BIO 1414 LAB Practical ^N2 PDF

Title BIO 1414 LAB Practical ^N2
Author Dani G
Course Biosciences II
Institution The University of Texas at San Antonio
Pages 7
File Size 412.4 KB
File Type PDF
Total Downloads 60
Total Views 160

Summary

STUDY GUIDE for bio 1414 lab Exam 2...


Description

BIO 1414 LAB PRACTICAL #2 REVIEW

LAB 5: PLANT PHYSIOLOGY -

-

-

-

Monocots (monocotyledons) o Have stomata on both sides of the leaves o Stomata are organized in rows o Have parallel leaf veins o One cotyledon in the embryo o Floral parts usually in multiples of three Dicots (dicotyledons) o Most stomata are on the lower epidermis o Stomata are in a random arrangement o Have branched veins MONOCOT DICOT o Two cotyledons in the embryo o Floral parts usually in multiples of four or five The Physical Process of Transpiration o Transpiration- the process of water movement through a plant and its evaporation from aerial parts o Uses capillary action- the ability of a liquid to flow in narrow spaces without the assistance of an external force, even against gravity o Since the diameter of the xylem tubes are small, the combination of surface tension of the water, cohesion forces between water molecules, and adhesive forces between the water and the xylem wall act to move the water upward to the leaves of the plant. o Also called the transpiration cohesion-tension mechanism 1. Water vapor diffuses out of the stomata 2. Water evaporated from mesophyll cell walls 3. Tension pulls water from the veins into apoplast surrounding the mesophyll cells a. The loss of water lowers the water potential of the cell 4. Which pulls water in the veins of the leaves upward and outward 5. Which pulls the water column in the xylem of the shoot and root upward 6. Cohesion between the water molecules form a continuous water column from the roots to the leaves 7. Water enters the roots of the plant by osmosis and is transported up in the xylem until it reaches the leaves Measuring transpiration rates

-

-

-

o Transpiration rates can be measured by: Final meniscus reading – Initial meniscus reading o Depending on the length of the experiment, conversions should be done ▪ Ex: initial meniscus reading is 0.12mL and the final meniscus reading after 15 minutes is 0.16mL. • Calculation: (0.16mL – 0.12mL) x 4= 0.16mL/hr • you multiple by 4 because 15 minutes x 4= 60 minutes Factors that affect the transcription rate o Temperature: Plants transpire more rapidly at higher temperatures because water evaporates more rapidly as the temperature rises o Humidity: The rate of transpiration is roughly inversely proportional to atmospheric humidity. As the outward diffusion of water vapors through stomata is in accordance with the law of simple diffusion, the rate of transpiration is greatly reduced when the atmosphere is very humid. As the air becomes dry, the rate of transpiration also increases o Direct sunlight: The increasing light intensity raises the temperature of leaf cells and increases the rate at which liquid water is transformed into vapors, thus increasing the transpiration rate o Wind speed: fast moving air currents continually bring fresh, dry masses of air in contact with leaf surfaces and thus maintain a high rate of transpiration Stomata density, calculations, and CO2 level o Stomata density is the number of stomata per unit area of the leaf o (# of stomata at high power)/ (Area at 400x) ▪ Ex: You take a leaf sample from two plants to calculate the stomata density. Using a 40x objective with a field diameter of 0.4 mm, you counted 4 stomata for sample 1 and 36 stomata for sample 2. Calculate the stomata density for the two samples. ▪ Steps to calculate stomata density: • First calculate the diameter; = (low power magnification/ high power magnification) (low power field diameter mm) o Diameter is given 0.4 mm • Calculate radius, = diameter/2 o 0.4/2 = 0.2 mm • Calculate area; =(pi)r^2 o A = pi (0.2)^2 = 0.126mm^2 • Use equation (# of stomata at high power)/ (Area at 400x) o (4 stomata)/ (0.126mm^2)= 32 stomata/ mm^2 • Perform the same calculation for sample 2 o Higher amounts of CO2 generally correlate to lower stomata density ▪ In deserts, plants tend to have fewer stomata to conserve water and in tropics, they tend to have more stomata for gas exchange Arabidopsis thaliana: member of the mustard (Brassicaceae) family o It has a small, diploid genome that has been completely sequenced since 2000.

-

o Extensive genetic and physical maps of all 5 chromosomes. o Its rapid life cycle - approximately 6 weeks from germination to mature seed o Prolific seed production and easy cultivation o Large number of mutant lines available for genetic research o Self-pollination (hermaphroditic) A, B, and C mutants o “A” genes are switched on in the sepal and petal whorls o “B” genes are switched on in the petal and stamen whorls o “C” genes are switched on in the stamen and carpel whorls o Sepals develop when only the “A” gene is active o Petals develop when both “A” and “B” genes are active o Stamens develop when both “B” and “C” genes are active o The carpel develops when only the “C” gene is active. o Male floral organs: Sepal, petal, stamen o Female floral organs: carpel

LAB 6: ECOLOGY -

-

Estimate biodiversity o Quadrat sampling- sampling method that involves counting organisms within a defined area of uniform size o Simpsons diversity index- measure of biodiversity of a habitat that considers both richness and evenness ∑𝑛(𝑛 − 1) ▪ 𝐷 = 1 − (  ) 𝑁(𝑁 − 1) ▪ n= total # of organisms of a particular species ▪ N= total # of organisms of all species ▪ Diversity index ranges between 0 and 1 • 1 represents an infinite diversity • 0 represents no diversity ▪ Richness: total # of different species present in a community ▪ Evenness: the measure of relative abundance of the different species making up the community Mark and recapture method o Only for mobile organisms 𝑀⋅𝑆 o Lincoln-Peterson Index: 𝑁= 𝑅

-

▪ N= population size estimate ▪ M= marked individuals released ▪ S= size of second sample ▪ R= marked animals recaptured Carrying capacity o The max population size that a habitat can hold (usually defined as the letter K) o In optimal conditions, populations tend to grow in a predictable pattern. Early growth is slow (“the lag phase”), then growth is extremely rapid (“the log phase”) o Factors that limit the growth of populations: weather, natural disaster, food, water, supplies, disease, predation, or migration o Two models used to describe population growth under different conditions





-

Exponential • Predicts the unlimited growth of a population because of no limitation on resources • J shaped curve Logistical • Predicts limited growth of a population due to limited resources • S shaped curve

Soil texture o Soils are distinguished by differences in their physical and chemical properties o There are three major components of soil texture ▪ sand: has a largest sized particle and feels gritty ▪ silt: has more moderate sized particles and has a smooth texture ▪ clay: has the smallest sized particles and feels sticky, controls the water holding capacity of soil and the exchange of ions between soil particles and soil solution o To calculate the percentage of the components, use the formula (component deposit volume)/ (total deposit volume) ▪ For example, the deposit volume for sand is 15mm, for silt 3 mm, and for clay 12 mm. Add the volumes together for the total deposit volume (which is 30mm). Now use the formula and plug in your values. For sand, 15 mm/

30 mm = 50%. For silt. 3 mm/ 30 mm = 10%. For clay, 12 mm/ 30 mm = 40 %. o We use the soil texture pyramid chart to determine what soil we have ▪ the triangle has three sides and each side represents the percentage of a component (sand, silt, and clay) ▪ determine the percent sand of your sample and find that number on the bottom of the triangle, draw a line along the diagonal line ▪ draw another diagonal line to correspond to the percentage of silt ▪ find the point where the two lines intersect and follow the line that corresponds to the percentage of clay, make sure that the percent clay on the chart is the same as a percentage calculated from the soil sample ▪ determine the soil type you have based on wherever your lines intersect -

Population distribution patterns o Uniform (or even): one in which the members of the population are spaced at relatively equal distances from one another ▪ often occurs in species that defend a defined territory ▪ may result from some form of negative interaction among individuals such as competition o Random: each individual's position is independent of the locations of other individuals o Clumped - the most common with most organisms in the population preferring to aggregate in the same areas

LAB 7: HABITAT SELECTION -

Growth and survival of an organism depends on physical conditions o Organisms who live in a particular habitat must be able to tolerate the conditions (both abiotic and biotic factors) of the habitat and find the resources that it needs to survive and reproduce ▪ abiotic factors: nonliving chemical and physical parts of the environment affecting living organisms in the functioning of ecosystems • temperature

-

-

• light • water • humidity • salinity • pH ▪ biotic factors: living chemical in physical parts of the environment that affect living organisms in the functioning of ecosystems • competition • predation • disease o Plants need to have access to nutrients, pollinators, dispersers, and protection to survive ▪ If a plant cannot tolerate its environment, then it would have evolved mechanisms that improve its ability to adapt to changing local environments o Animals need to have access to food, a mate, and shelter from adverse abiotic and biotic factors ▪ If an animal cannot tolerate its environments physical conditions then it would have to leave this habitat and search for a new, appropriate habitat to adapt and reproduce Habitats are determined by natural selection o The choice is not a conscious decision but determined by natural selection, how well the organism can change, adapt to the new environment quickly, and survive and reproduce Artemia: Brine shrimp or “Sea monkey” o Belongs to the phylum Arthropoda and class crustacea o Usually live in salt lakes or brine ponds worldwide, also found inland, as in the great Salt Lake in Utah o A female may live up to three months and produce as many as 300 live nauplii every four days, the young nauplii molts about 15 times before reaching adult size of 1 cm in length in 2-4 weeks o Males are identified by a large pair of claspers that used to hold onto female during mating o Females have a brute pooch, but males don’t o They feed on algae suspended in water o As the food supply decreases, salinity increases, dissolved oxygen decreases, temperature decreases, or a combination of these conditions occurs, the females will start producing cysts which are tough, coated eggs that may remain viable for many years if kept dry

-

-

o Brine shrimps prefer to live in a basic and brighter environment Swimming behavior of Artemia o Brine shrimps become solitary while swimming and don’t bump into each other o Female shrimps tend to have a faster swimming speed (may very) o Outside factors (abiotic and biotic) may affect their swimming behaviors Threatened habitats o Most ecological communities are in the state of changed to disturbances such as: ▪ Fire ▪ Earthquakes ▪ Volcanic eruption ▪ Weather ▪ Human activity o Recently, the loss of habitat due to human activity is a primary threat to the survival of plants and animals o There are three major kinds of habitat loss: ▪ Habitat destruction: occurs when natural habitats are no longer able to support the species present, resulting in the displacement or destruction of its biodiversity ▪ Habitat fragmentation: the spatial separation of habitats from a previous state of greater continuity. The "cutting up" of habitats into fragments is mainly caused by agricultural land conversion, urbanization, dams, water diversions, pollution, invasive species, and deforestation • Can also be caused by geological processes that slowly alter the layout of the physical environment ▪ Habitat degradation: the main cause of habitat degradation is pollution, invasive species, agricultural development, diminished resources (such as water and food), mining, and destructive fishing practices o The main drivers of habitat loss: ▪ Agriculture: oldest type of habitat last contributor many of the habitat loss from agriculture was done a long time ago when the settlers converted forest to cropland for plantation ▪ Land conversion for development: Many buildings such as housing developments, roads, parks, strip malls, parking lots are constructed ▪ Water development: dams and other waste diversions siphon off water in change water chemistry. ▪ Pollution: the pollutants are disturbing the biodiversity of organisms ▪ Global warming: emerging driver of habitat loss. Wildlife that need the cool temperatures of high elevations may soon run out of habitat. Coastal wildlife may find their habitat underwater at sea level rises...


Similar Free PDFs