Premium Essay

Biology of Organisms Tutorial 3

In: Science

Submitted By terrorsak
Words 989
Pages 4
Tutorial 3
Bio 207
1) Glomerular filtration rate (gfr) is a test used to check how well the kidneys are working. Specifically, it estimates how much blood passes through the tiny filters in the kidneys each minute. These tiny filters are called glomeruli.
Vasa recta is the hairpin shaped capillaries that serve the long loop of henle of juxtamedullary nephrons. Each ascending porion of the vasa recta lies next to the descending portion of a loop of henle and vice versa
Although the loop of henle and the vasa recta do not exchange materials directly, they function together as a counterpart of a countercurrent system that enhances nephron efficiency.
(source Campbell, Reece, et al., Biology 8th edition. 2009.
Renal clearance is a measurement that allows one to analyze the activity of the kidney. Renal clearance = urine concentration (M) x urine flow rate (ml/min)/plasma concentration (M)= UV/P RCGFR=secretion RC=GFR = nothing is secreted or absorbed.
(source: www.austince.edu/emeyerthclearancehtm.htm)
Single effect is produced by the use of metabolic energy between adjacent parts of the two oppositely flowing fluid streams in a counter current multiplier system (Source: Hill, Wyse and Anderson, Animal Physiology 2012)
2) Blood enters the kidneys through the bowman’s capsule where it enters a cluster of blood capillaries called the glomerulus. Fluid is driven through a filter in the glomerulus into the capsule’s lumen by the hydrostatic pressure of the blood. The fluid that accumulates in the lumen is called the capsular fluid. However large organic molecules of the size of 10,000 daltons are too large to enter the lumen and remain in the blood plasma. Molecules like glucose however enter capsular fluid and form the primary urine. Solutes and water flow down from the lumen, down the tubule of the nephron and much is reabsorbed here. Solutes like glucose and…...

Similar Documents

Premium Essay

Tutorial 3

...Arberesha Gashi                                                      10.02.12 Biology of organisms 207                                     lab tutorial 3 Kidney functions: 1) a) Glomerular filtration rate :   The glomerulus is known as a filtration center where majority of the blood vessels pump so they can take in nutrients and produce urine.  The glomerular filtration rate provides us with the rate of blood flow through the filters in the kidneys.  The rate is measured each minute, showing what has been filtered through. The GFR test informs people about their kidneys and if they are working well.  The test is conducted by taking a sample of blood, and evaluating how much creatinine is in the body.  The level of creatinine along with other factors measures the filtration rate. Healthy results should show that your body is filtering the creatinine well, because it is a waste that needs to be excreted. The GFR, about 125 mL/min (180 liters/day). b) Renal clearance:  relates to how much volume of the plasma that is cleared of a compound per unit of time.  An example would include, the kidney may filter and remove all of the urea in a 65ml of plasma in one minute. The kidneys help filter the blood and absorb nutrients in one flow, and release other parts as urea.  c) Vasa recta: This includes a thin blood vessels which aligns from the efferent arterioles, that leave each glomerulus in the kidney ( nephron). The vasa recta are is what forms the u shaped loops next to......

Words: 1231 - Pages: 5

Premium Essay

Biology of Organisms

...Tutorial #1 1) a) Density: Density is a physical property and is a measure of how heavy an object is in a given volume. It is also defined as mass per volume. It is expressed in g/ml or g/L. Example would be oil and water. Oil floats on water because it has less density. b) Solvent: Solvent is a liquid in which substances are dissolved to form a solution. Example is sugar and water. In this case, water is a solvent and sugar is a substance (solute). c) Amphipathic: A molecule that has a polar and non polar region. Polar being water soluble and have charges (+ve or –ve) whereas non polar being not water soluble and lack charges. A phospholipid molecule is an example of amphipathic. d) Hydrophobic: Generally hydrophobia means fear of water. It means that certain substances have the inability to dissolve in water. Again an example would be oil in water. 2) a) 3) a) The properties are called Colligative Properties. b) Four properties that depend upon effective particle concentration are freezing point, boiling point, water vapor pressure of a solution and osmotic pressure. 4) .85 concentration is closest to that of plasma (290 mOsm/Kg). 5) 6) 8) a) 1) When the red blood cell is placed in a fresh water, it would burst and swell because water molecules would enter into the cell. 2) When the red blood cell is placed in a salt solution at a concentration of 300mOsm, it would not change and no effect. 3) When the red blood cell is placed in a salt solution at......

Words: 350 - Pages: 2

Premium Essay

The Connection Between the Central Dogma of Molecular Biology/ Bioinformatics, Model Organism and Drug Designing.

...Report on the connection between the Central dogma of Molecular Biology/ Bioinformatics, Model Organism and Drug Designing. The basis of the central dogma of molecular biology is the expression of the genetic information in any call. It is a universal process that occurs in every cell. The genetic information is stored in the DNA. During gene expression DNA is transcript to RNA and these RNA are transcribed to proteins. Bioinformatics deals with the genetic information which involves collecting, analyzing, manipulating and predicting etc. For the functioning of bioinformatics it is essential to know the genetic information that is stored in DNA. Therefore sequencing of DNA, genes or genomes is the fundamental need in bioinformatics. Organisms that are used in biological experiments in laboratories are called ‘model organisms’, of which most genomes are sequenced at present (rat, yeast, Arabidopsis; plant model organism) These sequenced genomes could be analyzed using bioinformatics tools in order to identify genes of significance as in drought tolerance genes in plants etc. Information revealed from sequencing could be studied using bioinformatics tools to understand its underlying mechanisms and to generate models that could be used in further studies. This information could also be used in evolutionary studies, micro array analysis, identification of genetic disorders (Alzheimer’s disease, breast cancer, cystic fibrosis, spinal muscular atrophy......

Words: 414 - Pages: 2

Premium Essay

Bio Tutorial 3 Text

...Tutorial #3: Photosynthesis - The Basis of Life 1) Basics of photosynthesis: Review the anatomy of the leaf and the cellular locations and organelles involved in photosynthesis. Describe and explain the role of oxygen, carbon dioxide, and nitrogen in photosynthesis. Describe and explain the role of water in photosynthesis. Describe the two phases of the photosynthetic process. How do they differ in terms of their inputs, outputs, energy requirements, and spatial locations in the plant? Explain why Rubisco is possibly the most important protein on the planet. Explain how the eruption of a large volcano could lead to changes in the earth’s atmosphere, and consequently, life on earth. You are studying the carbon dioxide consumption at different levels of a forest. You have a portable leaf chamber and sensors attached to the data logger interface and can measure the photosynthetic rate of leaves on trees in a forest at different parts of the tree; high up, in the middle, and on lower branches. Below are the ‘raw’ data from your measurements. Assume all leaves completely cover the 9 cm2 chamber. Find the CO2 exchange (umole/m2/s) for each species at each level in the trees. If you present this to the class, prepare a clear explanation of how you arrived at your answers. Explain the differences between the values of CO2 exchange in the three levels. If you are presenting this to the class you should look for current studies about this topic and its......

Words: 484 - Pages: 2

Premium Essay

Tutorial 3

...Tutorial Question 3 James works at Challenger Ltd company. He and his colleagues feel that it is time for them to form a union at the company for currently, there is none. The management is aware of this and send a representative to discuss with James and his friends to persuade all workers to become members of the union in order to ensure the uniformity and ease in implementing policies and terms and conditions of employment. After the discussion, James proceed to apply for recognition for the union in order to commence collective bargaining with the management. Based on the Industrial Relations Act 1967, 1. Is the management’s action permissible? 2. In Malaysia, who has the right to accord recognition to the union? 3. Based on the case, would the union be granted recognition? Justify 4. Disscuss the conditions for union recognition. 1. The management’s action is permissible. Because they follow the procedure to form a union. For example, to form a union, the management allow James and his friends to persuade all worker s to become a member of the union. This means that, persuade can happen when the employer consent about the union. Another one is, the collective bargaining with employer after they had discus about the implementing policies and terms and the conditions of employment. 2. In Malaysian the person that has the right to accord recognition to the union is employer of the company. 3. Yes, because the management aware about that......

Words: 415 - Pages: 2

Free Essay

Aap Tutorial 3 Answer

...chapter 21       Test your understanding 3 ­ Hanford & Stopple Answer 1 (a) Hanford consolidated statement of financial position at 30  September 20X1 Non­current assets  Property, plant and equipment (W8) Goodwill (W3) $000 $000 109,510 6,850 –––––––  116,360 Current assets  Inventory (7,450 + 4,310) Accounts receivable (12,960 + 4,330 – 820  (W7)) Bank 11,760 16,470 520 –––––––  28,750 –––––––  145,110 –––––––  Equity and liabilities  Equity attributable to the equity holders of the  parent: Ordinary shares of $1 each (20,000 +10,000  (W6)) Reserves: Share premium (10,000 + 10,000 (W6)) Retained earnings (W5) 30,000 20,000 65,750 –––––––  Non­controlling interests (W4)  85,750 –––––––  115,750 6,950 –––––––  122,700 6,000 Non­current liabilities  8% Loan notes 20X4 Current liabilities  Trade accounts payable (5,920 + 4,160 – 620  (W7)) Bank overdraft Provision for taxation (3,070 + 2,180) 9,460 1,700 5,250 –––––––  Total equity and liabilities KAPLAN PUBLISHING 16,410 –––––––  145,110  403 Questions & Answers Workings (all figures in $000)  (W1) Group Structure            Hanford       | 75%     Stopple           6m/8m           Note: the unrealised profit on the sale of the plant was initially $400,000,  of this 10% i.e. $40,000 has been realised via Stopple's depreciation  charge, giving a net adjustment of $360,000 to both Hanford's profits  (W5) and the carrying value of the plant.     (W2) Net assets in subsidiary  At......

Words: 2081 - Pages: 9

Premium Essay

Biology Tutorial

...CENTRE FOR FOUNDATION STUDIES FOUNDATION IN SCIENCE MAY 2014 FHSB 1214 BIOLOGY I TUTORIAL 1 BASIC MOLECULES OF CELLS I Instructor’s Guide: 1. Standard answers are not encouraged. Discuss with the students and lead them to the correct answer. 2. Allow and push the students to express and explain the answers, at the same time, correct their mistakes or concepts. 3. If the tutorial questions are not yet covered in the lecture, it should be carried forward to the next tutorial. __________________________________________________________________________________ (Source: Final examination, Sept 2013) Q1. (a) Figure 1.1 shows the molecular structure of lactose (a disaccharide). [pic] Figure 1.1 i) Lactose can be broken down into its monomers by addition of water. Based on Figure 1.1, draw the molecular structure of the monomers. (2 marks) (ii) Name TWO (2) monomers referring above in part (i). (2 marks) (b) Figure 1.2 shows three different types of polysaccharides, B, C and D. Based on Figure 1.2, answer the following questions. [pic] Figure 1.2 i) Identify polysaccharides B and C. (2 marks) ii) Name the shape of polysaccharide D. (1 mark) iii) Compare and contrast polysaccharide C and D in terms of their monomer, shapes and functions. (3 marks) (Source: Final examination, April 2013) Q2. Sucrose is......

Words: 572 - Pages: 3

Premium Essay

Biology

...BIOLOGY I TUTORIAL 2 BASIC MOLECULES OF CELLS II (Source: Final examination, Dec 2012) Q1. (a) Amino acids are the building blocks of proteins. Figure 1.1 shows the dipeptide molecule. [pic] Figure 1.1 i) Briefly explain the formation of dipeptide from two amino acids. (3 marks) ii) Name the molecular unit that could determine the structure of a protein. (1 mark) (Source: Final examination, Apr 2012) Q2 (a) Figure 2.1 shows a representation of a phospholipid molecule and Figure 2.2 shows the detailed structure of a fat molecule. [pic] [pic] Figure 2.1 Figure 2.2 (i) Based on Figure 2.1, name the parts of the molecule labeled as A, B and C. (3 marks) (ii) Based on Figure 2.1 and 2.2, briefly describe the chemical structure of lipids that account for their insolubility in water. (1 mark) (iii) During a practical session, Ms. Chan demonstrated to the students that when water and oil were added to a test tube, the two liquids formed two separate layers. She then added a solution of phospholipids to the mixture of water and oil. With justification, predict where the phospholipids would have dissolved. (3 marks) (b) Figure 2.3 shows part of a DNA molecule. [pic] Figure 2.3 (i) Name the TWO components that make up the part of......

Words: 543 - Pages: 3

Premium Essay

Biology 160 Paper on Genetically Modified Organisms

...Biology 160 Term Paper Genetically Modified Organisms INTRODUCTION There is a lot of controversy surrounding genetically-modified organisms and whether they ultimately do more harm than good. Genetically-modified organisms are most commonly used to refer to “crop plants created for human or animal consumption using the latest molecular biology techniques” (Whitman). The goal of modifying these organisms is to enhance or introduce a desired trait, such as increased resistance to herbicides or improving the nutritional content of the particular organism. This is generally done in a lab using a “Gene Gun” or an Agrobacterium method as described in detail in the next section. However, as convenient and simple as this idea sounds, there are many concerns that will more than likely change the course of future modifications in organisms. First, I'll be describing the methods used to date to modify the genetic composition of an organism. Then, I'll be highlighting the benefits and the potential consequences of genetically-modifying organisms. SCIENCE BEHIND GMO'S Upon researching, I discovered an interesting fact about genetically modified organism's: it has apparently occurred in nature for millions of years through a species called Agrobacterium tumefaciens, which turns out is the cause of crown gall diseases in many ornamental and fruit plants (Keefer). Agrobacterium tumefaciens is a species of soil-dwelling bacteria that has the ability to infect plant cells with a piece......

Words: 2241 - Pages: 9

Free Essay

Tutorial 3 for Utar

...[pic] TUTORIAL 3 Task 1: The topic sentences below are followed by details. Identify the detail(s) which do not support the topic sentence. 1. Topic Sentence: Studying overseas offers students a chance to benefit from a refreshing insight into other cultures. Details: A. Students may get a culture shock at the beginning. B. Food overseas is often very different in taste and presentation. C. Fees can be very expensive. D. There are ample opportunities to make new friends from other ethnic groups and nationalities. E. Students get to know and be part of various festivals. F. Students learn to live and study independently. G. Students are not able to follow the lessons at times due to the difference in teaching methods incorporated. 2. Topic Sentence: The LRT system has proven to be a boon for commuters in the city. Details: A. It is an alternative to road transport. B. It is convenient and economical. C. The ticketing machines are sometimes a hassle but most people are now used to them. D. Commuters get to avoid problems with parking. E. The stations are located rather remote from residential and corporate areas. F. Commuters have to stand in long queues and wait a long time for the service. G. Facilities for the disabled are insufficient. 3. Topic Sentence: A part-time job not only gives students financial freedom but also benefits them...

Words: 680 - Pages: 3

Premium Essay

Biology Tutorial

...BIOL 0051 - Preliminary Biology TUTORIAL 2 1. Complete the table and determine the magnification for each of the drawings given below: Table 1: Calculation of magnification for drawings 1,2 and 3. Drawing Objective lens used Total magnification (mm) X40 Field of view Number of specimens across the field of view Size of each specimen Size of drawing 5 3 Magnification of specimen 10 (cm) (µm) 1 X4 2 X100 1 15 3 X10 10 25 2. The following micrograph highlights some of the structures found in a prokaryotic cell. www.dtc.pima.edu a. Explain how the presence of a capsule may give a prokaryotic cell a survival advantage over those cells that do not possess a capsule. b. Name two other structures that may be found in a prokaryotic cell and state their imortance. c. Compare the structure of the ribosome and the organization of the nuclear material in the prokaryotic cell with those found in a typical eukaryotic cell. 3. The diagram below represents the fluid mosaic model of the cell membrane: www.webanatomy.net a. Identify the structures labelled 1-10. b. Which of these structure(s) is/are responsible for cell-cell recognition? c. What is the purpose of the structure highlighted by the red circle? d. Explain what selectively permeable means and how it applies to the cell membrane. 4. With the aid of a diagram describe the structure of the plant cell chloroplast. 5. Explain the purpose of lysosomes in the animal eukaryotic cell. Which structure in the......

Words: 262 - Pages: 2

Premium Essay

Organismic Biology - Tutorial 4 (Photosynthesis: the Basis of Life)

...Tutorial #4 -- Photosynthesis: the basis of life ____________________________________________________________ ___________________ 1. Basics of photosynthesis: a. Review the anatomy of the leaf and the cellular locations and organelles involved in photosynthesis. A leaf consists of waxy cuticle layer with stoma surrounded by guard cells, xylem vascular bundle, and a mesophyll. The chloroplast is located within the mesophyll cells. The stroma and thylakoids are located inside the chloroplast. The thylakoids contain chlorophyll. The organelles that participate in photosynthesis are chloroplast, thylakoids and chlorophyll in photosystem I and II. b. Describe and explain the role of oxygen in photosynthesis. The role of oxygen in photosynthesis is that during the light reaction, the light that is absorbed by the chloroplast in photosystem II, excites an election, therefore making the electron travel through a series of electron carriers to produce ATP. So to replace the electron lost through this process, water molecules are split into oxygen and hydrogen. Oxygen is then released into the atmosphere. c. Describe and explain the role of carbon dioxide in photosynthesis. The role of carbon dioxide in photosynthesis is that during the Calvin cycle, CO2 that enters the cell is combined with RuBP. This molecule is extremely unstable and because of that it splits from a 6 carbon molecule to a 2 carbon molecule through series of reactions and with ATP and NADPH......

Words: 1925 - Pages: 8

Premium Essay

Biology

...Biology From Wikipedia, the free encyclopedia For other uses, see Biology (disambiguation). Biology deals with the study of the many varieties of living organisms. Clockwise from top left: Salmonella typhimurium, Phascolarctos cinereus, Athyrium filix-femina, Amanita muscaria, Agalychnis callidryas, and Brachypelma smithi Biology is a natural science concerned with the study of life and living organisms, including their structure, function, growth, origin, evolution, distribution, and taxonomy.[1] Biology is a vast subject containing many subdivisions, topics, and disciplines. Among the most important topics are five unifying principles that can be said to be the fundamental axioms of modern biology:[2] 1. Cells are the basic unit of life 2. New species and inherited traits are the product of evolution 3. Genes are the basic unit of heredity 4. An organism regulates its internal environment to maintain a stable and constant condition 5. Living organisms consume and transform energy. Subdisciplines of biology are recognized on the basis of the scale at which organisms are studied and the methods used to study them: biochemistry examines the rudimentary chemistry of life; molecular biology studies the complex interactions of systems of biological molecules; cellular biology examines the basic building block of all life, the cell; physiology examines the physical and chemical functions of the tissues, organs, and organ systems of an organism; and......

Words: 3394 - Pages: 14

Premium Essay

Additional Tutorial 3

...| | Course Title | : | Physics I | Year/ Trimester Session | :: | Year 1 / Trimester 12016/05 | | Lecturer | : | | Additional Questions 3: Kinematics 1. A balloon is 30.0 m above the ground and is rising vertically with a uniform speed when a coin is dropped from it. If the coin reaches the ground in 4.00 s, what is the speed of the balloon? Solution:- Initial velocity of coin = speed of balloon, v. by using the equation [Answer: 12.1 ms–1] 2. A car and train moves together along two parallel paths at 25.0 m s–1. The car then undergoes a uniform acceleration of -2.5 m s–2 because of a red light and comes to rest. It remains at rest for 45.0 s, then accelerates back to a speed of 25 m s–1 at a rate of +2.5 m s–2. How far behind the train is the car when it reaches the speed of 25 m s–1, assuming that the train’s speed has remained constant at 25 m s–1. Solution:- For the car to stop we used the equation v2=v02 + 2as and v = v0 + at m and s For the car to speed up again, m and time taken, s Total distance moved by car in that time = 125 m + 125 m = 250 m. Total distance travelled by the train = 25 × (10+45+10) = 1625 m Therefore the car is (1625 – 250) = 1375 m behind the train. [Answer: 1375 m] *3. A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts......

Words: 899 - Pages: 4

Premium Essay

Biology Lesson 3 Essay

...intermediate phenotypes. An example for directional selection can be found when it comes to the color of mice. Mice can range anywhere in color from white to dark brown, in directional selection, it appeared more mice that were brown were surviving so in turn, more brown mice were reproducing and white mice were becoming rare. Disruptive selection is when both the dark brown mice and the white mice are being favored over the lighter brown, causing the colors in between to be phased out. In the process of natural selection, organisms most fit for reproduction will survive. In the case of strains of bacteria, the bacteria will only be able to thrive and reproduce if the antibiotics are not able to kill it. Bacteria has a variability in that some strains are more unaffected by antibiotics than others. When certain bacteria are found to survive despite antibiotics, these bacteria will reproduce and pass on that trait because it makes then a more fit organism. The major evolutionary trends that allowed for aquatic vertebrates become terrestrial was the development of feet and legs rather than fins. When water was scares, aquatic animals would die out because the water was too shallow to survive, thus legs and feet were better adapted to get around in such environments. Aquatic plants, such as fern reproduce by having the sperm swim to another plant, plants became dependent on water in order to produce offspring. In order for this to be achieved, the eggs of the female had to......

Words: 385 - Pages: 2