Biology Lab Report

In: Science

Submitted By brittanybarnes
Words 2636
Pages 11

Experiment #5 - Archaea: Characterization of Halophiles

Student Name: Brittany Barnes

Student Number: 5408224

Lab Section: 35

Date Experiment Performed: January 13, 2014

Lab Partners:
Kayla Hutchinson
Fallon Fersaud
Ashley Maulucci


Halobacterium salinarum and Haloferax volcanii come from the domain called Archaea and are considered to be halophilic, because they thrive in extreme environments. In this lab experiment, several tasks took place. In order to be able to observe the colony morphology of both Archaea, they were grown on an agar surface that allowed them to form colonies of characteristic colour and appearance. In order to observe the physiology of both Archaea on the effect of salt concentration, pH, and temperature, they needed to be placed on agar plates and incubated for two weeks. Being incubated for two weeks, allowed the halophilic archaeal cultures to grow. The objective of this experiment was to determine the morphological and biochemical characteristics along with the growth requirements of the halophilic Archaeans; Halobacterium salinarum NRC-1 and Haloferax volcanii DS2.


The domain, Archaea, possesses prokaryotic cells and has a cell wall that contains no peptidoglycan. Archaea contain rRNA that is unique to the Archaea as indicated by the presence of molecular regions. Archaea usually live in extreme environments and include methanogens, extreme halophiles, and hyperthermophiles. One reason for this is that the ether-containing linkages in the Archaea membranes are more stable and are able to withstand higher temperatures and stronger acid concentrations. The other two domains of life are Bacteria and Eukarya. Unlike the Bacteria and the Eukarya domains, the Archaea domain has membranes composed of branched hydrocarbon chains that are attached to glycerol by ether…...

Similar Documents

Biology Lab Unit 2

...Name: Date: October 18, 2013 Instructor’s Name: Assignment: SCIE207 Phase 2 Lab Report Title: Animal and Plant Cell Structures Instructions: Your lab report will consist of the completed tables. Label each structure of the plant and animal cell with its description and function in the tables provided. When your lab report is complete, post it in Submitted Assignment files. 1. Animal Cell: Observe the diagram showing the components of an animal cell. Using the textbook and virtual library resources, fill in the following table: Animal Cell |Number |Cell Structure |Description and Function | |1 |Nuclear pore  |Microscopic channels that allow certain material in | | | |and out of the nucleus | |2 |Chromatin (DNA)  |Made up of DNA and protein, packages DNA into smaller| | | |volume to fit in the cell and serves as a mechanism | | | |to control expression and DNA replication | |3 |Nucleus  ......

Words: 606 - Pages: 3

Unit 5 Biology Lab

...Name: Date: November 10, 2013 Instructor’s Name: Assignment: SCIE207 Phase 5 Lab Report Title: Taxonomy Lab to Show Organism Relationships Instructions: You will need to fill out the data table and answer a set of questions. When your lab report is complete, post it in Submitted Assignment files. Part 1: Using the lab animation, fill in the following data tables to help you answer the questions that follow: Table 1: Samples 1–5 |Phylum/Division |Sample 1: Chrysophyta |Sample 2: Annelida |Sample 3: Arthropoda |Sample 4: Amphibia |Sample 5: Aves | |Common Feature | | | | | | |Nutrition |Autotrophic |Heterotrophic: Earthworms|Some are vegetarian, some|These are usually | | |How does the organism break down and absorb food? | |eat their way through |are carnivorous, and some|vegetarian as tadpoles | | | | |dirt, so they are ...

Words: 834 - Pages: 4

Biology 1020 Diffusion and Osmosis Lab Report

...Kristina Eskola BL 1020 L01 Diffusion and Osmosis Lab Report (Dialysis) Introduction: Dialysis Tubing is a membrane made of regenerated cellulose fibers formed into a flat tube. If two solutions containing dissolved substances of different molecular weights are separated by this membrane, some substances may readily pass through the pores of the membrane, but others may be excluded. We will be investigating the selective permeability of the tubing to reduce sugar, glucose, starch, and iodine potassium iodide. We will test this by placing a solution of glucose and starch into a dialysis tubing bag and then place this bag into a solution of iodine potassium iodide (I2KI). Prediction: The I2KI solution will turn blue when adding Benedict’s reagent. Hypothesis: The solution of water and I2KI will be the most permeable because they will mix and react with Benedict’s reagent and the heat so the cell membrane only allows certain molecules to enter and leave the cell Materials and Methods: In the experiment we will be using two tests. In the first test, we will be using I2KI to test for the presence of starch. When I2KI is added to an unknown solution, the solution will turn purple or black if starch is present. If there is no starch in the solution, it will remain pale yellow. In the second test we will be using Benedict’s test for reducing sugar. When Benedict’s reagent is added to an unknown solution and the solution is heated, it will turn green, orange or orange-red...

Words: 873 - Pages: 4

Lab Report

...Lab 6: How to Write a Lab Report (1) Atta, S., M. Ikbal, A. Kumar, and N. D. Pradeep Singh. 2012. Application of photoremovable protecting group for controlled release of plant growth regulators by sunlight. Journal of Photochemistry and Photobiology B: Biology 111:39-49. --This article shows that sunlight is the key requirement for plant growth. It explains how sunlight helps in releasing plant growth regulators in plants which enhances plant growth. This article is helpful for my lab report because we provided sunlight to our plants which makes their development faster. (2) Ikram-ul-Haq, Z. A., G. M. Taseer, M. U. D. Mukesh, and S. Ali. 2011. Effects of different fruit juices used as carbon source on cucumber seedling under in-virto cultures. African Journal of Biotechnology 10:7404-7408. --This article shows how fruit juices help plants to grow better because of the sugar that fruit juices contain. They used strawberry and apple juices, but they were not as good as orange and grape juices, regarding their effect on plant growth. This article is helpful for my lab report because we used orange juice to measure its effect on the growth of our plants. (3) Einset, J.W. 1978. Citrus tissue culture: stimulation of fruit explants cultures with orange juice. Plant Physiology 62:885-888. --This article shows the effect of orange juice on plant growth. It explains how using high concentrations of citric acid does not enhance the plant growth, but using...

Words: 299 - Pages: 2

Lab Report

...The use of 1 Microscopes Name: Khulud Abdulaziz Nazer ID: U00045236 September 23 2014 General biology lab Section: 13 T Instructor: Tasneem Obaid Introduction: Microscopes are instruments used to enlarge objects that are too small to be seen by one’s eyes. Microscopy is the science of the examination of small objects using microscopes. Technically we have two types of microscopes, the optical microscope which was first invented in late 1500s. It uses light and glass lenses (objectives) to magnify the image of an object up to 4X, 10X, 40X and a 100X. The second type is the electron microscope which was developed at early 1900s, it uses an electron beam in the place of light and electromagnets in the place of objectives to allow a much higher resolution up to two million times. In this experiment we will be using the optical microscope and not the electron microscope to examine specimens, as it is available in almost all the laboratories. In addition, the costs of the optical microscope is too low compared to that of the electron microscope. Moreover, the optical microscope can be stored in normal room temperature and pressure not requiring a vacuum as the electron one does. Materials: Compound light microscope and stereomicroscope. Prepared dry mount of the letter “e” . A plug. Method: First of all, we used a compound light microscope to assess the letter “e”. We followed the following procedure; A slide of dry mount of letter “e”......

Words: 935 - Pages: 4

Biology Lab Report 7

...Comparing The Resting Heart Rate and Recovery Time Of Males And Females After Physical Activity Devyn Jones 26 September 2014 Biology 140 Section 38 Dr. Maria Gainey Abstract In this study we tested to see whether or not males have a lower resting pulse rate and if males have a faster pulse recovery time after engaging in physical activity. The experiment groups are the males and females of the class. The experiment involves walking up and down a flight of stair s and recoding your pulse rate for a period of time. The average resting pulse for the males was 72 beats/min, and the average resting rate for females was 80. The average percent change was 35.7%, and the mean pulse recovery time is 4 minutes. The experiment only partially supported my hypothesis. Males have a lower resting heart rate than females, however females have a faster pulse recovery time than males. Introduction Cardiovascular fitness is frequently considered the most important aspect of physical fitness. In lieu of this statement, we tested which group, males or females, have a better cardiovascular fitness. I believe that males have a lower resting heart rate than females, and a faster recovery time after a period of exercise. In the experiment two groups will be tested, males, and females, they will undergo the step test, take a series of pulses, and record their results. If my hypothesis is valid then the results will show that males have a lower resting heart rate, and display a faster......

Words: 741 - Pages: 3

Lab Report

...Name: Gertrude Edwards Date: 12/1/2014 Instructor’s Name: Staci Lynn Assignment: SCI203 Phase 2 Lab Report TITLE: Speciation • Purpose o The purpose of this lab is to evaluate what would happen if a species within a population were suddenly split into 2 groups. • Introduction o If a population is divided indefinitely by a barrier members of the divided population will not have the opportunity to breed with each other, over years, the biotic and abiotic conditions on either side of the barrier will vary from one another. (M.U.S.E). • Hypothesis/Predicted Outcome o Based on what I’ve learned I expect that species will undergo changes if they were split into 2 different groups, some would adapt and some wouldn’t. • Methods o The methods I used in this lab came from M.U.S.E. The initial separation would consist of some species from the mainland reaching the isolated Island, then after that the isolated population would begin to diverge because of the genetic drift and natural selection, then after that overtime divergence may eventually become sufficient to cause reproduction isolation. (M.U.S.E). • Results/Outcome o As a result, Natural selection will cause different selective and adaptive pressures to occur between the two divided populations and they will evolve forever. Over time this will result in speciation which is the creation of two new species. (M.U.S.E). •......

Words: 309 - Pages: 2

Lab Report in the cytoplasm storing energy. The cells would also be packed together similarly due to the similarity of function (which is storing energy). By viewing the targeted cell in a microscope, it becomes then possible to discern the purpose of a cell in a larger complex multi-cellular organism. This can help in real life because then scientists can use the microscope and gather information on the cell's purpose by examining it's features and/or other characteristics. Most of the material in the slide was composed of spit and other food particles, whereas the actual human cells are very sparse since it is difficult to remove cells from their corresponding tissue. discussion .. 1. Why do we stain specimens?- -Stains, in general, are used to make certain parts of cells or cell types in tissues stand out against the background for purposes of illustration.  Iodine reacts specifically with starch, turning a deep purple colour therefore would be well suited for plant cells, due to plant cells using starch for food storage.  Methylene blue stains animal cells to make nuclei more visible and is useful for animal cells as they allow DNA and RNA to be viewed through a microscope. It also does not affect the functioning of the cell itself, being nontoxic and is not a permanent stain. methylene blue is necessary  because cells are......

Words: 2735 - Pages: 11

Biology Lab

...Introduction: The ability to isolate and quantify nucleic acids accurately and rapidly is a prerequisite for many of the methods used in biochemistry and molecular biology. The concentration of DNA or RNA in a sample, and its condition, are often estimated by running the sample on an agarose gel. Such concentration estimates are semiquantitative at best and are time-consuming. For a more accurate determination of the concentration of DNA or RNA in a sample, a UV spectrophotometer is commonly used. Spectrophotometry uses the fact that there is a relationship between the absorption of ultraviolet light by DNA/RNA and its concentration in a sample. The absorption maximum of DNA/RNA is approx 260nm. The purity of a solution of DNA can be determined using a comparison of the optical density values of the solution at various wavelengths. For pure DNA, the observed A260/A280 ratio will be near 1.8. Elevated ratios usually indicate the presence of RNA. The A260/A280 ratio is used to assess RNA purity. An A260/A280 ratio of 1.8-2.1 is indicative of highly purified RNA. The 260/280 ratio below 1.8 often signal the presence of a contaminating protein or phenol. Alternatively, protein or phenol contamination is indicated by 230/260 ratios greater than 0.5. Workflow Time 2 days before the lab session During lab session 1:30 pm Task Cell culture 2:00 pm RNA isolation 5:15 pm Spectrophotometric analysis of your sample Work done......

Words: 842 - Pages: 4

Biology Lab 4 Umuc

...Your Full Name: UMUC Biology 102/103 Lab 4: Enzymes INSTRUCTIONS: * On your own and without assistance, complete this Lab 4 Answer Sheet electronically and submit it via the Assignments Folder by the date listed in the Course Schedule (under Syllabus). * To conduct your laboratory exercises, use the Laboratory Manual located under Course Content. Read the introduction and the directions for each exercise/experiment carefully before completing the exercises/experiments and answering the questions. * Save your Lab 4 Answer Sheet in the following format: LastName_Lab4 (e.g., Smith_Lab4). * You should submit your document as a Word (.doc or .docx) or Rich Text Format (.rtf) file for best compatibility. Pre-Lab Questions 1. How could you test to see if an enzyme was completely saturated during an experiment? - Add more substrate and record the rate. If the rate of the reaction is constant, all the enzymes are saturated. 2. List three conditions that would alter the activity of an enzyme. Be specific with your explanation. * Temperature – Cold temperature will cause the enzyme to work slow, hot temperature will cause the enzyme to increase the movement making it less stable. * PH – Difference in range in the PH scale can alter the shape of the enzyme’s active site * Concentration Of Substrate – Less or more of enzymes to substrates ratio will affect the rate of collisions between the two affecting the number of reactions. ...

Words: 1006 - Pages: 5

Lab Report

...To do Lab Report for Single and Double Replacement reactions : This is a formal lab report. It must be typed or written by hand with blue or black ink. Make sure that you include the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used ( instructions were given to you already and they are in Moodle) 4. Pre-lab questions 5. Data Tables with the results obtained ( observations and predictions) 6. Post Lab: Part I : Single Replacement Reactions a. For every reaction that took place you must write the balanced chemical equation b. Which metal reacted the most? c. Rank your metals from more to least active Part II: Double replacement Reactions a. For every reaction where you observed precipitate, write the complete balanced molecular equation, the complete ionic equation and the net ionic equation; use the solubility rules to identify the precipitate and the states of matter of each substance participating in the reaction. b. Which cation produced the most number of precipitates? c. Write general rules of solubility that you observed. 7. Final Conclusion and error analysis To do Lab Report for Single and Double Replacement reactions : This is a formal lab report. It must be typed or written with blue or black ink. Make sure that you include the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used (......

Words: 374 - Pages: 2

General Biology Lab

...Keva Harris 25 February 2016 Biology Lab 27-13 Survey of the Kingdom Fungi Question 1: a. Are hyphae apparent? Yes b. Are the cells motile? Yes Question 2: a. How many species of mold are on the bread? Five b. Is pigment distribution uniformly in each mycelium? If not, where is the pigment concentrated in each mold? No, because the species is not all one color. Concentrated in the sporangium c. What is the adaptive significance of spores forming on ends of apright filaments rather than closer to the protective substrate? It can spread and disperse easier. Question 3: a. Is what structure is the dark pigment of Rhizopus concentrated? zygosporangium b. Is Rhizopus reproducing sexually as well as asexually in the same petri dish? How can you tell? Yes. There were stalks of Rhizopus (asexual) as well as the thick fuzz (sexual). Question 4: What is the relative size of Penicillium hyphae compared with Rhizopus hyphae? The Penicillium hyphae are much smaller. Question 5: a. Do you see chains of yeast cells produced by budding? Yes b. How is the structure of yeast hyphae different from that of molds? Yeast hyphae are short, round, and unicellular. Molds are long and multicellular. Question 6: What is the difference between dikaryotic and diploid cells? In a dikaryotic cell, there are 2 separate nuclei. In a diploid cell, there is one nucleus with 2 sets of chromosomes. Question 7: How many spores would you estimate......

Words: 1057 - Pages: 5

Biology Lab

...Melina Franco Lab 1 - Biology 1408 Metric Practice Kilo - Hecto - Deka - meter, liter, or gram - Deci - Centi - Milli --- Micro --- Nano Length Conversions: 1,000mm = 1m 10 mm = 1cm 1,000m = 1km 100cm = 1m 1,000,000 µm = 1 m 1,000,000,000 nm = 1 m Volume Conversions: 1,000ml = 1 Liter Mass Conversions: 1,000mg = 1 gram Convert the following: To convert from mm to cm you have to move left on the chart 1 unit or 1 decimal place. Therefore, 20.0 mm converts to 2 cm by moving 1 decimal place to the left. To convert from mm to meter you have to move left on the chart 3 unit or 3 decimal place. Therefore, 20.0 mm converts to 0.02 meters by moving 3 decimal places to the left. To convert from cm to mm you have to move right on the chart 1 unit or 1 decimal place. Therefore, 76.0 cm converts to 760 mm by moving 1 decimal place to the right. To convert from cm to meters you have to move left on the chart 2 unit or 2 decimal place. Therefore, 76.0 cm converts to 0.76 meters by moving 2 decimal places to the left. To convert from meters to cm you have to move right on the chart 2 unit or 2 decimal place. Therefore, 15.0 m converts to 1,500 cm by moving 2 decimal places to the right. To convert from meters to mm you have to move right on the chart 3 unit or 3 decimal place. Therefore, 15.0 m converts to 15,000 mm by moving 3 decimal places to the right. To convert from cm to mm you have to move right on the chart 1 unit or 1 decimal......

Words: 1244 - Pages: 5

Lab Report

...Lab Report Introduction: Natural selection relies upon the assumption that all organisms produce more offspring than can survive in an environment demonstration limited resources. The heritable traits that are best suited for a particular environment should be passed on at a greater frequency then those less adaptive traits. Natural selection is unique in that it leads to adaptive change, and these changes are likely to lead to increased survival and reproductive success. To sum up, natural selection is that organisms can survive only because they can suit to the changed environment which means those organisms also need to change with the environment all the time. Hypothesis: during the experiment, the forks out of all the “predators” should survive, and the peas out of all those “preys” will be survived. Material and Metnods: We will use 4 kind of beans (preys items) which are fava beans, white beans, red beans and peas. And we also have 4 mouthparts (predators) which are knives, forks, spoons and foreceps. First, we get 100 ml for each prey and spread them on the ground. And the students will be divided into 4 groups with those 4 different predators. Each group of predators include 6 student. And all of the predators will have a cup which will serve as the stomach into which the prey will be placed. We also have a timer and a security that will make sure the process is fair. Here we begin the game. The predators will be gathered, they will be instructed not......

Words: 806 - Pages: 4

Biology Lab Manual

...BIOLOGY 10 Introduction to Biology Laboratory Manual Prepared by: KLLabrador Table of Contents |Exercise |Title |Page No. | |1 |Observation and Description |2 | |2 |Formulation, Testing of Hypothesis, |6 | | |and Experimental Design | | |3 |The Use of Models and Controls |9 | |4 |Plant and Animal Tissues |14 | |5 |Cellular Respiration |22 | |6 |Photosynthesis |27 | |7 |Phylogeny and Systematics: Survey of Plant and Animal Families |30 | | ...

Words: 6127 - Pages: 25