Ecology of Bellarmine College Preparatory

In this lab, I looked at the habitats and ecosystems found on the Bellarmine Campus. During this lab, I classified and analyzed various organisms found on campus into their respective niche in the Bellarmine Ecosystem. On campus, I was capable of finding an entire food web, with numerous producers, primary, secondary, and tertiary consumers, as well as various decomposers. In addition, I found that Bellarmine is home to threatened and endangered species, several non-native species, and pollutants. Bellarmine has a very diverse biotic community.

Ecology, which derives from Greek oikos, meaning “study of our home” is the study of the interactions of the populations, communities, and ecosystems of organisms to their environments on Planet Earth. An organism’s environment consists of two factors, biotic and abiotic factors. Biotic factors are the living portions of an  environment, such as other organisms, while abiotic factors are the nonliving, chemical aspects of an organism’s environment, including sunlight, temperature, weather, humidity, soil type, amount of water, and rocks. 

The history of ecology began in 1866 when Ernest Hackel, a German biologist, coined the word “oekologie” to define the interactions between animals and their environment.  Then, in 1914, Victor Shelford, an animal ecologist, began to emphaise the importance of the relationships between plants and animals. Following Shelford’s breakthrough, ecologist A. G. Tansley theorized that plants and animals were parts of a system including  physical factors  as well, coining the term ecosystem. An ecosystem is defined by E.O. Wilson’s Life on Earth as “the natural unit consisting of all the habitats and communities in a given place, and the interactions that take place between them.” All ecosystems collectively form the biosphere. Later, Victor Shelford and F. E. Clements would introduce the concept  of biomes, defined as “major regional biotic communities of plants and animals together.” Examples of biomes include the tundra of Northern Canada, and the Tropical Rainforests of the Amazon River Basin. 

There are two forms of ecology, autecology and synecology. Autecology is the view of individual organisms and how they influence or are influenced by their environment, while synecology considers all interactions within an given ecosystem. For this lab, I primarily focused on autecology, but also implemented forms of synecology. The interpretations of synecology and autecology are shown in Figure 1.

Figure 1:  

Ecology is characterized by the numerous landscapes and environments on Earth. There are two kinds of biomes that encompass all of the ecosystems on planet Earth, terrestrial and aquatic biomes.

Terrestrial biomes include forests, grasslands, tundra, and deserts. 

There are three unique kinds of forests, tropical rainforests, temperate forests, and taiga, all of which amount to over a third of the Earth’s terrestrial land. 

Tropical forests are found in the regions between the Tropics of Capricorn and Cancer, located between 23 N and 23 S. Characterized by its warm, humid climate and large quantities of rain, tropical forests are the most diverse biome. Tropical forests also exhibit poor soil conditions, allowing epiphytes to adapt to retain the small amounts of nutrients they can recieve. A tropical rainforest in the Amazon River Basin is depicted in Figure 2.

Figure 2:

 Temperate forests, consisting of both deciduous and coniferous trees, grow in moderate climates. In areas with less rainfall, conifers dominate, while in wet areas, dedicuous trees thrive. The soils is of good quality, due to the decomposition of the leaves lost by dedicuous trees. A majority of the Eastern United States is temperate forests, as is the temperate forest in Figure 3.

Figure 3:
  
 Boreal forests, also categorized as taiga, is the largest biome on land. Comprised of evergreenns, they are found between 50° and 60° north, and form across North America , Europe, and  Siberia as in Figure 4. Taiga (“marshy forest”) have short wet summers, and  winters are cold.

Figure 4: 

  

Similar to forests, there are three kinds of grasslands: savannas, steppes, and praries. As the name suggests, grasslands are predominantly covered by grass, with scattered trees. In general, savannas and praries have longer grass than steppes.

Savannas, found throughout the continent of Africa,  teem with herds of wild animals. With a distinctive short wet season and a long dry season, fires rampage the plains of Africa, allowing for grasses to dominate local trees. Savannas account for around 20 percent of the Earth’s landmass.  An African savanna in the Serengheti National Park is depicted in Figure 5.

Figure 5:

 Prairies are found across North America and have been symbolized in numerous works of literature. Characterized by its tall grass and herds of roaming bison, the North American prairies account for almost all of the central United States. The praries where created  partially by the rain shadow of the Rocky Mountains. As seen in Figure 6, a prairie in rural Wyoming is characterized by its long grass.

Figure 6:  

Steppes are short grass grasslands found furthest away from the equator.  Steppes can be found in Patagonia to Mongolia, where they were used by the Mongols to create their vast Empire and Postal system. Steppes possess a semi-arid climate as well. As seen in figure 7, a Tibetian fox, navigates the Tibetian Steppes, walking between the  short grass.

Figure 7:  
Tundra, found across the northern parts of Canada, Scandavia, and Asia. “Tundra” derives from Finnish, meaning “treeless plain.” Hardy plants and animals endure the lack of water, poor soil, and cold climate found across the Arctic tundra. Large plant and animal life is prohibited by the hard permafrost frozen even throughout the Artic summer. As seen in Figure 8, the tundra is dominated by lowline shrubs and glacial ice.

Figure 8:  
Deserts, the most dry and barren biome, are found across the world. Desert dwellers must adapt to survive with the lack of water, food, and hot temperatures. Many desert communities are nocturnal, to take advantage of cooler temperatures.   As seen in Figure 9, the desert has much less vegetation compared to other biomes, and has a hot climate.
Figure 9:  

Ecologists recognize four branche of ecology, behavioral ecology,  physiological ecology, chemical ecology, and mathematical ecology.

Behavioral ecology, pioneered by ethnologist Konrad Lorenz, focuses on the interactions between animals. Lorenz famously discovered imprinting is ducklings, in which ducklings perceive the first thing they see and hear to be their mother.

Physiological ecology studies the affects of abiotic factors on organisms. Ecologist Justus Lingberg first discovered that plants need adaquete nutrients in the soil to survive. 

Chemical ecology studies the chemical communication, defense mechanisms, and smells of organisms. One example of chemical ecology is the defense mechanism of a skunk, spraying maladorous sulfur compounds at attackers.

Mathematical ecology involves the quantitative data of ecology, including population growth and average mass, gestation period, and life expectancy of organisms. From this data, scientists can establish potential predators in the area, parasites that may harm the organisms, or diseases that have an adverse affect on an organism. For example, as seen in Figure 10, matcha toca, ecologists can determine that the population of hares directly correlates to the population of lynx, as the fewer number of hares, the greater number of Lynx. 

Figure 10:

 

At Bellarmine, we live in a region comparable to a Chaparral. A Chaparral is a dry temperate forest, similar to the biome of the Mediterranean. The chaparral possess numerous scrubs  (as seen in Figure 11) and has plants with large leaves and animals associated with deserts and forests, such as coyotes, deer and jack rabbits.

Figure 11:

  

Materials:

• Camera
• Various Ecology Apps: (LeafSnap, Project Noah, Merlin Bird ID, iNaturalist)

Procedure: 

1. I located an organism or pollution source.
2. I identified the species and name of the organism using the various apps. If a plant, I used LeafSnap, if a bird, I used Merlin ID, and for other organisms, I used a combination of iNaturalist and Project Noah.
3. I identified plants, animals, and pollution sources to fit the following categories depicted in Figure 12.

Figure 12: 

  

 

Figure 13:

  

  
  

Common Name–Areca Palm Tree

Scientific Name–Chrysalidpcarpus Lutescens

As seen in Figure 13, a juvenile Areca Palm Tree (Chrysalidpcarpus Lutescens) is a  primary producer found in the Bellarmine Quad. Palm Trees, the only members of the family Arecacaea and order Arecales, generally grow in hot climates. There are over 3000 species of Palm trees, some of which are indigenous to the United States. Having a distinct appearance in the fossil record, palm-like monocots date back to 70 Million years ago. Ther are two major groups of Palm trees, pinnate feather leaved, like Acera Palms, and palmate and fan shaped leaved palms. Palm trees can grow in tropical to semitropical climates, and is located near the Statue of Mary in the Bellarmine Campus.

 Figure 14 

Common Name–Black Garden Ant

Scientific Name–Lasius Niger

As seen in Figure 14, a Black Garden Ant (Lasius Niger) is a primary consumer found on the Bellarmine campus. Like most ant colonies,  the Black Garden Ant has a colony centered around a sole queen ant. Having a fossil record dating back 92 Millennia. This particular Black Garden Ant was found on a discarded Orange outside of Sobrato, however Black Garden Ants can be found throughout the Northern Hemisphere.

Figures 15-16:

      

Common Name–Rock Dove

Scientific Name–Columba Livia

As seen in Figures 15-16, rock doves around Carney are secondary consumers, eating Incests. rock Soves can be found worldwide, but are indigineous to Europe and the Middle Easet. Rock Doves, or pigeons as they are often mislabeled, are gregarious, and occanassionally interact with other doves.  At Bellarmine, rock doves are found across campus, yet these birds were found outside Carney.

Figure 17

  

Common Name–Tabby Cat

Scientific Name–Felis catus

As seen in Figure 17, Cindy’s Tabby Cat (Felis catus) is a tertiary consumer. A variety of domestic cats with splotches and distinctive stripes, Tabby Cats originated from African Wild Cats with similar markings. The tabby pattern, in fact is not specific to a certain breed, and is in fact a wild type among cats. This tabby cat was found behind Bellarmine near Cindy’s but cats have been domesticated worldwide. 

Figure 18:

  

Common name: Dark Eyed Junco

Scientific Name: Junco Hyemalis

As seen in Figure 18, a Dark Eyed Junco, a seed eating herbivore  rests on the lawn behind Sobrato and the Robotics Lab. Characterized by a patch of black on their head and around their eye, the Dark Eyed Junco lives across North America. The Dark Eyed Junco can be found through campus, yet this bird was found behind Sobrato.

Figure 19:

  

Common Name: American Crow

Scientific Name: Corvus Brachyrhynchos

As seen in a figure 19, an American Crow (Corvus Brachyrhynchos) reads itself to fly behind Lokey. Common throughout much of North American the American Crow is Ominvorous, eating Incests, carrion, human food, seeds , and nestlings. They are covered in sleek black feathers and can be found around  Lokey and Liccardo, where they scavenge for food.

Figure 20:

  

Common Name: Jack Russell Terrier

Scientific Name: Canis Lupus Familiaris

As seen in Figure 20, the Jack Russell Terrier (Canis Lupus Familiaris) is a carnivorous, common breed of the domesticated dog found throughout the world. Jack Russell Terriers, originally used by Englishmen to hunt foxes, are now perform leisurely roles as family pets. Descended from wolves and other canines, dogs possess distinctive sharp tearing teeth and are quite agile. These Dogs where found in front of O’Donnell Hall.

Figure 21:

  
Common Name: Earthworm

Scientific Name: Lumbricus Terrestris

As seen in Figure 21, the common earthworm (Lumbricus Terrsrtris)squirms on the sidewalk outside Leontyne Chapel. Earthworms are decomposers, turning waste products back into soil. Earthworms are found on every continent except Antarctica, and can be found everywhere at Bellarmine, particularly after a rain storm. This earthworm was found in the garden outside Leontyne Chapel.

Figure 22:

  
Common Name: Bamboo

Scientific Name: Bambuseae

As seen in Figure 22, bamboo (Bambuseae)a threaten species, grows outside Sobrato. Bamboo is one of the fastest growing plants on Earth, but it is being threatened due to deforestation in the world, particularly in China. Bamboo is found throughout the world in areas with little frost, for the shoots perish in cold weather. At Bellarmine, juvenile and mature bamboo surrounds Sobrato.  

Figure 23:

  

 Common Name: Maidenhair Tree

Scientific Name: Gingko Biloba

As seen in Figure 23, the maidenhair tree (Gingko Biloba)an endangered species, grows outside Carney. It is the only species of tree in the division Gingkophyta, an indigenous to China. It is endangered due to deforestation, much like the Bamboo. At Bellarmine, Maidenhair Trees are found outside the front of Carney. 

Figure. 24:

  

Common Name: Japanese Maple

Scientific Name: Acer Palmatum

As seen in Figure 24, the Japanese Maple (Acer Palmatum)is a non-native species, indigenous to Japan. A deciduous tree, the Maple’s colorful leaves switch from purple to yellow in autumn. At Bellarmine, Japanese Maples are located in front of Carney, the flagpole, and behind Leontyne Chapel, where the maple in Figure 24 is located

Figure 25:

  
Common Name: Automobile

Scientific Name: —————

As seen in Figure 25, automobiles, a motorcycle and a car are pollution sources found at Bellarmine. Exhaust produced by these vehicles cause air pollution, thickening the ozone layer. Motorcycles and cars can be found surrounding campus in the numerous parking lots. Also, numerous golf carts drive through the quad, further emitting pollutants into the air.

*Define and differentiate between ecology and environmental science and discuss the Bellarmine campus in the context of both.

Ecology is the study of the interactions of the populations, communities, and ecosystems of organisms to their environments on Planet Earth, while Environmental Science is the study of the environment of organisms and the study of interactions among physical, chemical, and biological aspects of the environment. An organism’s environment includes both abiotic factors such as sunlight, climate, and rainfall, as well as biotic factors (other organisms that share its environment). While ecology focuses mainly on the physical aspects of nature and the interactions of organisms, environmental science focuses on the chemical and biological aspects as well, examining human affects on the environment, like pollution and deforestation. At Bellarmine, we have an ecology of the biotic and abiotic factors that call campus home, with animals being suited to possess a niche in the Bellarmine environment. The affects of humans at Bellarmine, as seen with students leaving trash on the ground and stepping on the plants effects the organisms in the Bellarmine ecosystem, an example of environmental science. 

 *Define and describe any population, community, ecosystem, biome and aquatic zone that you find on campus; and discuss the biotic and abiotic factors that contribute to that ecosystem.

A population a group of organisms of the same species that interact with one another, while a community is a group of organisms in a common habitat that interact with one another. On example of a population on campus is the number of Dark Eyed Juncos that congregate on the grass to feed, while an example of a community is the Juncos, Rock Doves, and Crows that interact with each other in their competition for food. An ecosystem is a community interacting with both additional biotic and abiotic factors. For example the Juncos, Rock Doves, and Crows, interact with other biotic factors, such as humans, predatory felines, and their prey, incests and plants. They also interact with abiotic factors, including the climate, rainfall, which would prohibit them from flying, obstacles, such as telephone wires, and sunlight. A biome is a group of ecosystems having similar abiotic conditions. At Bellarmine, we live in a biome comparable to a Chaparral. A Chaparral is a dry temperate forest, similar to the biome of the Mediterranean. The chaparral possess numerous scrubs  (as seen in Figure 11 above) and has plants with large leaves and animals associated with deserts and forests, such as coyotes, deer and jack rabbits. Although there are not such large animals on campus, numerous shrubs and other chaparral organisms including Juncos and Rock Doves dot Bellarmine’s campus. Furthermore, abiotic factors including similar !editerranean weather conditions are common between the Chaparral biome. This can be seen comparing Bellarmine’s campus to the ecosystem further North in Los Altos or Palo Alto, as well as abroad in the Mediterranean foothills. Aquatic Zones are biomes of water ecosystems, such as that of the San Francisco Bay Estuary. Although we do not have many aquatic zones on campus,masked from perhaps a teacher’s fish tank, a major aquatic zone is the San Francisco Bay. With the biotic algae and kelp, otters, fish, and grey whales interacting with the abiotic  water salinity, temperature, and depth, these factors encompass the ecosystem of the Bay.  

*Construct and discuss a food chain, a food web, and an ecological pyramid based on the trophic levels that you observe.

In an ecosystem, food chains play an major role in the interactions between organisms. Predation, one of the most important relationships between species is described  in a food chain. A food chain at Bellarmine is depicted in Figure 26. 

Producer: Orange Tree->Primary Consumer:Ants->Secondary Consumer: Rock Dove-> Tertiary Consumer: Tabby Cat.

Figure 26:

  
A food chain describes the who-eats-who in an ecosystem, following the path of nutrients in the ten percent rule. As a predator increases per tropic level, only ten percent of the nutrients from the prey is obtained. This allows for there to be more primary consumers than tertiary consumers, for the latter only receive 1% of the energy stored by the primary consumer.   Organisms are grouped into tropic levels, divided into producers, consumers, and decomposers. Producers are photo autotrophs and produce their own food via photosynthesis, yet consumers must obtain food by consuming another organism. Consumers can be herbivores (primary consumers), omnivores, or carnivores. Decomposers rid of the waste and carrion and turn it into soil.  In the Bellarmine Food Chain, the Orange tree and Palm tree are producers, the Ants are primary consumers, the secondary consumers are the Rock doves, and the tertiary consumers are the tabby cats. 

Food webs are the interconnection of food chains for an entire ecosystem as opposed to a food chain, demonstrating a single chain for a community. A Bellarmine Food Web is depicted in Figure 27, with Palm trees and roses as consumers, bees, ants, and Jercos as primary consumers, crows and pigeons as secondary consumers, and dogs and cats a tertiary consumers.
Figure 27:

  

Figure. 28

  

 As seen in Figure 28, the trophic levels of animals in Bellarmine’s ecosystem is compared to that of a chaparral. 

Trophic Levels are several levels of hierarchy grouping organisms that share the same function and diet in a food chain. For example, at Bellarmine, both cats and dogs are tertiary consumers, for they both can eat birds, such as Rock doves and crows. Rock Doves and Crows are also on the same secondary consumer trophic level, for they both eat herbivorous incests and are eaten by similar animals. 

*Investigate and discuss any endangered, threatened, and invasive species on campus.

Endangered species are organisms at risk of going extinct because of climate change, human activity, and predation, while threatened species are organisms at risk of going extinct to a lesser extent of endangered species. At Bellarmine, the endangered Maidenhair Tree, is located outside of Carney. As Gingko Biloba is the only member of the clase Gingkophyta at is not extinct, Maidenhair trees are classified as endangered. The plant has been used in traditional medicine and in culinary applications, and is subject to deforestation, endangering its existance on the Earth. Also at Bellarmine, bamboo, is a threatened species, also due to deforestation. Bamboo is harvested for many practical purposes and bamboo forests are being cleared worldwide for architectural and agricultural purposes. The elimination of bamboo forests has led to the threatened classification for bamboo, as well as similar classifications for animals that feed on bamboo, including giant pandas. Invasive species are organisms not native to a region that dominate over local communities. One example of a non native species at Bellarmine is the Japanese Maple. Originating in East Asia, the Japanese Maple has taken root across Bellarmine, dominating several flowerbeds, in its non-native habitat. Other invasive species include Burmese Pythons in Florida, African Killer Bees in the United States, and Zebra Quagga Mussels in the Great Lakes.  

*Define pollution, and describe and discuss the various types that you observe on campus. 

Pollution is the introduction of contaminants into the atmosphere that having an detrimental affect on the a Earth. Pollution comes in numerous forms: air pollution, water pollution, and land pollution, all of which were apparent at Bellarmine. Air pollution is the build up of substances in the atmosphere that can produce adverse affects on living matter. Major sources of air pollution include powerplants, transportation, and burning of solid wastes. The major types of air pollutants include carbon monoxide, hydrocarbons, nitrogen oxides, particulates, sulfur dioxide, and photochemical oxidants. At Bellarmine, examples of air pollution include noise pollution from the airplanes going to San Jose Norman Mineta Airport, the exhaust from the motorized vehicles (As seen in figure 29) at Bellarmine, and the large power generator near the robotics lab. Air pollution further fuels global warming, for greenhouse gases emitted in forms of air pollutants trap the sun’s energy, warming the Earth. 

Figure 29:

 

Water Pollution is the contamination of groundwater, lakes, rivers, and oceans. Water Pollution can be deadly to humans, as contaminated water kills over 14,000 people daily. At Bellarmine, water pollution can be found in the usage of chemicals outside as pesticides, that when washed down the drainpipes contaminate the water sources of various organisms. Land Pollution is the degrading of land directly resulting from man made activities. At Bellarmine, fertilizers damage the soil used to grow plants, and littering pollutes the ground with trash. Other examples of air pollution include Mercury deposits from mining expeditions and the usage of DDT in the early to mid 1900s.
If I were to do this experiment in the future, I would compare the ecosystems of other nearby schools, to see if slight differences in loaction had an adverse affect of communities.furthermore, I would examine and contrast different areas of the school, to compare how abiotic factors had an affect on plant growth, such as sunlight allowing plants to grow taller. 

Works Cited:

Oregon State University Mathematics:

University California Clermont Ecology

http://biology.clc.uc.edu/courses/bio303/history.htm

Toledo University Ecology

http://www.eeescience.utoledo.edu/Faculty/Heckathorn/EEES%202150/Lecture%2018%20-%20ecology.ppt

Stanford University Ecology

http://plato.stanford.edu/entries/ecology/

EO Wilson’s Life On Earth Book 7 

National Geographic

http://animals.nationalgeographic.com/animals/facts/

Chi Squares and Corn: A Match Made for Science

In this lab, I used Indian Corn to explore Mendelian Genetics, Independent Assortment, and Chi Square Analysis. During the lab, I counted and classified the number of dihybrid cross corn kernels in the following four categories: purple and smooth, purple and wrinkled, yellow and smooth, and yellow and wrinkled. In this cross, our corn had an approximate 9:3:3:1 ratio, making it a perfect selection for a Chi Square analysis. In our monohybrid cross, I classified corn kernels as yellow smooth or yellow wrinkled, finding a constant and expected 3:1 ratio. After comparing our data to the Chi Square, we found that our value of 0.400827 with a variance factor of three gave a 90-95% chance that the slight differences were due to random chance.

Indian Corn has thrived in the Americas since before the arrival of Europeans to the new world. In fact, Native Americans taught settlers how to cultivate Indian Corn. Once this corn was brought to the old world, sweet corn became the corn of choice, but Indian corn was still grown as an ornamental crop. Today, corn serves many purposes: in the culinary arts, in livestock feed, and in the production of paper, fuel, and even alcohol.

In our lab, we used a particular variety of Indian corn as it has many visible phenotypes, namely purple vs. yellow and wrinkled vs. smooth. Furthermore, Indian corn provides a large sample size to determine the expected ratio. I hypothesized that in the monohybrid corn, a 3:1 ratio of smooth to wrinkled kernels would be apparent, and in the dihybrid cross a 9:3:3:1 ratio would be apparent.

This pattern of inheritance of phenotypes and genotypes was established by monk Gregor Mendel, whose work with pea plants allowed him to discover Mendelian Genetics. The concept that genes do not blend, but instead are dominant and recessive, is widely accepted today and applies to the Indian Corn kernels in our lab.

In order to determine expected ratios of crosses, I used a Punnett Square. A Punnett Square is a chart used by geneticists to predict the outcome of certain genotypes and phenotypes in genetic crosses. In the case of our lab, for the monohybrid cross, our Punnett Square is depicted in figure 1.
Figure 1

As seen in Figure 1, this Punnett Square depicts the monohybrid cross between our two heterozygous yellow smooth corn ears, using S to represent the dominant smooth phenotype and s to represent the recessive wrinkled phenotype. Thus, the Punnett square and our hypothesis predicts a 3:1 ratio of smooth kernels to wrinkled kernels.
In case of our lab’s dihybrid cross, I used the Punnett Square depicted in Figure 2.
Figure 2:

As seen in Figure 2, this Punnett Square depicts the dihybrid cross between our two heterozygous corn ears, again using S to represent the dominant smooth phenotype and s to represent the recessive wrinkled phenotype. I also used P to represent the dominant purple phenotype, and p to represent the recessive yellow phenotype. Thus, the Punnett square predicts a 9:3:3:1 ratio of purple smooth kernels purple wrinkled kernels to yellow smooth kernels to yellow wrinkled kernels, which is also our hypothesis.

To determine if my hypothesis is valid, I used a Chi Square. A Chi Square is a test that compares observed data with the expected data from a hypothesis. A null hypothesis is when there is no difference between the observed and expected data. The Chi Square determines the probability of deviations from the expected data being due to random chance. To use the Chi Square Chart, one must determine the Chi Square Value, which is found via the equation in Figure 3.
Figure 3:

While determining the Chi Square value, one finds the degrees of freedom, or the number of classes observed minus one. Then the value and degree of freedom are plotted on the Chi Square graph, which corresponds to a percentile that represents the probability that deviations were caused by random chance. A Chi square is depicted below in Figures 4.1 and 4.2
Figures 4.1 and 4.2:

Materials:
-Indian Corn, one with a Monohybrid Cross (Smooth vs. Wrinkled), one with a Dihybrid Cross (Smooth vs. Wrinkled and Yellow vs. Purple)
-Calculator
-Chi Square Table
-Paper or Note Taking App

Procedure:
Monohybrid Cross:
1.) I counted the number of smooth kernels and wrinkled kernels in five rows of corn kernels and noted the total numbers.
2.) I found the expected value of the smooth and wrinkled kernels using the total number of kernels and using the 3:1 expected ratio.
3.) I found the Chi square number by the Chi Square Equation described above, and compared it with the Chi Square with a variance of one.
Dihybrid Cross:
1.) I counted the number of purple smooth, purple wrinkled, yellow smooth and yellow wrinkled kernels in five rows of corn kernels and noted the total numbers.
2.) I found the percentages of each phenotype of corn out of the total number of kernels.
3.) I calculated the expected value of the kernels using the total number of kernels and using the 9:3:3:1 expected ratio.
4.) I again found the Chi square number by the Chi Square Equation described above, then added the values together.
5.) I compared the Chi Square Number with the Chi Square with a variance of three.

Figure 5:

As depicted in Figure 5, a monohybrid cross Indian Corn contains both yellow smooth and yellow wrinkled kernels. I calculated that this particular cob had a total of 215 kernels in our five row test area, 166 of them yellow smooth, accounting for 77.8 percent, and 49 of them yellow wrinkled, accounting for 22.2 percent of the total kernels.

Figures 6 and 7:

As seen in Figures 6 and 7, two circle graphs represent the expected quantities of yellow smooth and wrinkled kernels based on our Punnett Square Analysis and the results from our lab. Although my Punnett Square predicted that 161.25 kernels would be yellow smooth and 53,75 would be yellow wrinkled, actually 166 yellow smooth and 49 yellow wrinkled were observed.

Figure 8:

As depicted in Figure 8, a monohybrid cross of purple smooth and yellow smooth is present. In five row of corn kernels, I observed a total of 254 kernels, more than that of the yellow smooth/wrinkled variety. In this ear of corn, 206 kernels were purple smooth and 49 kernels were yellow smooth. The purple kernels accounted for 81.7 percent of the total kernels, while yellow kernels only accounted for 19.3 percent.

Figures 9 and 10:

As seen in Figures 9 and 10, two circle graphs represent the expected quantities of purple smooth and yellow smooth kernels based on our Punnett Square Analysis and the results from our lab. Although my Punnett Square predicted that 190.5 kernels would be purple and 63.5 would be yellow, actually 205 purple smooth and 49 yellow smooth were observed.

Figure 11:

As seen in Figure 11, this corn cob represents a dihybrid cross, with both smooth and wrinkled, yellow and purple kernels. The Punnett Square predicts a 9:3:3:1 Ratio of Purple Smooth to Purple Wrinkled to Yellow Smooth to Yellow Wrinkled. In five rows of this cob, I totaled 244 kernels, 137 purple smooth, 47 purple wrinkled, 43 yellow smooth, and 17 yellow wrinkled. This approximately yields 56.55 percent purple smooth, 19.22 purple wrinkled, 17.22 yellow smooth, and 6.97 percent yellow wrinkled.

Figures 12 and 13

As seen in Figures 12 and 13, two circle graphs represent the expected quantities of purple smooth, purple wrinkled, yellow smooth, and yellow wrinkled kernels based on our Punnett Square Analysis and the results from our lab. Although my Punnett Square predicted that 137.25 kernels would be purple smooth, 45.75 would be purple wrinkled and yellow smooth and 15.25 would be yellow wrinkled, but 137 purple smooth, 47 purple wrinkled, 43 yellow smooth and 17 yellow wrinkled were observed.

Figure 14:

As seen in Figure 14, the results of our Chi Square analysis is shown, we had a combined total Chi Square Number of 0.400728, with a variance factor of 3.
Figure 15:

As seen in Figure 15, I plotted the Chi Square Number on a Simplified Chi Square. My number of .400728 yields that there was a greater than 90% chance our minor discrepancies were due to random chance, making our dihybrid corn cob a good test subject.
Figures 16 and 17:

As seen in Figures 16 and 17, additional tables represent the total number of kernels observed in both the monohybrid and dihybrid crosses, with the respective percentages for each corn phenotype.

Discussion Questions:
Monohybrid Cross:
1. Which phenotypes appear to be dominant?
*In the yellow wrinkled/yellow smooth corn cob, the smooth phenotype, appears to be dominant, while in the yellow/purple corn cob, the purple appears to be dominant.
2. What are the probable genotypes of the parents with regard to coloration? (Use P)
*In the yellow smooth, yellow wrinkled corn cob, the genotypes of its parents where pp and pp, while in the yellow/purple corn cob, it exhibited characteristics of Pp and Pp.
3. What are the probable genotypes of the parents with regard to texture? (Use S)
*In the yellow smooth, yellow wrinkled corn cob, the genotypes of its parents where Ss and Ss, as it exhibited a 3:1 ratio of smooth to wrinkled kernels. while in the yellow/purple corn cob, it displayed characteristics of SS and SS.
4. What is the expected % of purple to yellow?
*In a cross of two heterozygous purple corn ears, and expected 3:1 ratio purple to yellow or 75% to 25% (purple to yellow)
5. Does your observed data match what is expected?
*In both of our monohybrid cross corn ears the observed data does not exactly correlate with the expected data, particularly in the purple/yellow corn cob, which is off by an expected 14.5 kernels.
Dihybrid Cross:
6.Did you obtain a 9:3:3:1 ratio?
*In our dihybrid cross corn ear, I did not obtain an exact 9:3:3:1 ratio, but it was very close. In our Chi Square Analysis, I determined that our minor discrepancies had an over 90% chance of being caused by random chance.
7. Explain what it means to have a “good fit” or a “poor fit”. Does your chi square analysis of real corn data support the hypothesis that the parental generation was PpSs x PpSs?
*A good fit on a Chi Square Analysis is considered to be a sample size with a greater than 5 percent probability that discrepancies in results were due to random chance, while poor fits are considered the probability of random chance at 5 percent or less. Poor fits are considered bad sample sizes and the results contradict the hypothesis. Our data, with minor discrepancies and a greater than 90% chance of them being due to random chance, is definitely a “good fit.”
8. List two reasons why data might have a poor chi square fit.
*Two reasons for a poor Chi Square Fit include a poor sample selection, in our case the corn cob, or an inaccurate hypothesis, in which the expected total is significantly different from that of the actual total.
Sample Questions:
1.
Problem: A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet.
Your Tentative Hypothesis: This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.
Objective: Test your hypothesis using chi square and probability values.
To find the expected total number of kernels of a dihybrid cross, I used the following Punnett Square, where P is the dominant purple and S the dominant starchy:
Figure 18:

e
By finding the 9:3:3:1 ratio, I can calculate the expected quantities for each:
Purple Smooth: 9/16 x 433 = 243.5625
Purple Wrinkled: 3/16 x 433 = 81.1825
Yellow Smooth: 3/16 x 433 = 81.1825
Yellow Wrinkled: 1/16 x 43 = 27.0625
Next, using the equation displayed in Figure 19, I can determine the individual Chi Square Values for each phenotype,
Figure 19

This results in the following equation:

3.5579+0.8247+4.0723+0.0417=8.4965
With a variance of three, it yield a value between 1 and 5 percent making this a poor fit.
Figure 20:

2.
Problem: In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yellow-eyed lizards.Your Tentative Hypothesis: The black eyed parents were Bb x Bb.Objective: Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined, the degrees of freedom (df) is 1. SHOW ALL WORK!
in a dihybrid cross of two heterozygous black eyed lizards,,the expected totals are 75 black eyed and 25 white eyed by the following Punnett Square:
Figure 21:

From this, I can derive the following equation:
Figure 22:

This results in 0.36+0.12, or 0.48.
This is classified as a good fit, on the Chi Square Below (Figure 23), with a variance of 1
Figure 23:

3. Problem: A sample of mice (all from the same parents) shows 58 Black hair, black eyes 16 Black hair, red eyes 19 White hair, black eyes 7 White hair, red eyes Your tentative hypothesis: (what are the parents?)
Objective: Use a chi square analysis to support your hypothesis
I hypothesis that the parents are heterozygous black hair, black eyes.
The following Punnett Square, in which H is hair color and E is eye color supports my hypothesis:
Figure 24:

There are 100 mice, so
Black Hair and Black Eyes 9/16 of 100 = 56.25
Black Hair Red Eyes 3/16 of 100 = 18.75
White Hair Black Eyes3/16 of 100 = 18.76
White Hair Red Eyes 1/16 of 100 = 6.25
The following Chi Square equation yields the Chi Square Number:
Figure 25:

This results in:
.05 + .40 + .003 + .09 = .54
This correlates to a good fit in the 90th chance percentile:
Figure 26:

In this lab, I was able to apply the Chi Square and Mendelian Genetics to a part of the real world, observing the genotypes and phenotypes of Indian Corn.
In my dihybrid crossed corn kernels the following four categories were present: 137 purple and smooth, 47 purple and wrinkled, 43 yellow and smooth, and 17 yellow and wrinkled. In this cross, our corn had an approximate 9:3:3:1 ratio, slotting above the 90% category on the Chi Square Analysis. In the monohybrid cross, I classified 166 yellow smooth and 49 yellow wrinkled, finding a near expected 3:1 ratio. After comparing our data to the Chi Square, rye dihybrid’s value of 0.400827 with a variance factor of three yielded over a 90% chance the slight differences were due to random chance.
The following are possible sources of error in this experiment: miscalculating the total number of corn kernels, miscalculating the Chi Square Analysis, misidentifying corn kernels, miscounting corn kernels, using the inaccurate Chi Square Equation depicted in Figure 27, and not using Punnett Squares to anticipate the expected numbers of each variety of corn kernel.
Figure 27:

If I were to repeat this experiment, I would use different types of Indian Corn, and examine different phenotypes, such as number of kernels in a row, or whether or not the kernel has a “dent” in it, using it to make a trihybrid or quad-hybrid cross. I would also look at other kinds of specimens, such as flies, or peas, or different varieties of tomatoes, to examine their phenotypes and respective genotypes. Given more time, I would analyze the genetics of a human being by creating a monstrous Punnett Square for human genotypes and phenotypes, and then applying it to Chi Square Analysis.

University of California Berkeley-Biology
http://evolution.berkeley.edu/evolibrary/article/history_13
Dennis O’Niell Palomar University Anthropology
http://anthro.palomar.edu/mendel/mendel_2.htm
Encyclopedia Brittanica
http://www.britannica.com/EBchecked/topic/137741/corn
Stanford University Statistics
http://mips.stanford.edu/courses/stats_data_analsys/principles/chi_table.html
North Dakota State University Biology
http://www.ndsu.edu/pubweb/~mcclean/plsc431/mendel/index.htm
University of South Carolina Orangeburg
http://physics.scsu.edu/~dscott/gen/CornLab.html
University of Indiana Bloomington
http://www.indiana.edu/~l113/labs/fly_genetics/Chi-Square.pdf
Graphs: Create a Graph
nces.ed.gov/nceskids/createagraph/

NEW Meiosis Stop Motion Video

Here is a revised meiosis stop motion video, to accompany the lab below.

Meiosis Stop Motion Video

This past week, I explored the process of meiosis via the production of a stop motion video.

What is the function of meiosis?
*The function of meiosis to creat haploid gamete cells to facilitate sexual reproduction. Meiosis allows for genetic diversity, which via the theory of natural selection, can help a species evolve and survive.

What events promote genetic variation during meiosis?
*There are three events in meiosis that promote genetic variation. The first event occurs during prophase 1, where crossing over occurs. This process of the sharing and reorganizing of genetic materials creates a genetic variation constant throughout meiosis. The second event, random alignment, occurs during metaphase I, in which the tetrads randomly align with genetic information split between the two future cells. Half the genetic material will account for one cell, the other half for the other. This process of random alignment happens again in metaphase II.

What causes non-disjunction?
*Non-disjunction occurs during anaphase I, when a pair of homologues fail to separate. This can be caused by the failure of separation of the tetrads in anaphase 1.
Panda bears have 42 chromosomes compared to 74 chromosomes found in most bears. How could this occur? Explain in terms of non-disjunction.
*Non disjunction is the reason panda bears have 42 chromosomes, instead of 74. After multiple generations, if cells with fewer chromosomes due to non-disjunction continued to divide in meiosis, eventually, a new species of bear would appear.

How could this lesson be improved?
*I believe this lesson could be improved by examining real cells that divide via meiosis, instead of creating them with pipe cleaners. Other options for demonstrating meiosis might include computer graphics, 3D printing, and drawing.

Stop Motion:
https://m.youtube.com/watch?v=2QaxIXjzdGI

A Talk With a Cancer Patient

This past week, I had the privilege to talk with and learn from a survivor of cancer. My interviewee was a seventy-seven year old man who had prostate cancer. While he is in remission, I could tell this was still a very sensitive and painful topic to discuss. Two years prior, he had joined a new health care system, and had gone to a routine medical checkup for the first time in a few years. On a routine blood test, a PSA level was checked, and it was determined that it was elevated. The PSA, or prostate specific antigen was eleven. He told me this meant the chance of him having prostrate cancer was over fifty percent. He then underwent a CT Scan, a bone scan, and eventually a prostate biopsy. The prostate biopsy “was uncomfortable,” he remarked. He joked, “he had no symptoms before the biopsy, only after.” After it was discovered that he had cancer, he alternated between denial and “I am going to beat this” attitude. Admittedly, he told me, that he was scared of the dreaded c-word. After he was diagnosed, he told me that he was given a choice; surgery, or radiation and hormone therapy. He was reluctant to have surgery in fear of lifelong incontinence, so he chose hormone therapy with radiation. My survivor received external beam radiation, five days a week for a nine week period and he started a hormone therapy called Lupron therapy. He told me that the radiation therapy targeted the cancer cells in his prostate, and the Lupron stopped the release of testosterone. This slows or stops the growth of cells including his cancer cells which are testosterone dependent. Initially, his travel plans with his wife had to be altered to allow for the nine weeks of radiation therapy. He associated the treatment with his extreme fatigue, which prohibited him from enjoying his everyday lifestyle. My interviewee said that he was grateful for his wife in helping him, and for all the support from his family and friends, while struggling with not wanting to be a burden to them. He grew closer to a friend who had previously survived prostate cancer, but did not feel comfortable talking about it to most of his other friends. He focused more on the present tense than the future, enjoying each moment. Most of his difficultly centered upon the “what ifs” and struggling to deal with the possibility of the worst possible outcome. In regard to addressing common misconceptions about cancer, he replied, “people believe that you have no voice in your options for treatment. However, the most important thing is to become educated in your treatment options, and discuss them openly with your physician. The goal is to work with your doctor, “not follow blindly.” He shared personal advice as well: “appreciate every day; allow yourself to be inspired and inspire others. No matter what circumstance you are in, keep a positive outlook on the future and don’t dwell on the ‘what-ifs.'”

After talking with a cancer patient, I developed a strong admiration for all of those who endure the hardships of cancer. His views on life drastically changed mine, reminding me to focus on the present and not dwell on the past, or the future. Reading the text book and classroom discussions gave me a third party detailed view on cancer, but actually speaking to somebody who had experienced cancer firsthand deepened both my knowledge and perception of cancer, as well as being an august experience. I am grateful to my interviewee for all that he shared with me, and I will remember it forever.

About Prostate Cancer

According to E.O. Wilson’s Life on Earth cancer is “a disease in which abnormal cells divide uncontrollably and invade other tissues. A cancer can occur anywhere in the human body, and then spread to other tissues via metastasis.

My cancer patient developed prostate cancer, which affects the prostate gland in males. The prostate gland is a part of the male reproductive system located between the bladder and the rectum. Prostate cancer, if it becomes metastatic, can spread throughout the body, even affecting vital organs such as the liver, lung, brain and heart. Luckily, my interviewee’s cancer did not become metastatic, and was contained to the prostate.

The cause of changes that lead to prostate cancer is not known by medical professionals, however, medical researchers now understand the changes in DNA that can create cancerous prostate cells.

Contrary to popular belief, genetics are only responsible for approximately 5-10% of all prostate cancers. In these cases, genetic inheritance of three subsets of gene mutations lead to a higher rate of prostate cancer.

One of these gene mutations is RNASEL, derived from HPC1. HPC1 kills tumors when they appear in the prostate glands. The mutation of RNASEL allows the irregular cells to last longer, increasing the chance of developing prostate cancer.

Another two genes that can harmfully mutate include BRCA1 and BRCA2. These genes suppress tumors by fixing mistakes in DNA synthesis. Although mutations in these genes are uncommon, they account for a number of prostate cancers, as well as breast and ovarian cancers.

MSH2 and MLH1 are another two genes that correct irregularities in DNA. Men who inherit mutations in this gene develop the condition Lynch syndrome, also called hereditary nonpolyposis colorectal cancer (HNPCC). Men with Lynch syndrome are at a higher risk at developing prostate, colorectal, small intestine, stomach, gallbladder, and brain cancers.

In addition to inherited genes that may cause cancer, some scientists hypothesize that DNA mutations or high production of substances can attribute to increased risk of developing cancer. Increased androgen level, such as large amounts of testosterone, have been proven to promote cell growth, perhaps contributing to the cell growth of cancerous cells in the prostate.

Another hormone, insulin-like growth factor-1 (IGF-1), perhaps can correlate to an increased risk of developing prostate cancer. IGF-1 is another hormone that regulates cell growth that some studies have related to cancer growth, however, additional studies are yet to connect IGF-1 and prostate cancer.

In prostate cancers, the cell cycle plays an important role. When DNA replicates during the S stage of interphase, mutations in genes described above can pass through the cell cycle, producing new cancerous cells, which can then uncontrollably divide as tumors. Other proteins that interact with cells during different stages of the cell cycle affect, modulating ARs. These include “G0 (RB), G1 to S phase (cyclin D1, cyclin E, Cdk6), or G2 (Cdk1)” (Balk-Knudsen). Errors in the modulation of ARs could possibly cause a cancer to develop.

To treat prostate cancer, medical professionals use many techniques. The patient usually can consider which path to take, dependent upon the patient’s age and life span, other health problems, the stage of the cancer, and medical opinions.
Patients with stage one prostate cancer are recommended to undergo “active surveillance”, and treatments include radiation therapy involving either external beams or brachytherapy, and radical prostatectomy.

Patients with stage two prostate cancer have the same treatment options as stage one prostate cancer patients, however, a majority of radiation therapies also include hormone therapy as well.

Patients with stage three prostate cancer usually also have the same options as stage two patients, but it is recommended by medical professionals to have radiation, then a radical prostatectomy.

Finally, patients with stage four prostate cancer unfortunately usually cannot partake in the standard cancer treatments. Treatment for stage four patients include hormone therapy, surgery to remove pain in the bladder or rectum. Bone metastatic development treatment, such as denosumab, a drug also used to cure osteoporosis, or “radiopharmaceutical such as strontium-89, samarium-153 or radium-223.” (Cancer.org)

As we do not know a definite cause for prostate cancer, there are not strong evidence of prevention, however many other ways to prevent cancers remain true for prostate cancer.
The American Cancer Society suggests that one eats 2 1/2 cups of assorted fruits and vegetables daily, one maintains a healthy weight and diet as well. Additional studies have shown that some medicines can also decrease the rate of developing prostate cancer. Early studies show that supplements of Vitamin E and selenium lower the rate of prostate cancer in males. Furthermore, “5-alpha reductase inhibitors [treat]!prostatic hyperplasia (BPH), a non-cancerous growth of the prostate” (cancer.org) These medicines include Finasteride, marketed as Proscar, and Dutasteride marketed as Avodart may decrease the risk of developing prostate cancer. Other studies suggest that prolonged aspirin intake could also decrease the rate of prostate cancer, but further studies are required.

In 2014, it is estimated that another 233,000 men in the U.S. will develop prostate cancer, and approximately 1,112,000 people will be diagnosed in the world. Of these numbers, 29,480 men in the U.S. Will die of prostate cancer.

Physicians whom help treat prostate cancer include:
–Urologists, or a surgeon that specializes in surgery of the urinary tract and the male reproductive system.
–Radiation oncologists, a physician who treat cancer with radiation therapy, and
–Medical oncologists, a doctor who treats cancer with chemotherapy or hormone therapy.

Sources
National Cancer Institute
http://www.cancer.gov/cancertopics/types/prostate
National Cancer Association
http://www.cancer.org/cancer/prostatecancer/detailedguide/prostate-cancer-treating-general-info
http://www.cancer.org/cancer/prostatecancer/detailedguide/prostate-cancer-treating-by-stage
http://www.cancer.org/cancer/prostatecancer/detailedguide/prostate-cancer-prevention
http://www.cancer.org/cancer/prostatecancer/detailedguide/prostate-cancer-what-causes
World Cancer Research Foundation
http://www.wcrf.org/int/cancer-facts-figures/worldwide-data
Genetics Home Reference
http://ghr.nlm.nih.gov/condition/lynch-syndrome
Balk, Steven and Knudsen, Karen : National Center For Biotechnology Information
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2254330/#__sec6title

Mitosis Lab

Abstract

In this experiment, I investigated slides of root tips and analyzed the phases of mitosis and the relative duration of each phase in the meristem root tissue. An organism’s growth is delicately controlled by regulating the cell cycle. Mitosis is the cell cycle process in which nuclear division and cytokinesis form two identical daughter cells.

Introduction and Background

In the early 1880s, the German anatomist Walther Flemming observed the fibrous structures of spindle fibers and created the term mitosis, from the Greek word for thread, to describe the actions of spindle fibers in a dividing cell.
To study the process of mitosis, I examined cells from the meristem root tissue of an onion, in which many cells are undergoing the process of mitosis. Mitosis is the process of nuclear division and cytokinesis to form two identical daughter cells.
Mitosis consists of the following stages: interphase (although not technically considered part of mitosis, it is the precursor to mitosis), in which the chromosomes duplicate and the cell readies itself for mitosis; prophase, in which chromosomes condense and spindle fibers form; prometaphase, when the nuclear membrane disintegrates; metaphase, when the chromosomes line up one the middle of the spindle fibers with the centromere attached to the spindle; anaphase, in which the chromosomes separate and migrate towards the asters; and telophase, when nuclear membranes begin to form around the chromosomes on both sides of the cell. During telophase, the chromosomes condense and the cell prepares for cytokinesis, in which the cell splits into two identical daughter cells. Interphase and the stages of mitosis are seen in Figure 1 below.

Figure 1:

I used onions in this experiment due to the high percentage of cells that undergo mitosis in an onion’s apical meristem root tissue. This tissue is located towards the end of the root, near the root stem. The location of the apical meristem root tissue is displayed in Figure 2 below.

I used whitefish blastula in this experiment because of its rapid rate of nuclear division from fertilization. This rapid division allows one to visualize spindle fibers very clearly in the cell.

Figure 2:

The duration of the stages of mitosis can be determined by counting the number of cells in mitosis at a given time and multiplying that value by the average duration of mitosis in an onion, which is approximately twenty-four hours.

I hypothesized that a majority of cells would be in interphase, because of the time required to synthesize new DNA molecules during the S stage.
Materials and Procedure

Materials List

–1 Light Microscope, outfitted with a 10X eyepiece and 4X, 10X, and 40X objective lens
–Prepared Microscope Slide of 3 Onion Root Tips
–Prepared Microscope Slide of Whitefish Blastula

Procedure

Part One: Observing Mitosis in Plant and Animal Cells

1. I brought the microscope into focus on the lowest 4X setting with the onion root tip in view, making sure the diaphragm was at its widest setting.
2. I located the meristem root tissue of the onion root tip and switched the microscope to the 40X measure, in which cells were visible.
3. I identified and classified the cells based on their stage in mitosis: interphase, prophase, prometaphase, metaphase, anaphase, and telophase, as depicted below in Figure 3 and Figure 4
Figure 3:

Figure 4:

4. I looked in all three onion root tips to find all of the stages of mitosis.
5. Steps 1-4 were repeated using the whitefish blastula, which is an animal cell.

Part Two: Determining the Rate of Mitosis in Plant and Animal Cells.

1. I focused the microscope on the 4X setting with the onion root tip in view, making sure the diaphragm again was at its widest setting.
2. I located the meristem root tissue of the onion root tip and changed to the 40X high power lens, so the cells of the onion root were visible.
3. Using the first onion root’s meristem root tissue, I found a large collection of cells in many stages of mitosis.
4. I counted the number of cells in each stage and recorded the values in my data table.
5. I repeated steps 3 and 4 twice, each time using a different onion root tip to locate the cell clusters.
6. I totaled the number of cells in each stage in the three microscope views, and determined the percentage of the total number of cells in each stage.
7. I multiplied the percent of the total number of each cell in a phase by 1440, the number of minutes in a day, which is the time an onion cell takes to undergo mitosis. This product is approximately equal to the time a cell spends in each stage of mitosis.

Results

Figure 5

As seen in Figure 5, I observed many cells in an onion root tip under the 40X lens. A majority of the cells are in interphase; however, several are in prophase, metaphase, and anaphase.
Figure 6

As seen in Figure 6, I observed the cells in another slide of onion root tips. These cells were stained blue and red, in contrast to the cells in Figure 5. In this slide, a majority of cells are in interphase, and a notable telophase is seen in the center.
Figure 7

As seen in Figure 7, I identified the cells by stage of the cell cycle, with black representing interphase, green prophase, red metaphase, yellow anaphase, and blue telophase. A majority of the cells (376) are in interphase, representing ninety-six percent of the total cells in view. I noticed slight differences among the interphase cells, which represent the many stages of interphase: G1, S, and G2.
Graph 1

As seen in Graph 1, a pie chart represents the total percentage of cells in each stage of the cell cycle. Interphase accounted for 95.31 percent of the cells, followed by prophase with 1.58 percent, metaphase with 1.27 percent, telophase with 0.95 percent, and anaphase with 0.87 percent.
Graph 2

As seen in Graph 2, this pie chart represents the total number of minutes per cycle a cell spends in each stage. Cells spend most of the time in interphase, a total of 1372.52 minutes, or 22 hours and 52 minutes.

Table 1

Table 1 tabulates the total number of cells counted in part two of the experiment. I counted 1200 in interphase (95.31%), 20 in prophase (1.58%), 16 in metaphase (1.27%), 11 in anaphase (0.87%), and 12 in telophase (0.95%).

Discussion Questions and Conclusion
1.) Why is it more accurate to call mitosis nuclear replication rather than cell division?
*It is more accurate to call mitosis nuclear replication instead of cell division as the actual division of cells occurs during cytokinesis, not mitosis. Only the nucleus is replicated during mitosis, and it migrates to the ends of the cell. However, the cell has not yet divided.
2.)Explain why whitefish blastula and onion root tips are selected for a study of mitosis.
*The whitefish blastula is used because the rapid growth of cells and frequent cell division allows for observation of the various phases of mitosis. A blastula is a spherical arrangement of cells, like a zygote, that results from fertilization. Onion root tips are used because the growth of the roots occur in the root tips, which results in many cells going through mitosis.
3.) If your observations had not been restricted to the area of the root tip that is actively dividing, how would your results differ?
*If I observed areas of the root that were not actively dividing, I would have found few or no cells in prophase, metaphase, anaphase, or telophase. Instead, the cells would have been growing during interphase. In addition, my measurement of time spent in the cell cycle phases would have been less accurate due to the small fraction of dividing cells.
4.)Based on the data in table 3.1 what can you infer about the relative length of time an onion root-tip cell spends in each stage of cell division?
* Based on the data in table 3.1, I can determine that a cell spends a majority of its time in cell division during interphase. In mitosis, a cell spends most of its time in prophase, followed my metaphase, telepath and anaphase, respectively.

In this experiment, I was able to study mitosis and the stages of the cell cycle by observing and identifying the stages under a microscope. Calculating the total amount of time that cell spends in each phase of the cell cycle helped me understand the relative time a cell spends in mitosis and interphase. In addition, I proved my hypothesis correct, as a majority of the cells I identified were in interphase.

As this experiment involved observing cells, possible sources of error include the misidentification of cells. I found it difficult to distinguish cells in anaphase from cells in telophase, which may have affected my results. Furthermore, due to the large number of cells in interphase, it is possible to incorrectly count the interphase cells in the field of view. Other possible errors include viewing the onion root tip in a region other than the apical meristem, in which one would observe almost all cells in interphase due to the low rate of mitosis.

If I were to repeat this experiment, I would try to determine if different varieties of onions had different rates of mitosis in their apical meristem tissue. Another possible experiment would be to use the roots of an unwatered onion plant, with roots that lacked the nutrients needed to grow, to see if poor nutrition affects the rate of mitosis. I would also like to experiment with meiosis using a chromosome replication kit. From this experiment, I could compare my results from both processes, and determine the similarities and differences between the two. A final experiment is to use different varieties of plant and animal cells with different rates of growth in order to determine how long it takes their cells to complete the cell cycle.

References

University of Arizona-Biology
http://www.biology.arizona.edu/cell_bio/tutorials/cell_cycle/main.html
Marietta Univeristy, Marietta, Georgia

Click to access Onion%20root%20mitosis.pdf

Photosynthesis Lab

In this experiment, we determined the starting substrates that leaves need to perform photosynthesis. We analyzed ivy leaves in light and dark environments, and we measured photosynthesis by analyzing the production of oxygen by the leaves. We observed that exposure to light causes leaf discs with oxygen deprived spongy mesophyll to float in water, presumably because new oxygen is produced within the discs. Our hypothesis is that light is required for photosynthesis to occur becuase leaves exposed to CO2 and light will form oxygen and glucose by photosynthesis via the equation shown in Figure 1:

Figure 1:

The leaves that are exposed to light float to the surface of a cup filled with water. The leaves deprived of light cannot initiate photosynthesis and produce oxygen, and thus do not float.

In the 1770s, Joseph Priestley demonstrated that plants release a substance that allows combustion to occur. He showed this by burning a candle in a closed system until the flame died out. He then added a spring of mint and after several days he showed that the candle can burn again. We now know that the substance released by the mint leaves is oxygen. In 1845, Julius von Mayer described photosynthesis as a process of converting light energy into chemical energy. The overall reaction of photosynthesis is:

There are two parts of photosynthesis: the light reaction and the dark reaction. Most of the proteins used in the light reaction are inside the chloroplast in the thylakoid membrane. The two parts of photosynthesis are displayed in Figures 2 and 3 below.
Figure 2

Figure 3

In this experiment we studied the light dependent phase of photosynthesis by indirectly measuring the oxygen produced in floating leaf capsules.

Materials
-0.2% sodium bicarbonate (baking soda)
-Hedera helix leaves
-600 ml Distilled Water
-Dishwashing detergent
-Plastic cups
-Beaker
-Aluminum foil
-100-watt lamp
-Thermometer
-Stopwatch
-Paper towels
-Camera
-Syringe without needle

Lab Set Up:

1. We took some of the leaves and hole-punched them, and we collected 100 of the resulting leaf disks.
2. Meanwhile, we mixed 0.6 g of baking soda and one drop of detergent with 300 ml of distilled water to create a 0.2% sodium bicarbonate solution.
3. We repeated step two, as we would use one for the part of the experiment that required light and another for the part that required total darkness.
4. We used the thermometer to make sure both solutions had the same temperature.

Figure 4: Hole Punching the Leaves

Removing air from the spongy mesophyll:

1. We then determined if our syringe could hold a vacuum by pressing one finger on its tip, pulling back on the plunger, and letting go to see if it popped back.
2. We vacuumed the air out of the spongy mesophyll of half of our leaves by placing them into the syringe chamber with a small amount of the solution, then repeating the vacuum step from step one.
3. We put this half of our disks into one cup of sodium bicarbonate solution and covered it with aluminum foil to prevent any light from coming in.
4. We then took the remaining disks and put them all in the syringe. After that, we drew up 2 milliliters of the sodium bicarbonate solution with the syringe.
5. With the leaf disks and the solution both in the syringe, we pulled the plunger back, sealed the tip with his finger to create a vacuum, and rocked it back and forth. After that, we pulled out the plunger.
6. We put the disks in the cup of sodium bicarbonate solution, and placed it under the lamp, which was powered on.

Performing the experiment:

1. The timer was turned on.
2. We waited patiently for one disk to float as a result of oxygen being produced within the disk.
Figure 5 and 6:

We waited patiently for our leaf disks to rise.

Figure 7:

As seen in Figure 5, after six minutes, leaves began to tilt upwards from the bottom of the cup, as oxygen was produced by the leaves via photosynthesis, which made them more buoyant.

Figure 8:

As seen in figure 6, leaf disks eventually rose to the top of the solution. These leaf disks were placed in solution with sodium bicarbonate and were exposed to light. This triggered the photosynthesis reaction, which produced oxygen in the spongy mesophyll and caused the leaf to become buoyant and rise to the surface.

Graph 1

As seen in Graph 1, the y axis is the number of leaves that rose in water and the x axis is time. The black line represents leaves that had no light exposure. In this group, no leaf disks became buoyant, which indicates that no photosynthesis occurred and therefore no oxygen was formed. The yellow line represents leaf disks that were exposed to light. Between the 8 and 10 minute interval, the leaf disks began to rise. This indicates that these leaves initiated photosynthesis and produced oxygen at that time. Leaf disks continued to rise throughout the remainder of the experiment.
Table 1:

Table 1 shows the corresponding data that is illustrated in Graph 1. From this data we concluded that light is required for plants to initiate photosynthesis

1.) How does the floating-disk method measure photosynthesis? How else could you measure photosynthesis?

*The floating disk method indirectly measures photosynthesis by measuring oxygen formation, which is one product of photosynthesis. The oxygen formation was measured by observing the leaf disks in water. When photosynthesis occurred, oxygen was produced in the leaves, causing them to rise in water. Other ways to measure photosynthesis are:

1. Measuring the uptake of CO2
2. Measuring the production of carbohydrates
3. Measuring the increase in dry mass

2.) Which experimental condition yielded the highest rate of oxygen production? Why?

The experimental condition that yielded the highest rate of oxygen production was the experiment with leaves exposed to light after being mixed with dish water soap (which disrupted the lipid outer layer of the leaf) and carbon dioxide to start the photosynthesis. This facilitated access to all reactants needed for photosynthesis to occur, and led to a higher rate of oxygen production.

3.) Which experimental condition yielded the lowest rate of oxygen production? Why?

The condition that yielded the lowest rate of oxygen production was the experiment that had no light source. The leaf disks in this group had one of the necessary substrates for photosynthesis (C02) but it did not have light energy. The experimental conditions did not support photosynthesis and therefore no oxygen was produced.

4.) Do you think the rate of photosynthesis is proportional to the surface area of the disks? How would you test your hypothesis?

We think the rate of photosynthesis is proportional to the surface area of the disks. The greater the surface area, the greater the amount of starting product, light energy and CO2 that is available for photosynthesis. We could test this hypothesis by making different size discs and comparing their buoyancy over time.

5.) Light and temperature and pH are environmental variables. What are some biological variables that could impact the rate of photosynthesis? How would you test the effect of a biological variable in photosynthesis?

Biological variables that affect the rate of photosynthesis include the health of the plant, age, whether it flowers, the size of the leaves, content of minerals, leaf species; number of chloroplasts, color of leaves, and enzyme activity. We could obtain diffecnt leaves and compare them in the same experimental design to measure oxygen production based on these biological factors.

In this experiment we were able study the rate of photosynthesis using leaf disks. Leaf disks normally float, however, we removed the air from the leaf discs and removed the air from the spongy mesophyll layer, which caused the disks to sink. Sodium bicarbonate was used because it provided a source of CO2, one of the required substrates for photosynthesis. Dishwater soap was added to disrupt the lipids in the cuticle layer of the leaf, which allowed for easier diffusion of the CO2 into the stomata of the leaves. As photosynthesis occurred, oxygen was released in the interior of the leaf, and the leaves became buoyant and rose to top of the water.
Our data shows that the leaf disks that were exposed to light energy along with CO2 produced oxygen The leaves that were not exposed to light energy did not produce oxygen and consequently we observed no change in buoyancy.

If we were to do this experiment again, we would use a constant intensity light source. Our lamp varied in intensity during the experiment, which may have altered our data.

It would be interesting to explore photosynthesis using different wavelengths of light. From the graph below of wavelength and photochemical efficiency, it appears that only the far ends of the visible part of the electromagnetic spectrum are used by plants in photosynthesis. It would be interesting if we could demonstrate that using our experimental design, by using different colored light bulbs. It would also be interesting to explore mechanisms designed to increase the rate of photosynthesis, which might also increase crop production. This year in the journal Nature (Nature 513, 547–550 (25 September 2014), scientists reported successful genetic engineering of a tobacco plant to include faster enzymes to allow it to conduct photosynthesis at a higher rate than conventional tobacco leaves.

http://www.ftexploring.com/photosyn/chloroplast
http://biology.clc.uc.edu/courses/bio104/photosyn.htm
http://www.life.illinois.edu/govindjee/paper/gov.htm
http://hyperphysics.phy-astr.gsu.edu/hbase/biology/ligabs.html
Nature 513, 547–550 (25 September 2014)

Catalase Lab

In this lab, I studied the effects of the concentration of catalase on enzymatic activity. I found that a greater concentration of enzyme leads to a greater rate of reaction. The lab measured the rate of the formation of oxygen via the breakdown of hydrogen peroxide. The catalase reaction is very important in cells because it stops the accumulation of hydrogen peroxide, which is lethal to the cell. I hypothesized that as the concentration of the enzyme increased, so would the quantity of the products.
Enzymes such as catalase are catalysts that execute thousands of chemical reactions that occur in living cells. In an enzyme catalyzed reaction, the substrate (the substance to be acted upon), binds to the active site of the enzyme. The enzyme then changes the substrate into the reaction products. The product is then released into solution and the unchanged enzyme is ready to make another enzyme-substrate compound. Both the enzyme concentration and the substrate concentration can affect the reaction rate. The presence of more molecules increases the probability of collisions and enzyme substrate coupling. In this lab, the enzyme catalase accelerated the breakdown of hydrogen peroxide, which is summarized in this reaction.
Figure 1

This reaction is critical in living cells, because it stops the accumulation of hydrogen peroxide, which is lethal to the cell. Hydrogen peroxide is a harmful byproduct of many normal metabolic reactions. In order to prevent damage to the cell, it must quickly be changed into less harmful substances. Therefore, the presence of catalase is important because it breaks down a potential toxin.
Catalase is also used in the production of cheese, as hydrogen peroxide is a byproduct of the cheese-making process, and it must be removed so that the cheese is not spoiled. In addition, some studies have shown that the graying of human hair results from a buildup in hydrogen peroxide due to decreasing amounts of catalase in the aging body.

Materials used in the experiment:
-50 mL beaker
-10 mL and 50 mL graduated cylinders
-Fresh 3% Hydrogen Peroxide Solution
-Forceps
-Water pan
-Catalase solution
-Filter Paper, cut into round dots with whole puncher
-Reaction chambers (Drosophila vials with 1- hole stoppers)
-Stopwatch
-Ice

Setting up the experiment:
1. I put a small amount of the catalase solution into the 50-ml beaker
2. I placed four catalase-soaked filter paper disks high on one interior sidewall of the reaction chamber. (They will stick to the sidewall). To prepare a disk for use in the reaction chamber, I held it by its edge with a pair of forceps and dipped it into the catalase solution for a few seconds. I thin drained the disk against the sidewall of the beaker transferring it to the reaction chamber.

Preparing the reaction chamber:
1. I placed the reaction chamber upright and added 10 mL of 3% hydrogen peroxide solution, without allowing it to touch the filter paper disks.
2. I put a stopper in the chamber so that it fit tightly.
3. I filled a pan (almost full) with water.

Conducting the experiment:
1. I laid the 50-mL graduated cylinder on its side in the pan so that it filled completely with water. As an air bubble was present, I carefully worked it out by gently shaking it.
2. I turned the cylinder upside down into an upright position while keeping it filled with water.
3. I carefully placed the reaction chamber and contents on its side in the pan of water, making certain that the side with the disks faced upward.
4. I moved the graduated cylinder into a position so that its mouth came to lie directly over the tip of the stopper. I held it in this position for the duration of the experiment.
5. I rotated the reaction chamber 180 degrees on its side so that the hydrogen peroxide solution came into contact with the catalase-soaked disks.

Recording data:
1. I measured gas levels in the graduated cylinder at 30-second intervals during a 5 minute period.

Figure 2

Figure 2 illustrates the production of gas bubbles in the test tube, which rose to form a pocket of gas in the tube.

Figure 3

Figure 3 illustrates my table of results for the experiment. From this data, I concluded that when a higher concentration of enzyme is present, the rate of reaction is faster. A notable exception is seen in the discrepancy in the 3 dot category. The one dot solution increased by an average of 0.31 mL per 30 second interval, the two dot solution increased by 0.9 mL every 30 seconds, the three dot solution increased by 0.35 mL every 30 seconds, and the four dot solution increased by 1.7 mL every 30 seconds.

Figure 4

Figure 4 depicts a line graph of the trends of the solutions during the five minute experiment. In general, as noted above, when a higher concentration of enzyme is present, more of the product (oxygen) is produced.

Discussion Questions:
-(Graph Above) Although the amount of oxygen produced eventually slowed down, the action of the catalase remained the same. As catalase is both a reactant and product in the equation (as noted above), the amount of catalase available is constant, and it continued to react with the substrate (hydrogen peroxide) until the substrate was depleted.
-(Table above) Based on the data, I concluded that as the concentration of enzyme increases, so does the quantity of products.
-For optimum enzymatic activity, the surrounding conditions must be set at an appropriate temperature, pH, substrate concentration, and enzyme concentration. Enzymes function well only a specific environment, and within a narrow range of conditions that may vary with the structure of the enzyme. For example, catalase, the enzyme used in this experiment, worked well at room temperature (70 degrees), in water (pH 7), and in a solution in which the available substrate exceeded the active sites on the enzyme.
In this experiment, I studied the effects of the quantity of catalase on enzymatic activity. I proved that my hypothesis was correct, as my data reveals that when a greater concentration of an enzyme is present, a larger amount of products are produced. This is because the greater quantity of enzyme provided more active sites with which the substrate could react, which resulted in a higher probability of a successful reaction. This can be visualized in the following figures:
Figure 5

Figure 6

As illustrated in Figure 4, the four dot solution produced the most oxygen (17 mL after five minutes), followed by the two dot solution (9 mL after 5 minutes), the three dot solution (3.5 mL after 5 minutes), and the one dot solution (3.1 mL after 5 minutes).
Of note, there was an unanticipated result in the three dot solution, as less oxygen was produced than was expected. The amount of oxygen visible during the reaction produced in that reaction remained constant; however, some of the gas was inadvertently not captured in the test tube due to human error. In addition, one of the dots in that solution was not completely soaked in catalase, which likely affected the results by reducing the amount of enzyme available to facilitate the breakdown of hydrogen peroxide.
If I were to repeat the experiment, I would position the test tube so that all of the oxygen is collected. I would also vary the ambient temperature, pH, and salinity in order to determine the effect of changing those variables on the activity of catalase. I would also determine the effect of a competitive inhibitor of catalase on the rate of decomposition of hydrogen peroxide. An interesting potential application of catalase is to restore natural hair pigments, as white hair may result from an increase in hydrogen peroxide in aging hair follicles, and exogenous catalase might reverse pigment degradation. Catalase may also have a role in cancer treatment, as tumor cells could be targeted with a chemical inhibitor of catalase, which might allow naturally produced hydrogen peroxide to preferentially destroy those cells.

http://www.sciencedaily.com/releases/2009/02/090223131123.htm
http://www.princeton.edu/~achaney/tvme/wiki100k/docs/Catalase.html

Interstellar

Last week, my biology honors class and I went to an early showing of IMAX Insterstellar, a film about Cooper, an ex-member of the decommissioned NASA, working with some of his old colleagues to find a new home for people as the Earth becomes uninhabitable.
In the film, the crew of the spacecraft use a wormhole near Saturn to reach a distant galaxy in search for members of an earlier expedition. The wormhole is thought to have two points set at different places and time in the galaxy. Although we have no visual evidence for their existence, the theories of relatively as discussed in the film, suggest they do exist. Even though time travel is considered to be science fiction fantasy popularized by Captain Kirk and Picard, humanity may achieve the ability to travel into the dimension of time.

Time travel revolves around relatively. If wormholes do exist this would point to the possibility of time travel. By accelerating one end of a wormhole to a velocity higher than the other, it would be possible to travel through time, as long as one brought it back to normal velocity during the journey. However, this theory may not be possible due to unknown quantum effects. Furthermore, for humans to travel through time, we would need to develop a way to travel at faster speeds in space. At current speeds, it takes about 300 days to reach Mars, and 80000 to reach Proxima Centauri, our nearest star system. Finding these wormholes could require traveling far beyond Promixa Centauri.
As discussed after the film, Earth is eventually headed into the direction possible destruction. Although this will not occur for thousands of years, the emission of greenhouse gases, the abundant use of Earth’s natural resources, and overpopulation could contribute to such a world. If humanity can change our actions, and become more Eco-friendly, a change in the outcome of our future could arise.
In summary, the film Interstellar raises the possibility of the existence of wormholes and the capability and possible need for time travel.

http://www.nasa.gov/centers/glenn/technologiy/warp-scales_prt.htm

An Egg-Citing Egg-Speriment By William H. Vince F. and Nolan D.

In this lab, we studied the osmosis of water through the semi-permeable membrane of an egg. Osmosis occurs when a solvent moves from an area of high concentration to one of low concentration via a semi-permeable membrane. To illustrate the process of osmosis, we dissolved the eggshell via an acid-base reaction by combining calcium carbonate (a base) and acetic acid, and we then placed the denuded egg in water. The amount of water that passed through the semipermeable membrane was determined by weighing the egg at five minute intervals. We concluded that the egg membrane is semi-permeable, and we determined that the weight of the egg increased by approximately 1.6% of its original mass every five minutes.
In this experiment, we observed the process of osmosis through the semi-permeable membranes of a chicken egg. The semi-permeable membrane is located under the hard calcium carbonate shell, as illustrated in Figure 1.
Figure 1

The outer shell of a chicken egg is composed of 94-97 percent CaCO3. We first dissolved the outer shell by means of an acid-base reaction with vinegar, which contains acetic acid. When calcium carbonate mixes with acetic acid (CH3CO2H) the following chemical reaction occurs: Initially, the carbonate (CO3–) part of calcium carbonate is protonated by acetic acid to make carbonic acid (H2CO3),and the calcium and acetate form calcium acetate.
2 CH3COOH + CaCO3 = H2CO3 + Ca(CH3COO)2.
Next, the carbonic acid breaks down to form carbon dioxide and water.
H2CO3 = H2O + CO2.
The overall reaction can be written as the sum of two reactions:
2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.

After the egg shells are dissolved, the egg’s outermost layer is composed of two semi-permeable membranes. These membranes play a vital role in the development of a chick by allowing air and moisture to pass through to the developing embryo. The membranes are partially made of keratin, and they provide the embryo with a defense against bacterial invasion.

The de-shelled eggs were placed in water in order to simulate the behavior of a cell, which is also surrounded by a semi-permeable membrane, when placed into a hypotonic solution. We predicted that water would flow by osmosis through the semi-permeable membranes down its concentration gradient. Since water is a relatively hypotonic solution, which means that it has a lower concentration of solute when compared to the interior of the egg, water flows from an area of low solute concentration to an area of high solute concentration: into the egg. The absorption of water by the egg can be measured by weighing the egg and observing an increase in circumference.

In this lab, we used the following materials:
-3 eggs of relatively the same size
-white vinegar (apple cider or rice wine vinegar will work too)
-large beaker
-three large plastic cups
-triple beam balance scale
-plastic tray
-300ml of distilled water
-timer

Preparing the eggs:
1.I placed the eggs in the beaker gently without cracking the shell. I covered the eggs in the vinegar and let them sit overnight.
2. After the eggs were soaked, I removed them from the vinegar solution, and I gently rubbed off the white calcium acetate from the egg.

Starting the experiment:
1. I filled a cup with 100 ml of distilled water.
2. I marked each egg with a number (1-3), and I weighed egg 1with the triple beam balance, after placing the egg in the plastic tray.
3. I added egg 1 to the cup.
4. I let the egg sit in the water for five minutes, and I then weighed it again.
5. I repeated steps 1-4 using eggs 2 and 3.

Measuring the rate of osmosis:
1. I weighed egg 1 in the plastic tray after five minutes in the water bath, and I returned it to the bath as soon as possible.
2. After a second five minute interval, I re-weighed the egg 1 and returned it to the water for a third five minute period.
3. I removed egg 1 and weighed it a final time.
4. I repeated steps 1-3 for eggs 2 and 3.
5. I recorded my data in the table.
Note: When recording data, I accounted for the weight of the plastic tray.

Figure 2

Figure 2 illustrates the egg’s CaCO3 shell being dissolved by the surrounding vinegar. Carbon dioxide bubbles surround the egg, because CO2 is a product of this reaction as described previously.
Figure 3

Figure 3 depicts the egg after the shell has disintegrated. A layer of calcium acetate, another product of the reaction, is visible on the egg. After the shell dissolved, the egg became rubbery to the touch.
Figure 4

Figure 5

Figures 4 and 5 depict the eggs being weighed on the triple beam balance. I found that the weight of egg 1 increased from 71.5 g to 74 g after five minutes, it remained at 74 g after ten minutes, and it increased to 75 g after 15 minutes. The weight of egg 2 increased from 76.5 g to 78 g, 79g, and 80 g after five, ten and fifteen minute intervals respectively. Finally, the weight of egg 3 increased from 86.8 g to 88 g, 89.3 g, and 91g after sequential five minute immersions.
Figure 6

The graph in Figure 6 illustrates the increase in weight of each egg (expressed as percent weight gain) during the time they were immersed in distilled water.
Figure 7

The graph in Figure 7 depicts the increase in actual weight of the eggs during the time they were immersed in distilled water. It is apparent that the weight of the eggs increased with a longer duration of immersion in water.
Figure 8

Figure 8 depicts the actual weight gain and percent increase in the weight of each egg during the experiment. As discussed previously, the weight of each egg progressively increases during the experiment, as the eggs absorb water by osmosis through the semi-permeable membranes.
Figure 9

Figure 9 depicts an egg that ruptured after the experiment was complete. At the top of the photo, the gray semipermeable membrane lies next to the egg yolk. We noted that the thickness of the egg membrane is quite small in comparison to the size of an intact egg.

Discussion Questions:
*When the eggs were placed in the vinegar, they swelled because the reaction of vinegar and the eggshell (2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.), results in water, which moved into the egg by osmosis.
*In the first part of the experiment, water moved into the egg. We know that water has moved into the egg, because they gained weight during the immersion in distilled water.
*Chicken egg membranes are semi-permeable because water was able to move into the egg by osmosis but no egg albumin (which is a large molecule) was able to diffuse through the membrane and out of the egg.
*It is possible to calculate an egg’s volume by measuring the amount of water displaced by the egg. If a beaker is filled with water and an egg is submerged in the water, the amount of displaced water can be measured and is equal to the volume of the egg. A change in weight indicates that more water has moved into the egg, and this would likely be accompanied by a change in volume. This is because the egg expands as water passes into it.
*Another possible experiment is to place the de-shelled egg in solutions of different tonicity (isotonic, hypertonic, and hypotonic) in order to determine the effect of those solutions on the water content of the egg. I anticipate that the egg would shrink when placed into a hypertonic solution, and an isotonic solution would have no effect of the egg. It would also be interesting to study active transport of molecules across an egg membrane.
*Saline solution is used to expand blood volume, as it is isotonic. Hypotonic distilled water enters red blood cells via osmosis and causes them to burst open (see Figure 10).
Figure 10

In this experiment, we observed an acid base reaction, osmosis through a semipermeable membrane and the effect of a hypotonic solution on the weight of a chicken egg.
The first part of the experiment involved a chemical reaction between a base (CaCO3), which is found in egg shells, and an acid (acetic acid), which is found in vinegar.
The chemical reaction is:

2 CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2.

We observed carbon dioxide bubbles, the dissolution of the hard calcium carbonate shell, and the formation of calcium acetate. We also noted that the egg became rubbery after immersion in vinegar. This may have occurred because the acid started to denature the albumin found in egg whites. Denaturation of proteins can occur through exposure to acids, bases or high temperature.
The next part of our experiment was to observe the osmosis of water through the semipermeable membranes of the egg, which lie under the hard shell. These membranes are important in the development of the embryo because they are permeable to water and air, and they provide a barrier to dust and bacteria. After we soaked the egg in water (a hypotonic solution), the egg began to enlarge. We weighed the egg at five minute intervals, and we noted an average increase in weight of 2.2% over the first five minutes, then an average increase of 0.8% over the next five minutes, then an average increase of 1.2% during the final five minutes (see Figure 7). The increase may have slowed as the experiment progressed because the concentration gradient decreased over time.
Next, we observed the effect of the tonicity of the solution on the egg as a model for the reaction of a cell to its environment (or a bathing solution). The egg swelled because the hypotonic solution resulted in an osmotic gradient that caused water to flow from a low concentration of solute to a high concentration. In this case, the high solute concentration was provided by egg proteins.
If a cell is in an isotonic environment (the external osmolarity is equal to the internal osmolarity), the cell volume remains unchanged. If a cell is in a hypertonic environment (the external osmolarity is higher than the internal osmolarity), the cell will shrink. Finally, if a cell is in a hypotonic environment (the external osmolarity is less than the internal osmolarity), the cell will swell.
The relative osmolarity of a solution is determined by the number of solute particles, and not by type of particles. An increase in temperature will speed up the motion of molecules and will also increase the rate of diffusion or osmosis. Although diffusion occurs in both directions, the overall net movement will be down the concentration gradient.
If we were to repeat this experiment, I would use solutions with different osmolarity, and I would vary the temperature to see if the rate of osmosis would be affected. Finally, I would study the egg membrane’s ability to let air and water flow osmotically while preventing the movement of bacteria and large molecules. It is possible that eggs could be analyzed for an intact shell and cell membranes to decrease the possibility of eating eggs that are contaminated with salmonella and other disease causing organisms. If one could develop an inexpensive semi-permeable membrane, it might be useful in the desalination of water for use in developing countries.

https://www.exploratorium.edu/cooking/eggs/eggcomposition.html
World’s Poultry Science Journal/Volume 61/Issue 1/March 2005, pp 71-86
Van.physics.illinois.edu/qa/listing.php?id=461
imaginationstationtoledo.org/content/2011/04/how-to-make-a-naked-egg