds the spores are borne in sori beneath the outer margins of the pinnules. Fronds are killed by frost each winter and new fronds grow in spring. Dead fronds form a mat of highly flammable litter that insulates the below-ground rhizomes from frost when there is no snow cover. This litter also delays the rise in soil temperature and emergence of frost-sensitive fronds in the spring.
• Rhizomes are the main carbohydrate and water storage organs (87% water). Rhizomes can be up to 1" diameter and branching is alternate. The rhizome system has two components. The long shoots form the main axis or stem of the plant. They elongate rapidly, have few lateral buds, do not produce fronds, and store carbohydrates. Short shoots, or leaf-bearing lateral branches, may be closer to the soil surface. They arise from the long shoots, are slow growing, and produce annual fronds and many dormant frond buds. Transition shoots start from both short and long shoots and may develop into either.
• Roots thin, black, brittle extending from the rhizome to over 20" inches into the soil.
• Brackenfern is a large, coarse, perennial fern that has almost horizontal leaves and can grow 1½ to 6½ feet tall (sometimes up to 10 feet). Unlike our more typical broadleaf perennials, this primitive perennial lacks true stems. Each leaf arises directly from a rhizome (horizontal underground stem), and is supported on a rigid leaf stalk. In addition, brackenfern does not produce flowers or seeds. Instead, it reproduces by spores and creeping rhizomes. This species often forms large colonies.
• Root system - The black, scaly, creeping rhizomes (horizontal underground stems) are ½ inch thick, and can grow as much as 20 feet long and 10 feet deep. Stout, black, wide-spreading roots grow sparsely along the rhizomes.
• Seedlings & Shoots - The curled leaves (fiddleheads) emerging from rhizomes in the spring are covered with silvery gray hair.
• Stems - The leaf stalk (not a true stem) is tall (about the same length as the leaf), smooth, rigid and grooved in front. It is green when young, but turns dark brown later in the season.
• Leaves - The leaf stalk supports a broad (3 feet long, 3 feet wide), triangular, dark green, leathery and coarse-textured leaf that often bends nearly horizontal. The leaf is divided into 3 parts, 1 terminal and 2 opposite. Each of the leaf parts is triangular and composed of numerous oblong, pointed leaflets, which are in turn composed of narrow, blunt-tipped subleaflets.
• Fruits & Seeds - A continuous line of spore cases (spore-producing structures) is formed along the underside edge of leaflets, but the spore cases are partially or completely covered by inrolled leaf margins and are difficult to see. Spore cases produce minute, brown spores.
• Biology: Spores of brackenfern are produced August through September. Brackenfern is one of the earliest ferns to appear in spring or after a fire. It sometimes forms large colonies of nearly solid stands. In the fall, it is one of the first plants to be killed by frost, resulting in large patches of crisp, brown foliage.
• Brackenfern is resistant to many herbicides and is tolerant of various forms of mechanical control. However, effective control has been obtained by repeated removal of aboveground growth, which eventually exhausts the food reserves in the rhizomes.
Identification: • Distinguished from other large North Country ferns by the large three part leaf atop a tall stalk.
• Field Marks
o broad triangular leaf held almost parallel to the ground
o smooth, grooved, rigid stalk about as long as the leaf
o narrowed tip to leaflets
Distribution: • Global; throughout the world with the exception of hot and cold deserts
Habitat: • Fossil evidence suggests that bracken fern has had at least 55 million years to evolve and perfect antidisease and antiherbivore chemicals. It produces bitter tasting sesquiterpenes and tannins, phytosterols that are closely related to the insect moulting-hormone, and cyanogenic glycosides that yield hydrogen cyanide (HCN) when crushed. It generates simple phenolic acids that reduce grazing, may act as fungicides, and are implicated in bracken fern's allelopathic activity. Severe disease outbreaks are very rare in bracken fern.
• Grows on a variety of soils with the exception of heavily waterlogged sites. Efficient stomatal control allows it to succeed on sites that would be too dry for most ferns, and its distribution does not normally seem limited by moisture. Grows best on deep, well-drained soils with good water-holding capacity, and may dominate other vegetation on such sites.
• Rhizomes are particularly effective at mobilizing phosphorus from inorganic sources into an available form for plant use. Bracken fern contributes to potassium cycling on sites and is associated with high levels of potassium.
• In northern climates bracken fern is frequently found on uplands and side slopes, since it is susceptible to spring frost damage. Fronds growing in the open or without litter cover are often killed as crosiers by spring frost damage, since the soil warms earlier and growth begins sooner. The result is that fronds appear earlier in shaded habitats.
• A shade intolerant pioneer and succession species that is sufficiently shade tolerant to survive in light spots in old growth forests.
• Light, windborne spores allow colonization of newly vacant areas.
• Despite production of bitter-tasting compounds, chemicals that interfere with insect growth, and toxic chemicals, bracken fern hosts a relatively large number and variety of herbivorous insects.
• Competition: Invades cultivated fields and disturbed areas, effectively competing for soil moisture and nutrients. Rhizomes grow under the roots of herbs and tree or shrub seedlings, and when the fronds emerge, they shade the smaller plants. In the winter dead fronds may bury other plants and press them to the ground. On some sites shading may protect tree seedlings and increase survival.
• Allelopathy: Bracken fern's production and release of allelopathic chemicals is an important factor in its ability to dominate other vegetation. Farther north no allelopathic chemicals are released from the green fronds but are readily leached from standing dead fronds. Herbs may be inhibited for a full growing season after bracken fern is removed, apparently because active plant toxins remain in the soil.
Fire: • A fire-adapted species throughout the world. Not merely well adapted to fire, it promotes fire by producing a highly flammable layer of dried fronds every fall. Repeated fires favor Bracken.
• Primary fire adaptation is deeply buried rhizomes which sprout vigorously following fires before most competing vegetation is established. Windborne spores may disperse over long distances.
• Fire removes competition and creates the alkaline soil conditions suitable for its establishment from spores
• Fuel loading in areas dominated by bracken fern can be quite high.
Associates: • Shrubs: Bunchberry (Cornus canadensis), Twinflower (Linnaea borealis)
• Herbs: Wild Sarsaparilla (Aralia nudicaulis), Large Leaf Aster (Aster macrophyllus), Blue Bead Lily (Clintonia borealis), Gold Thread (Coptis trifolia), Bedstraws (Galium ssp.), Oak Fern (Gymnocarpium dryopteris), Canada Mayflower (Maianthemum canadense), Bishop's Cap (Mitella nuda), One Flowered Pyrola (Moneses uniflora), One Sided Pyrola (Pyrola secunda), Rose Twisted Stalk (Streptopus rosea), Starflower (Trientalis borealis), Kidney Leaf Violet (Viola renifolia), Violets (Viola spp.)
• Mammals: Palatability is usually nil to poor
History: • Considered so valuable during the Middle Ages it was used to pay rents.
• Used as roofing thatch and as fuel when a quick hot fire was desired.
• The ash was used as a source of potash in the soap and glass industry until 1860 and for making soap and bleach. The rhizomes were used in tanning leathers and to dye wool yellow.
• Bracken still used for winter livestock bedding in parts of Wales since it is more absorbent, warmer, and easier to handle than straw.
• Also used as a green mulch and compost
Uses: • Most commonly used today as a food for humans. The newly emerging croziers or fiddleheads are picked in spring and may be consumed fresh or preserved by salting, pickling, or sun drying. Both fronds and rhizomes have been used in brewing beer, and rhizome starch has been used as a substitute for arrowroot. Bread can be made out of dried and powered rhizomes alone or with other flour. American Indians cooked the rhizomes, then peeled and ate them or pounded the starchy fiber into flour. In Japan starch from the rhizomes is used to make confections. Bracken fern is grown commercially for use as a food and herbal remedy in Canada, the United States, Siberia, China, Japan, and Brazil and is often listed as an edible wild plant. Powdered rhizome has been considered particularly effective against parasitic worms. American Indians ate raw rhizomes as a remedy for bronchitis
• Bracken fern has been found to be mutagenic and carcinogenic in rats and mice, usually causing stomach or intestinal cancer. It is implicated in some leukemias, bladder cancer, and cancer of the esophagus and stomach in humans. All parts of the plant, including the spores, are carcinogenic, and face masks are recommended for people working in dense bracken. The toxins in bracken fern pass into cow's milk. The growing tips of the fronds are more carcinogenic than the stalks. If young fronds are boiled under alkaline conditions, they will be safer to eat and less bitter.
• Bracken fern is a potential source of insecticides and it has potential as a biofuel. Bracken fern increases soil fertility by bringing larger amounts of phosphate, nitrogen, and potassium into circulation through litter leaching and stem flow; its rhizomes also mobilize mineral phosphate. Bracken fern fronds are particularly sensitive to acid rain which also reduces gamete fertilization. Both effects signal the amount of pollutants in rain water making bracken fern a useful indicator.
• Fronds may release hydrogen cyanide (HCN) when they are damaged (cyanogenesis), particularly the younger fronds. Herbivores, including sheep, selectively graze young fronds that are acyanogenic (without HCN) Lignin, tannin, and silicate levels tend to increase through the growing season making the plants less palatable. Cyanide (HCN) levels fall during the season as do the levels of a thiaminase which prevents utilization of B vitamins.
• Toxicity: Known to be poisonous to livestock throughout the US, Canada, and Europe. Simple stomach animals like horses, pigs, and rats develop a thiamine deficiency within a month. Acute bracken poisoning affects the bone marrow of both cattle and sheep, causing anemia and hemorrhaging which is often fatal. Blindness and tumors of the jaws, rumen, intestine, and liver are found in sheep feeding on bracken fern.
• Toxicity: All parts of brackenfern, including rootstocks, fresh or dry leaves, fiddleheads and spores, contain toxic compounds, and are poisonous to livestock and humans. Consumption of brackenfern causes vitamin B1 deficiency in horses, and toxins can pass into the milk of cattle. Young leaves of brackenfern have been used as a human food source, especially in Japan, and may be linked to increased incidence of stomach cancer. Humans working outdoors near abundant stands of the plant may be at risk from cancer-causing compounds in the spores.
• Facts and Folklore:
It was once thought that, if the spores of the brackenfern were gathered on St. John's Eve, it would make the possessor invisible.
In the 17th century, live brackenfern was set on fire in hopes of producing rain.
Brackenfern fiddleheads have been used as a food source; however, their consumption has been linked to various types of cancer in humans.
Reproduction: • Reproduces by spores and vegetatively by rhizomes
• Most regeneration is vegetative. Many have searched for young plants growing from spores, but few have found them. However, spores do germinate and grow readily in culture.
• Young plants produce spores by the end of the second growing season in cultivation but normally do not produce spores until the third or fourth growing season. A single, fertile frond can produce 300,000,000 spores annually. Spore production varies from year to year depending on plant age, frond development, weather, and light exposure. Production decreases with increasing shade. The wind-borne spores are extremely small. Dry spores are very resistant to extreme physical conditions, although the germination of bracken fern spores declines from 95-96% to around 30-35% after 3 years storage. The spores germinate without any dormancy requirement. Under favorable conditions, young plants could be found 6 to 7 weeks after the spores are shed. Under normal conditions the spores may not germinate until the spring after they are shed.
• Sufficient moisture and shelter from wind are important factors in fern spore germination. Bracken fern spore germination appears to require soil sterilized by fire. On unsterilized soils spores may germinate, but the new plants are quickly overwhelmed by other growth. Temperatures between 59º and 86º F are generally best for germination, although bracken fern is capable of germination at 33º-36ºF.
• A pH range of 5.5 to 7.5 is optimal for germination. Germination is indifferent to light quality; it is one of the few ferns that can germinate in the dark. Despite limitations on spore germination, genotype analysis in the Northeast indicates that many stands of bracken fern represent multiple establishment of individuals from spores.
• When spores germinate, they produce bisexual, gamete-bearing plants about ¼" in diameter and one cell thick. These tiny plants have no vascular system and require very moist conditions to survive. The young spore-bearing plant which develops from the fertilized egg is initially dependent on the gametopyte until it develops its first leaf and roots. The first fronds are simple and lobed. They develop into thin, delicate fronds divided into lobed pinnae. They do not look like adult plants and are frequently not recognized as bracken fern. Cultivated plants begin to resemble adult fern after 18 weeks. The rhizomes begin to develop after there are a number (up to 10) of fronds and a well-developed root system or in the fifteenth week of growth under optimal conditions.In the first year rhizomes may grow to 86 inches long. By the end of a second year the rhizome system may exceed 6' in diameter.
• Aggressive rhizome system gives it the ability to reproduce vegetatively and reduces dependence on water for reproduction. The rhizomatous clones can be up to 400' in diameter and hundreds of years old; some clones alive today may be over 1,000 years old.
• Rhizomes have a high proportion of dormant buds. When disturbed or broken off, all portions of the rhizome may sprout, and plants growing from small rhizome fragments revert temporarily to a juvenile morphology.
• Shaded plants produce fewerspores than plants in full sun
• Bracken fern is a survivor. The fronds are generally killed by fire, but some rhizomes survive. The rhizomes are sensitive to elevated temperatures. During fires the rhizome system is insulated by mineral soil. Depth of the main rhizome system is normally between 3½" and 12" short rhizomes may be within 1½" of the surface and some rhizomes may be as deep as 40".
• Well known postfire colonizer in eastern pine and oak forests. Fire benefits bracken by removing competition while it sprouts profusely from surviving rhizomes. New sprouts are more vigorous following fire, and bracken fern becomes more fertile, producing far more spores than it does in the shade
• Spores germinate well on alkaline soils, allowing them to establish in the basic conditions created by fire.
Propagation: • Division most successful method
Cultivation: • Hardy to USDA Zone 3 (average minimum annual temperature -40ºF)
• Characteristically found on soils with medium to very rich nutrients.
• Cultivated and shaded plants produce fewer, thinner but larger fronds than open-grown plants
Population Genetics and Evolution
Introduction:
In 1908, G.H.Hardy and W. Weinberg independently suggested a scheme whereby evolution could be viewed as changes in frequency of alleles in a population of organisms. In this scheme, if A and a are alleles for a particular gene locus and each diploid individual has two such loci, then p can be designated as the frequency of the A allele and q as the frequency of the a allele. For example, in a population of 100 individuals ( each with two loci ) in which 40% of the alleles are A, p would be 0.40. The rest of the alleles would be ( 60%) would be a and q would be equal to 0.60. p + q = 1 These are referred to as allele frequencies. The frequency of the possible diploid combinations of these alleles ( AA, Aa, aa ) is expressed as p2 +2pq +q2 = 1.0. Hardy and Weinberg also argued that if 5 conditions are met, the population's alleles and genotype frequencies will remain constant from generation to generation. These conditions are as follows:
• The breeding population is large. ( Reduces the problem of genetic drift.)
• Mating is random. ( Individual show no preference for a particular mating type.)
• There is no mutation of the alleles.
• No differential migration occurs. ( No immigration or emigration.)
• There is no selection. ( All genotypes have an equal chance of surviving and reproducing.)
The Hardy-Weinberg equation describes an existing situation. Of what value is such a rule? It provides a yardstick by which changes in allelic frequencies can be measured. If a population's allelic frequencies change, it is undergoing evolution.
Estimating Allele Frequencies for a Specific Trait within a Sample Population:
Using the class as a sample population, the allele frequency of a gene controlling the ability to taste the chemical PTC (phenylthiocarbamide) could be estimated. A bitter taste reaction is evidence of the presence of a dominant allele in either a homozygous (AA) or heterozygous (Aa) condition. The inability to taste the PTC is dependent on the presence of the two recessive alleles (aa). Instead of using the PTC paper the trait for tongue rolling may be substituted. To estimate the frequency of the PTC -tasting allele in the population, one must find p. To find p, one must first determine q ( the frequency of the non tasting allele).
Procedure:
1. Using the PTC taste test paper, tear off a short strip and press it to your tongue tip. PTC tasters will sense a bitter taste.
2. A decimal number representing the frequency of tasters (p2+2pq) should be calculated by dividing the number of tasters in the class by the total number of students in the class. A decimal number representing the frequency of the non tasters (q2) can be obtained by dividing the number of non tasters by the total number of students. You should then record these numbers in Table 8.1.
3. Use the Hardy-Weinberg equation to determine the frequencies (p and q ) of the two alleles. The frequency q can be calculated by taking the square root of q2. Once q has been determined, p can be determined because 1-q=p. Record these values in Table 8.1 for the class and also calculate and record values of p and q for the North American population.
Table 8.1 Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles
Phenotypes Allele Frequency Based on the H-W Equation
Tasters (p2+2pq) Non Tastes(q2) p q
Class Population #= %= #= %=
North American Population 0.55 0.45
Topics for Discussion:
1. What is the percentage of heterozygous tasters (2pq) in your class? ______________________.
2. What percentage of the North American population is heterozygous for the taster allele? _____________
Case Studies:
Case 1 ( Test of an Ideal Hardy-Weinberg Community)
The entire class will represent a breeding population, so find a large open space for its simulation. In order to ensure random mating, choose another student at random. In this simulation, we will assume that gender and genotype are irrelevant to mate selection.
The class will simulate a population of randomly mating heterozygous individuals with an initial gene frequency of 0.5 for the dominant allele A and the recessive allele a and genotype frequencies of 0.25AA, 0.50Aa, and 0.25aa. Record this on the Data page at the end of the lab. Each member of the class will receive four cards. Two cards will have A and two cards will have a. The four cars represent the products of meiosis. Each "parent" will contribute a haploid set of chromosomes to the next generation.
Procedure:
1. Turn the four cards over so the letters are not showing, shuffle them, and take the card on top to contribute to the production of the first offspring. Your partner should do the same. Put the cards together. The two cards represent the alleles of the first offspring. One of you should record the genotype of this offspring in the Case 1 section at the end of the lab. Each student pair must produce two offspring, so all four cards must be reshuffled and the process repeated to produce a second offspring.
2. The other partner should then record the genotype of the second offspring in the Case 1 section at the end of the lab. Using the genotypes produced from the matings, you and your partner will mate again using the genotypes of the two offspring. That is , student 1 assumes the genotype of the first offspring, and student 2 assumes the genotype of the second offspring.
3. Each student should obtain, if necessary, new cards representing their alleles in his or her respective gametes after the process of meiosis. For example, student 1 becomes the genotype Aa and obtains cards A,A,a,a; student 2 becomes aa and obtains cards,a,a,a,a. Each participant should randomly seek out another person with whom to mate in order to produce offspring of the next generation. You should follow the same mating procedure as for the first generation, being sure you record your new genotype after each generation in the Case 1 section. Class data should be collected after each generation for five generations. At the end of each generation, remember to record the genotype that you have assumed. Your teacher will collect class data after each generation by asking you to raise your hand to report your genotype.
Allele frequency: The allele frequencies, p and q, should be calculated for the population after five generations of simulated random mating.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA _____________ X 2= _______________ A alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ A alleles
p = Total number of A alleles =
Total number of alleles in the population
In this case, the total number of alleles in the population is equal to the number of students in the class X 2.
Number of a alleles present at the fifth generation
Number of offspring with genotype aa _____________ X 2= _______________ a alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ a alleles
q = Total number of a alleles =
Total number of alleles in the population
1. What does the Hardy-Weinberg equation predict for the new p and q?.
_____________________________________________________________________
_____________________________________________________________________
2. Do the results you obtained in this simulation agree? __________ If not, why not?
_____________________________________________________________________
_____________________________________________________________________
3. What major assumption(s) were not strictly followed in this simulation?
_____________________________________________________________________
_____________________________________________________________________
Case 2 ( Selection )
In this case you will modify the simulation to make it more realistic. in the natural environment , not all genotypes have the same rate of survival; that is, the environment might favor some genotypes while selecting against others. An example is the human condition sickle-celled anemia. It is a condition caused by a mutation on one allele, in which a homozygous recessive does not survive to reproduce. For this simulation you will assume that the homozygous recessive individuals never survive. Heterozygous and homozygous dominant individuals always survive.
The procedure is similar to that for Case 1. Start again with your initial genotype, and produce your "offspring" as in Case 1. This time, However, there is one important difference. Every time your offspring is aa it does not reproduce. Since we want to maintain a constant population size, the same two parents must try again until they produce two surviving offspring. You may need to get new allele cards from the pool.
Proceed through five generations, selecting against the homozygous offspring 100% of the time. Then add up the genotype frequencies that exist in the population and calculate the new p and q frequencies in the same way as it was done in Case 1.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA _____________ X 2= _______________ A alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ A alleles
p = Total number of A alleles =
Total number of alleles in the population
In this case, the total number of alleles in the population is equal to the number of students in the class X 2.
Number of a alleles present at the fifth generation
Number of offspring with genotype aa _____________ X 2= _______________ a alleles
Number of offspring with genotype Aa _____________ X 1= ________________A alleles
Total = ____________ a alleles
q = Total number of a alleles =
Total number of alleles in the population
1. How do the new frequencies of p and q compare to the initial frequencies in Case 1?
_____________________________________________________________________
_____________________________________________________________________
2. How has the allelic frequency of the population changed?
_____________________________________________________________________
_____________________________________________________________________
3. Predict what would happen to the frequencies of p and q if you simulated another 5 generations.
_____________________________________________________________________
_____________________________________________________________________
4. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Hardy-Weinberg Problems
1. In Drosophila, the allele for normal length wings is dominant over the allele for vestigial wings. In a population of 1,000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
2. The allele for the ability to roll one's tongue is dominant over the allele for the lack of this ability. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
3. The allele for the hair pattern called "widow's peak" is dominant over the allele for no "widow's peak." In a population of 1,000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?
4. In a certain population, the dominant phenotype of a certain trait occurs 91 % of the time. What is the frequency of the dominant allele?
Data Page:
Case 1 ( Hardy-Weinberg Equilibrium )
Initial Class Frequencies:
AA ________ Aa________ aa_________
My initial genotype :_______________
F1 Genotype ______
F2 Genotype ______
F3 Genotype ______
F4 Genotype ______
F5 Genotype ______
Final Class Frequencies:
AA ________ Aa________ aa_________
p _________ q __________
Case 2 ( Selection )
Initial Class Frequencies:
AA ________ Aa________ aa_________
My initial genotype :_______________
F1 Genotype ______
F2 Genotype ______
F3 Genotype ______
F4 Genotype ______
F5 Genotype ______
Final Class Frequencies:
AA ________ Aa________ aa_________
p _________ q __________
Biology 198
PRINCIPLES OF BIOLOGY
Hardy-Weinberg practice questions
POPULATION GENETICS AND THE HARDY-WEINBERG LAW
The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:
1. No mutations must occur so that new alleles do not enter the population.
2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
3. Random mating must occur (i.e. individuals must pair by chance)
4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
5. No selection can occur so that certain alleles are not selected for, or against.
Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place:
ANSWERS TO THE QUESTIONS
Remember the basic formulas:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
1. PROBLEM #1.
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
A. The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself.
B. The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
C. The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
D. The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.
2. PROBLEM #2.
Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene? Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).
3. PROBLEM #3.
There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:
A. The frequency of the recessive allele. Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%).
B. The frequency of the dominant allele. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%).
C. The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).
4. PROBLEM #4.
Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:
A. The percentage of butterflies in the population that are heterozygous.
B. The frequency of homozygous dominant individuals.
Answers: The first thing you'll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency of homozygous dominant individuals? That would be p2 or (0.37)2 = 0.14.
5. PROBLEM #5.
A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:
A. The allele frequencies of each allele. Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don't have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355.
B. The expected genotype frequencies. Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above).
C. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0.458 x 953 = about 436.
D. The expected phenotype frequencies. Answer: Well, the "A" phenotype = 0.126 + 0.458 = 0.584 and the "a" phenotype = 0.416 (you already knew this from part A above).
E. Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? Answer: Simply put, The "A" phenotype = 0.584 x 1,245 = 727 tan-sided and the "a" phenotype = 0.416 x 1,245 = 518 red-sided ( or 1,245 - 727 = 518).
6. PROBLEM #6.
A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.
7. PROBLEM #7.
After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island? Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.
8. PROBLEM #8.
You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:
BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY
M MM 490 0.49
MN MN 420 0.42
N NN 90 0.09
Using the data provide above, calculate the following:
A. The frequency of each allele in the population. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3.
B. Supposing the matings are random, the frequencies of the matings. Answer: This is a little harder to figure out. Try setting up a "Punnett square" type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:
MM (0.49) MN (0.42) NN (0.09)
MM (0.49) 0.2401* 0.2058 0.0441
MN (0.42) 0.2058 0.1764* 0.0378
NN (0.09) 0.0441 0.0378 0.0081*
C. Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three "unique" combinations. Thus, three of the possibilities must be doubled.
D. MM x MM = 0.49 x 0.49 = 0.2401
MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
MN x MN = 0.42 x 0.42 = 0.1764
MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
NN x NN = 0.09 x 0.09 = 0.0081
E. The probability of each genotype resulting from each potential cross. Answer: You may wish to do a simple Punnett's square monohybrid cross and, if you do, you'll come out with the following result:
MM x MM = 1.0 MM
MM x MN = 0.5 MM 0.5 MN
MM x NN = 1.0 MN
MN x MN = 0.25 MM 0.5 MN 0.25 NN
MN x NN = 0.5 MN 0.5 NN
NN x NN = 1.0 NN
9. PROBLEM #9.
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
A. The frequency of the recessive allele in the population. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
B. The frequency of the dominant allele in the population. Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%).
C. The percentage of heterozygous individuals (carriers) in the population. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.
10. PROBLEM #10.
In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)? Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.
11. PROBLEM #11.
The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies. Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. You already know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 = 0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and 0.495 and these should equal 1.0 or very close to it. This type of problem may be on the exam.
12. PROBLEM #12. (You will not have this type of problem on the exam)
What allelic frequency will generate twice as many recessive homozygotes as heterozygotes? Answer: We need to solve for the following equation: q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or another way of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q, we can substitute 1 - q for p and, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 - 4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, we get -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is 0.8 =q. I cannot imagine you getting this type of problem in this general biology course although if you take algebra good luck
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