Saturday, November 25, 2006



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.
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.
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.
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.

• 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

• 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.)

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.

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
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).


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 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

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).

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.
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
Hardy-Weinberg practice questions


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:


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:
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


Mossin' Annie said...

Great overview of bryophytes. Thanks for sharing with others. Go Green With Moss! Mossin' Annie

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