Big Idea 2 - Review

8/10/2019 Big Idea 2 - Review

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Big Idea 2

Biological systems utilize free energy and molecular building blocks to grow, to reproduceand to maintain dynamic homeostasis.

Living systems require both free energy and matter to maintain order, grow and reproduce.

Organisms employ various strategies to capture, use and store free energy and other vitalresources. Energy deficiencies are not only detrimental to individual organisms; they also cancause disruptions at the population and ecosystem levels.

Biological systems must both capture free energy and then transform the energy into usableforms. Autotrophic cells capture free energy through photosynthesis and chemosynthesis.Photosynthesis traps free energy present in sunlight that, in turn, is used to producecarbohydrates from carbon dioxide. Chemosynthesis captures energy present in inorganicchemicals. Cellular respiration and fermentation harvest free energy from sugars to produce freeenergy carriers, including ATP. The free energy available in sugars drives metabolic pathways incells. Photosynthesis and respiration are interdependent processes.

Cells and organisms exchange matter with the environment. For example, water and nutrients areused in the synthesis of new molecules; carbon moves from the environment to organisms whereit is incorporated into carbohydrates, proteins, nucleic acids or fats; and oxygen is necessary formore efficient free energy use in cellular respiration. These processes release matter to theenvironment as waste products. For example, cellular respiration will release carbon dioxide. Inaddition, programmed cell death (apoptosis) plays a role in normal development anddifferentiation (e.g., morphogenesis). Differences in surface-to-volume ratios affect the capacityof a biological system to obtain resources and eliminate wastes.

Membranes allow cells to create and maintain internal environments that differ from external

environments. The structure of cell membranes results in selective permeability preventingmolecules with certain characteristics from passing through the membrane while allowing othersto pass. Processes that maintain dynamic homeostasis by allowing the movement of moleculesacross membranes include osmosis, diffusion and active transport. In eukaryotes, internalmembranes partition the cell into specialized regions. Each region provides a localization ofchemical reactions allowing the cell processes to operate with optimal efficiency.

Feedback mechanisms maintain dynamic homeostasis within an organism by regulatingresponses to changes in both internal and external environments. Negative feedback loopsmaintain optimal internal environments while positive feedback mechanisms amplify responses.Changes in a biological systems environment, particularly the availability of resources,

influence an organisms responses and activities. Organisms use various means to obtainnutrients and remove wastes. Homeostatic mechanisms across phyla reflect both continuity dueto common ancestry and change due to evolution and natural selection. Examples ofhomeostatic mechanisms that have evolved in plants and animals include defense mechanisms aswell as the timing and coordination of developmental, physiological and behavioral eventregulation. These mechanisms increase the fitness of individuals and long-term survival of

populations.


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Units in AP Biology Course

Be sure to know the following:

01A Viruses and Cells How organelles use and produce ATP. How organelles maintain homeostasis within cells.

01B Homeostasis in Cells All the notes in the unit. How to use the water potential formula to calculate water potential values.

01C Replication of Viruses and Cells The cell cycle and regulation of the cell cycle via check points (cyclins, CDKs, MPFs,

and PDGFs) Three phases of interphase and alternation of interphase with mitosis The sequence of events in mitosis (replication, alignment, separation)

How the reduction division of meiosis followed by fertilization ensures genetic diversity How viral replication differs from other reproductive strategies and generates geneticvariation via various mechanisms.

Mechanism of viral transduction creating a lytic or lysogenic (latent) infection andmechanisms for creating increased pathogenicity

01D Chemical Components of Viruses and Cells How molecules and atoms from the environment are cycled (water, carbon, nitrogen, and

phosphorus cycles) through organisms and a used as building blocks within organisms forthe synthesis of carbohydrates, lipids, proteins and nucleic acids.

How living organisms depend on the properties of water resulting from its polarity andhydrogen bonding (Cohesion, Adhesion, High specific heat capacity, Universal solventsupports reactions, Heat of vaporization, Heat of fusion, Waters thermal conductivity)

Negative feedback mechanisms maintain dynamic homeostasis for a particular condition(variable) by regulating physiological processes, returning the changing condition back toits target set point.

Structure and function of polymers are derived from the way their monomers areassembled. (amino acid primary, secondary, tertiary and quaternary structure;amphipathic nature of phospholipids, lipids non-polar; comparison of cellulose vs. starchstructure.

Directionality determines function (nucleic acid 5, 3 carbons in sugar backbone;Proteins with amino (NH 2) end and carboxyl (COOH) end; carbohydrate subunit bondingdetermines secondary structure of carbohydrate

Change in the structure of a molecular system may result in a change of the function ofthe system (enzyme active site substrate interaction and interaction with cofactors orcoenzymes)

Methods to measure enzyme activity(rate) Variations within molecular classes provide wider range of functions (phospholipids

variability, hemoglobin varieties, MHC proteins, variety of chlorophyll)


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01F Cellular Communication and Interaction How positive feedback mechanisms amplify responses in biological organisms (e.g.,

immune system) and how disruption negatively affects the organisms (e.g., HIV) How plants, invertebrates and vertebrates have multiple, nonspecific immune

responses. How mammals use both the cell mediated and humoral specific immune responses

triggered by natural or artificial agents that disrupt dynamic homeostasis How cells communicate with each other through direct contact with other cells or

from a distance via chemical signaling. How organisms exchange information with each other in response to internal changes

and external cues, which can change behavior and this behavior(learned or innate)leads to differential reproductive success via natural selection

Variation in molecular units (e.g., antibodies) provides cells with a wider range offunctions.

01G Cellular Communication in Multicellular Organisms How energy is involved in contraction of muscle cells.

How energy is necessary in the maintenance of membrane potential in nerve cells andmuscle cells (sodium potassium pump).

02A Laws of Energetics All living systems require constant input of free energy to maintain a highly ordered

system offsetting disorder and entropy. Without high order which death occurs The second law of thermodynamics(entropy increases over time) is not violated because

order is maintained by coupling cellular processes that increase entropy (negativechanges in free energy) with those that decrease entropy (positive changes in freeenergy).

Energy input must exceed free energy lost to entropy to maintain order and power

cellular processes. Energetically favorable exergonic reactions, such as ATPADP, that have a negativechange in free energy can be used to maintain or increase order in system by beingcoupled with reactions that have a positive free energy change

Living systems have a variety of processes to capture free energy using different types ofelectron acceptors (e.g., NADP +, Oxygen).

Know how the electron transport chain captures free energy from electrons in a series ofcoupled reactions that establish an electrochemical gradient across membranes.

Know that free energy becomes available for metabolism by the conversion ofATPADP, which is coupled to many steps in metabolic pathways.

02B Respiration Organisms share metabolic pathways and features and demonstrate evolutionary

relatedness. The energy-related pathways in biological systems (Krebs cycle, Glycolysis, Calvin

cycle, Fermentation) are sequential and may be entered at multiple points in the pathway. Heterotrophs capture free energy present in carbon compounds produced by other

organisms.


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04A Origin of Different Species Know that energy efficiency is a key selecting factor for selection of characteristics and

the ultimate survival of an individual.

04B Plant, Animal and Fungal Evolution Know that organisms with adaptations are specialized for efficiency. Know that ATP energy is coupled with reactions that are necessary for specialized

adaptations and process. Know that maintaining order and structure associated with specialized adaptations goes

against the trend for increasing entropy in the universe, and that energy is expended tomaintain order.


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AP Lab 4 Osmosis and Diffusion AP Lab Manual p S51

Diffusion is the process through which materials move through membranes and throughout thecytoplasm without the expenditure of energy. The plasma membrane and organelle membranesare selectively permeable, limiting the movement of materials through them. Phospholipids fatty

acids of the plasma membrane with their hydrophobic characteristics limit the movement of polar substances such as water. Aquaporins (specialized protein channels for the movement ofwater) increase the rate of osmosis, the diffusion of water. Other channel proteins facilitate themovement of other materials such as ions while transport proteins facilitate the movement oflarger molecules (e.g., carbohydrates)

In diffusion, solutes move from an area of high concentration to an area of low concentration.Water also moves down its concentration gradient. Water moves from areas of high potential(high free water concentration) and low solute concentration to areas of low potential (low freewater concentration) and high solute concentration. Solutes decrease the concentration of free

water because water molecules surround the solute molecules.

The following terms are used to describe solutions separated by selectively permeable membranes:

Hypertonic higher solute concentration (and lower water potential) compared to theother solution. Water will move into this solution.Hypotonic lower solute concentration (and high water potential) compared to the othersolution. Water will move to the other solution.Isotonic identical solute concentration (and water potential) compared to the othersolution

In cells that have a cell wall (e.g., fungal and plant cells), water movement is affected by soluteconcentration and turgor pressure . Turgor pressure is a resistance to water movement that

builds up as water moves into the cell and pushes the cell membrane against the cell wall. In thiscase the cell wall prevents the cell from bursting.

Water Potential () can be calculated to predict the movement of water into and through planttissues such as roots, shoots and leaves . Water potential is the free energy per mole of water.Water moves from an area of higher water potential (higher free energy) to an area of lowerwater potential (lower free energy). Water potential can be used as a measure of the tendency ofwater to diffuse from one compartment to another.

There are two major components in the calculation of water potential:

solute potential (S) or osmotic potential - dependent on solute concentration.

pressure potential (P) - results from the exertion of pressure either positive ornegative (tension) on a solution.


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The water potential of pure water in an open beaker is zero ( = 0) because both thesolute and pressure potentials are zero (S = 0; P = 0).

An increase in positive pressure raises the pressure potential and the water potential.

The addition of solute to the water lowers the solute potential and decreases the water potential. This means that a solution at atmospheric pressure has a negative water potential due to the solute potential.

Comparing solutions with similar concentration but with different ionization constants.

1) What is the water potential of the following solutions?

a. A 0.15 M solution of sucrose at atmospheric pressure in an open container (P =0) and 25C . Sucrose does not ionize (i = 1)

= P + S

= 0+ S

(S) = iCRT

= - 1(0.15mole/L)( 0.0831 liter bars/mole-K)(298K)

= - 3.7 bars

(1 bar is a metric measure of pressure equal to 1 atmosphere at sea level)

The Formula for Water Potential

= P + S

Water Potential = Pressure Potential + Solute Potential

The solute potential (S) = iCRT,

where i is the ionization constant,

C is the molar concentration,

R is the pressure constant (R = 0.0831 liter bars/mole-K),

T is the temperature in K (273 + C).


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2) A 0.15 M NaCl solution at atmospheric pressure in an open container (P = 0) and 25C . NaCl contains 2 ions, Na+ and Cl-; therefore i = 2

= P + S

= 0+ S

= 0 + ( iCRT)

= 0 + (- 2(0.15mole/L)( 0.0831 liter bars/mole-K)(298K))

= - 7.4 bars

When Cells have Walls

When a cells cytoplasm is separated from pure water by a selectively permeable membrane,water moves from the surrounding area, where the water potential is higher ( = 0), into the cell,where water potential is lower because of solutes in the cytoplasm ( is negative).

The movement of water into the cell causes the cell to swell, and the cell membrane pushesagainst the cell wall. Turgor Pressure increases in the cell counteracting the diffusion of waterinto the cell. Eventually, because of increased turgor pressure, the water potential of the cellequals the water potential of the pure water outside the cell ( of cell = of pure water = 0). Atthis point, a dynamic equilibrium is reached and net water movement ceases

When Solute is Added Outside the Cell

When solute is added to the water surrounding the plant cell, the water potential of this solutiondecreases. If enough solute is added, the water potential outside the cell will equal the water

potential inside the cell, and there will be no net movement of water. This does not mean that thesolute concentrations inside and outside the cell are not equal. Remember, the water potentialinside the cell results from the combination of both the turgor pressure (P) and the solute

pressure (S).

If additional solute is added to the water surrounding the cell the water potential of this solutionwill continue to decrease and water will leave the cell (toward the lower water potential). The

water loss causes the cell to lose turgor and may cause the cell membrane to shrink away fromthe cell wall. This is known as plasmolysis.

This Set of Laboratory Investigations consists of three parts

1) Using artificial cells to study the relationship of surface area and volume2) Create models of living cells to explore osmosis and diffusion3) Observe osmosis in living cells.


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Terms

Kinetic energy - the energy of motion of a body Potential energy - the energy of a body or system as a result of its position Osmosis - diffusion of water through a semipermeable membrane from a solution with a

low solute concentration to a solution with a higher solute concentration

Gradients - a series of progressively increasing or decreasing differences Diffusion - intermingling of the particles of two or more substances as a result of randomthermal motion

Convolutions - a form or part that is folded or coiled.

Procedure 1: Surface Area and Cell Size

In this procedure you measure the surface area of drawn cubes of different sizes and calculate thesurface area to volume ratio of the three different sized cubes.

The following formulas are available from the AP formula sheet to use for calculating theseratios:

These formulas can beused for creatingsolutions for this lab.


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With these calculations and your understanding of diffusion and osmosis you predicted therelative rate of diffusion in each of the different sized cubes.

Using different sized cubes made with 2% agar, 0.1 M NaOH, and the pH indicator dye phenolphthalein you were able to observe the relative rates of diffusion in the different sizedcubes. Each cube was placed in a 0.1 M HCl solution. As the HCl diffused into the cube the

pink cube would turn colorless.

You reflected upon the relationship of cell size and shape as important factors determining therate of diffusion. In particular you reviewed cells with specialized functions such as epithelialcells that line the small intestine and plant root hairs.

Small cells have a greater surface area to volume ratio than do larger cells. This allows the smallcells to more efficiently move material in and out of the cytoplasm and maintain homeostasis.

Similarly, cells with convolutions have a greater efficiency than do cells without convolutions.

This is important in the nutrient procurement of epithelial cells that line the small intestine and plant root hairs.


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Procedure 2: Modeling Diffusion and Osmosis

In this experiment, you created models of living cells using dialysis tubing. Dialysis tubing ismade from a material that is selectively permeable to water and some solutes. You filled themodel cells with different solutions and determine the rate of diffusion. Rate of diffusion anddirection of diffusion was determined by weighing the model cells before and after placing each

cell into a beaker filled with one of a variety of solutions. Percent change was then determined

The solutions that you placed in the model cell or the beaker were:

o 1 M Sucrose (a disaccharide)o Watero 1 M NaClo 1 M Glucose (a monosaccharide)o 5% Ovalbumin (a polypeptide)

Factors that affected diffusion include: disaccharides and polypeptides are too large to fit

through the pores in the dialysis tubing. Monosaccharaides can fit through the pores within thedialysis tubing as can NaCl and water. NaCl ionizes and therefore has an ionization constant of 2while glucose does not and has an ionization constant of 1.

Procedure 3: Observing Osmosis in Living Cells

In this exercise you observed red onion cells, an aquatic plant ( Elodea) or a moss plant ( Mniumhornum ) under the light microscope. You then placed a solution from the previous procedure on

the cells to observe osmosis. You observed the following:

Cell Membrane

Hypertonic Solution

Cell Wall
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If you applied saltwater to the roots of a plant, water would diffuse out of the root cells to thearea with lower water potential. In order for a plant to control turgor pressure it must regulate theflow of solute particles or ions across its cell membrane.

In order to determine the water potential of plant cells you placed pieces of potato cut with a cork borer into known sucrose solutions with varying concentration. You measured the percent

change in mass of the potatoes and by graphing the data determined the solute concentration ofthe potato cells.

Using this value and the following formula (22 oC)

(S) = iCRT

= 1(0.32)(0.0831 liter bar/mole K)(295K)

= -7.8 bars

You calculate the solute potential. With the solute potential calculated and knowing that the pressure potential is zero you can determine the water potential of the potato cells

= P + S

= 0+ S

Additionally, you can determine the concentration of unknown sucrose solutions. In order to dothis place the unknown solution in cells (bags) made from dialysis tubing. Weigh each bag and

place each cell in beakers with different but known sucrose concentrations. Following thirtyminutes the percent change in mass can be determined. Graphing the percent change in eachsolution can facilitate determining the isotonic concentration (where there is no percent changein mass)

Percent Change in Mass of Potato Cores in SucroseSolutions


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AP Lab 5 Photosynthesis

Rate of photosynthesis can be determined by measuring the

amount of carbon dioxide consumed over time.

number of redox reactions over time.

amount of oxygen produced over time.

Note Rate of photosynthesis can also be determined by measuring sugar (starch) production over time.

Detection of CO 2 Consumption

With a water plant or algae, as carbon dioxide is consumedin the dark reaction of photosynthesis, bromthymol blue will change from yellow to green .

When CO 2 is dissolved in water, it forms H 2CO 3 (carbonic acid) and causes the pH of thesolution to drop. As CO 2 is removed from the water, the pH of the solution increases.

When CO 2 is removed from a solution containing bromthymol blue, it changes fromyellow to green, and the color change may be qualitatively observed or quantitativelymeasured using a spectrophotometer .

Spectrophotometer: Wavelength Yellow Absorption will Increase / Transmission willDecrease

Detection of Redox Reactions

When chloroplasts are placed in a sucrose solution with theindicator DPIP (dichlorophenol-indolphenol), the indicatorwill change from blue to colorless when it is reduced inthe light reactions of photosynthesis.

This color change may be qualitatively observed or quantitatively measured using aspectrophotometer .

Spectrophotometer: Wavelength Green (any wavelength except blue) Absorption willDecrease / Transmission will Increase

Bromthymol Blue

Yellow Green Blue(pH 6) (pH 7.5)

DPIP

Blue Colorless Reduction

(gains electrons)


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Detection of O 2 Production

Oxygen can directly be observed by counting bubbles orcollecting gas in an inverted tube. If the tube is graduated,quantitative data can be obtained.

As well, if leaf disks are infiltrated with carbonated fluid will rise as oxygen is produced and the air spaces within the leaf fillup oxygen gas.

L

Leaf disks are made by using a hole punch.

Leaf discs are infiltrated using avacuum created inside a syringe.

The solution used for infiltratingdiscs consists of water, baking soda(sodium bicarbonate forms CO 2),and dish soap (surfactant).


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AP Lab 6 Cellular Respiration AP Lab Manual - p S71-S73

Rate of cellular respiration can be determined by measuring the .

amount of carbon dioxide produced over time.

amount of oxygen consumed over time.

Note Rate of respiration can also be determined by measuring heat production or redoxreaction over time.

Detection of CO 2 Production

When an aquatic animal produces carbon dioxide,bromthymol blue in the solution around it will changefrom blue to yellow .

When CO 2 is dissolved in water, it forms H 2CO 3 (carbonic acid) and causes the pH of thesolution to drop.

As CO 2 is added to a solution containing bromthymol blue, it will change from blue toyellow, and the color change may be qualitatively observed or quantitatively measuredusing a spectrophotometer .

Spectrophotometer: Wavelength Blue Absorption will Increase / Transmission willDecrease

Detection of O 2 Consumption

A respirometer allows O 2 consumption to be monitored. As O 2 is consumed water or a soap bubble will move down the pipette leading toward the chamber of the respirometer. The amountof movement represents the amount of O 2 (mL) consumed by the organism in the chamber sinceall CO 2 in the chamber is removed by KOH in the cotton in the bottom of the respirometer.

Bromthymol Blue

Blue Green Yellow (pH 7.5) (pH 6)

Respirometer measuresO2 consumption and canonly be used in respirationexperiments.
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Experimental Design

Controls

Experimental Set-up

Control Set-up

Both set-ups must be in the same size tubes .

Both set-ups must have the same amounts of absorbent and non-absorbent cotton .

Both set-ups must have the same amount of KOH .

Both set-ups must have the same volume of gas . Therefore, the volume of peas and beadsmust be the same. The number of beads does not matter, but the volume of the beads must match the volume of the seeds.

Pea seeds do NOT carry out photosynthesis. They consist of an embryonic plant that onlyrespires along with the cotyledon (2n) that serves as a food source. The embryo andcotyledon are surrounded by the testa (seed coat).

Corrected Data

Pressure Changesdue to TemperatureFluctuation

P = Pressure n = number of moles of gasV = Volume R = gas constant

T = temperature

As the temperature increases, the pressure and/or volume of the gas increases.

A negative control should notshow any change and it isknown that glass beads do notrespire.

If the volume in negativecontrol does change, the datavalues must be changed to

reflect a situation where thereis no change in the control.


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Question Is there a statistically significant difference in respiration rates of germinatingand non-germinating and non-germinating pea seeds?

Three respirometers will be used to accomplish to accomplish this goal.

7. Let the respirometers equilibrate for 10 minutes before submerging under water. This allowstime for the CO2 to be removed from the respirometers and for the respirometers to adjust to thetemperature of the water bath.


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Measurement of Oxygen Consumption

Since the control may change in volume due to temperature fluctuations and pressurechanges, the experimental data must be corrected (corrected difference).

Negative Germinating Peas (Wet) Non-germinating Peas (Dry) Control

Tempo C

Time(min)

Readingat time

X Diff. Readingat time

X Diff. Corrected

Diff.

Readingat time

X Diff. Corrected

Diff.

25 Initial-0

14.4 13.9 14.2

25 0 to 5 14.1 0.3 13 0.9 0.6 14.1 0.1 -0.2

25 5 to10

14.0 0.4 11.1 2.8 2.4 13.9 0.3 -0.1

25 10 to15

13.9 0.5 10.3 3.6 3.1 13.7 0.5 0.0

25 15 to20

13.9 0.5 8.8 5.1 4.6 13.5 0.7 0.2

Collect the standardized data (corrected differences) from the multiple trials carried out by all thelab groups in your class and calculate the average data .

Non-Germinating Peas Average Data from Multiple Trials

Time(min)

Group1

Group2

Group3

Group4

Group5

Group6

Group7

Group8

AverageData

(Mean)

50.01 0.01 0.01 0.04 0.03 0.00 0.01 0.00 0.014

100.01 0.02 0.02 0.01 0.04 0.01 0.025 0.00 0.017

150.02 0.03 0.04 0.01 0.05 0.02 0.04 0.00 0.020

200.03 0.03 0.05 0.00 0.06 0.02 0.05 0.01 0.031


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Germinating Peas Average Data from Multiple Trials

Time(min)

Group1

Group2

Group3

Group4

Group5

Group6

Group7

Group8

AverageData

(Mean)

50.02 0.06 0.08 0.09 0.05 0.09 0.04 0.05 0.060

100.06 0.12 0.14 0.06 0.10 0.18 0.075 0.09 0.103

150.09 0.19 0.23 0.07 0.15 0.27 0.11 0.13 0.155

200.13 0.25 0.24 0.07 0.19 0.38 0.135 0.17 0.196

Use the data collected from multiple trials to create a graph with standard error bars .

Plot Mean Values Use Mean to Calculate Standard Deviation Use Standard Deviation to Calculate Standard Error Use Standard Error to Determine Standard Error Bars Calculate the Slope of Each Line (Best Fit) to Determine Respiration Rate

Use a t-test to determine if there is a statistically significant difference between the rate ofrespiration in germinating and non-germinating pea seeds.

Pea Seed Respiration

-0.1-0.05

00.05

0.1

0.150.2

0.250.3

0 5 10 15 20 25

Time (min)

A v e r a g e

O x y g e n

C o n s u m p

t i o n

( m L

Germinating Peas

Non-germinatingPeas


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Inferences

Imagine that you are given 25 germinating pea seeds that have been placed inboiling water for five minutes. You place these seeds in a respirometer andcollect data. Predict the rate of oxygen consumption (cellular respiration) for

these seeds and explain your reasons.Enzymes necessary for respiration will denature at high temperatures and most likely will notrenature correctly when cooled. As a result, boiled peas are not expected to respire.

Rare

Imagine that you are asked to measure the rate of respiration for a 25g reptileand a 25g mammal at 10oC. Predict how the results would compare and justify your prediction.

Reptiles are cold blooded animals (ectotherms or poikilotherms) and mammals are warm blooded animals (endotherms). The respiratory rate within the cells of the reptile willoccur more slowly than the respiratory rate of the cells within the mammal when theenvironmental temperature is 10 oC.

Warm blooded animals maintain a relatively constant internal body temperature andconvert ATP energy produced during respiration into body heat. As well, they haveadaptations such as fur or hair to slow the loss of body heat. As a result, warm bloodedanimals carry out respiration within their cells at a relatively constant rate.

Cold blooded animals do not maintain a constant internal body temperature. As theenvironment cools, their body temperature drops and the rate of respiration within theircells slows down significantly. Cold blooded organisms may change their behavior toadjust their body temperature and often sit in the sun to warm up on cold days.


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What difficulties would there be if you used a living green plant in thisinvestigation instead of germinating seeds?

Leaves of plants carryout photosynthesis and produce oxygen gas. This oxygen gas isdirectly used in respiration within leaf cells.

As a result of oxygen production within leaf cells, the decrease in the oxygen level within therespirometer would be lower than the actual amount of oxygen used in respiration.

Note Some oxygen would be consumed because not all plant tissues are photosynthetic.Roots, non-green stems, and flowers do not carry out photosynthesis. As well, the cotyledons(seed leaves) inside of seeds do not carry out photosynthesis.

Note Even though carbon dioxide is removed from the respirometer by KOH, the lightreactions of photosynthesis that produce oxygen by lysing water molecules most likely willcontinue during the duration of the experiment. As well, stored carbon dioxide within the leafcan be used to keep the dark reactions operational for a short period of time.

Further Investigation

Design an experiment to test the effect of a variable such as pH, light intensity, temperature,or level of CO 2 on the rate of respiration.

Compare the photosynthetic rates of seeds from different species of plants.

Compare the photosynthetic rates of different species of insects.

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