AP Biology Laboratory Review

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In this lesson, the laboratory review, we are going to discuss the laboratory0002

component of an AP biology course starting out with lab 1: diffusion and osmosis.0007

In the first part of this laboratory, you explored the process of diffusion.0015

And during this lecture today, I am going to briefly review the concepts behind the lab.0019

But, in earlier lectures, we have already gone through each of these concepts in detail.0025

And then, I am going to talk about what you should focus on for the laboratory.0030

So, just to briefly review the concept of diffusion, in diffusion, substances move down their concentration gradients.0034

So, they move from a region of higher concentration to a region of lower concentration, and this is the result of the random movement of molecules.0044

That information is given in more detail in the lecture on transport across cell membranes.0056

And we talked in that lecture about different processes of transport across cell membranes including diffusion.0061

Now, in this lab, you used a bag made from dialysis membrane.0068

And what is important to know for the lab is that dialysis membrane is semi-permeable,0075

meaning that small solutes in water can cross it, whereas, larger solutes cannot.0086

So, in this lab, what you did is you took dialysis tubing and tied it off to form a bag.0107

Inside the bag, so you have this bag made out of dialysis tubing, and inside it, you are going to place a solution of glucose plus starch.0113

And to confirm the presence of glucose, you tested it with either Tes-Tape or another strip that allows you to test for the presence of glucose.0127

So, you confirm that there is indeed glucose in the solution.0137

Next, what you should have done is filled a beaker about two thirds full with distilled water, and to that, you added the solution of IKI.0141

IKI is iodine-potassium iodide, and you add the IKI to the beaker.0161

And it is going to turn a yellowish color, sort of, a more of a brown or orangish-yellow, a darker yellow.0182

You also tested this IKI water solution for glucose, and it would have tested negative.0192

I will put a negative here, negative for glucose, whereas, in the dialysis bag you did test, and there was indeed glucose.0202

Next, what you did is you would tie off this bag, and at this point, the fluid in this bag is clear.0212

And you placed the bag inside this beaker, and so we will make that clear.0220

Then, you wait 30 minutes. After 30 minutes, what you will notice is a color change.0231

So, after 30 minutes, the solution inside the bag turns a bluish black color. Now, why is this?0239

Well, it turns out that in the presence of starch, iodine turns this color, so iodine plus starch turns a bluish black color.0249

Now, thinking about the overall concept behind the lab, what has occurred, well, remember that this is a semi-permeable membrane.0270

And small molecules in water can move across the membrane, larger molecules cannot.0280

And at the beginning of the lab, what you did is you tested for glucose.0286

And we found no glucose inside the bag - excuse me - glucose inside the bag, no glucose in the water that was just in the beaker.0290

Well, at the end of the experiment, you, again, are going to test the solution in the beaker.0296

And this time, you are going to find that there actually is glucose in the beaker.0301

So, the results: the fluid in the bag turned bluish black, and in the beaker, we now have glucose.0307

So think about what happened.0321

Where did the molecules go? Where did they move to?0323

Well, in order for the solution inside the bag to turn bluish black, starch had to be there and so did iodine, so iodine moved into the bag.0325

And we started out with glucose in the bag but not in the beaker, so glucose moved across the semi-permeable membrane into the beaker.0339

Starch, however, is too large to cross the semi-permeable membrane, and how do I know that there is not starch out here?0350

Well, remember that there is this iodine solution out here, and if starch had managed to cross the semi-permeable membrane0361

and ended up in the beaker, then, the solution out here would have turned bluish black.0370

So, this was the first part of the lab on diffusion and osmosis, so that was the diffusion part; and the second part of this lab was on osmosis.0379

Let's now focus on osmosis. Recall that osmosis is the diffusion of water down its concentration gradient.0389

And water is going to move, therefore, from a region of higher water concentration to a region of lower water concentration.0398

Another way to look at this is in terms of solutes.0406

So, if I have a container, and inside the container are two solutions, and they are separated by a semi-permeable membrane0409

where water can cross but solutes cannot, and on one side, I have a higher solute concentration than the other,0421

what I am going to see is that the water is going to move from the region of lower solute concentration to the region of higher solute concentration.0431

And again, this is a concept that we covered earlier in the course when we talked about transport across cell membranes.0440

So, what you did in this part of the lecture is that again, you used dialysis tubing, and you made six bags from the tubing.0448

So, in the osmosis component of the lab, you made six dialysis bags, and you filled one with distilled water0456

and then, the others with various concentrations of sucrose solution 0.2M, 0.4, 0.6, 0.8 and on and on up to 1M sucrose.0471

So, you filled these bags, and you, then, weighed the bag, so you filled the bag with distilled water.0489

Another bag was 0.2M sucrose solution, 0.4M sucrose solution, and you weighed the bag. You determine the mass of each bag.0499

Now, sucrose cannot cross the dialysis membrane, so you have this bag filled with the sucrose solution, and sucrose cannot cross it but water can.0509

Then, what you did is you placed the bags in beakers filled about 2/3 full of water.0524

So, here you have water in the beaker, and inside the bag, either sucrose solution or distilled water.0535

You, then, waited 30 minutes and weighed the bags the second time, and you should have left room in the bag so that water could enter the bag via osmosis.0545

So, again, you have six bags filled with different solutions, either just regular distilled water or various sucrose solutions.0553

You weighed those bags. You place the bags in beakers partly filled with distilled water.0560

You waited 30 minutes, and then, you weighed the bags again.0565

What you are going to see is that the change in mass was greater for bags with a higher molarity of sucrose.0570

And the reason is that more water would enter the bag if it had a higher concentration of sucrose.0581

So, just like I showed up here, water is going to move from areas of lower solute concentration to higher solute concentration.0588

And that is what is going to happen with these bags, and the more water enters the bag, the greater the mass will be.0596

Now, recall some terminology, when we were talking about cells earlier on and taking a cell and sticking it in some type of solution,0601

we said that if the concentration of the solution is hypertonic, we are saying that it has a higher solute concentration than in the inside of the cell.0611

And water is going to leave the cell. The cell will shrink.0625

In a hypotonic solution relative to the cell, that means that there is a lower concentration of solutes outside the cell than inside.0629

In this case, water is going to move into the cell, and the cell will increase in volume.0653

Finally, recall that in an isotonic solution, there is going to be no net movement of water because...0658

There would be movement of water, but there is no net movement.0669

So, overall, the cell will remain the same in size, in volume because the concentration0671

of solutes inside the cell is the same as the concentration of solutes outside the cell.0677

So, hypertonic, hypotonic and isotonic solutions- these are concepts you should definitely be familiar with.0681

Finally, for this part of the lab, what you did is you graphed your data.0690

And on the X axis, you put the molarity of the solution, and on the Y axis, you graphed the change in mass.0695

And again, this will help you visualize that the change in mass was greater for solutions in which the molarity of the sucrose is greater.0702

As part of lab 1, first, to review the concept of water potential, water potential determines the direction for movement of water.0714

And remember that water is going to move from an area of high water potential to a region of lower water potential.0723

and water potential is given as the Greek letter Psi, and it is determined by a combination of solute potential and pressure potential.0732

Now, increase in solute potential, so increased solute concentration - I will just put increase in solutes -0743

is going to raise the water potential - excuse me - lower the water potential.0752

It is going to lower the water potential- correction.0757

So, we started out looking at pure water that has a potential of 0.0763

Therefore, if I add solutes, I am going to decrease the water potential, and it is actually going to become negative.0769

It is going to be a negative number.0777

Now, what is the effect of the pressure potential? Well, if you increase the pressure, increase pressure potential translates to increased water potential.0779

So, let's think about the situation where we would have a lower water potential.0792

This would be a situation where I have got a lot of solutes, a high solute concentration, so a high solute potential and low pressure.0798

These are the conditions that are going to give me a lower Psi, a nice negative number, and water is going to move in that direction.0808

It is going to move from an area of higher to lower solute concentration.0817

You have to look at the water potentials of the two solutions that you are looking at.0822

And then, the water was going to want to move in the direction of this lower water potential.0828

So, in this experiment, what you did is you explored the concept of water potential.0834

And you did that by cutting some potatoes into identical pieces and then, weighing them.0838

So, you cut potatoes. You weighed those pieces, and then, you placed one potato piece in each of several beakers.0846

And one of the beakers just had distilled water in it. The others had sucrose solutions, 0.2M, 0.4, 0.6 and on up usually to 1M sucrose.0853

So, you put your potato slices in there. You left them overnight.0867

You came back the next day, and you weighed those slices, and what you found is some of the slices increased in mass.0870

Other slices may have decreased in mass, and one of the slices may have had no change in mass.0880

and what determined whether the mass changed and in what direction it changed0887

is the water potential in the potato relative to the water potential outside the potato.0894

And remember that water is going to move from an area of higher water potential to an area of lower water potential.0899

So, if let's say I had a higher water potential here and a relatively lower water potential here,0906

then, water is going to move into the potato, and mass of the potato would increase.0915

Now, what you were supposed to determine and one way to do this is actually just by plotting it out and looking at a graph0920

and seeing that if you graph the percent change in the mass of the potato over here on the Y axis,0926

and then, on the X axis, you graph the molarity of the solution, and then, you see where this line crosses the X axis,0943

that is going to be the point at which the sucrose solution had the same water potential as the potato.0955

And therefore, there was no net movement of water into, out of the potato, and the potato, therefore, did not lose or gain mass.0964

The final concept that you studied was plasmolysis and closely-related to the other concepts in this lab but investigated in a slightly different way.0973

Now, if we look at plasmolysis, first, we want to think back to what it is.0983

Remember that plant cells have a cell membrane, and the cell membrane does pull away from the cell wall when water leaves the cell.0992

So, if you place a cell in a hypertonic solution, so if a hypertonic solution, what is going to happen is water is going to leave the cell, and the cell shrinks.1004

Well, that is what is going to happen in an animal cell.1023

In a plant cell, the cell membrane will pull away actually from the cell wall because the cell wall, itself, is a more rigid structure.1025

It is not going to shrink away.1037

And this process of the cell membrane pulling away from the cell wall - cell membrane pulls away from cell wall - is called plasmolysis.1038

And what you did is, in your lab, you observed this process.1055

So, you took a piece of onion or a tissue from another plant, and you added a sodium chloride solution; and this is a hypertonic solution.1059

So, water left the cell. The cell membrane pulled away from the cell wall, and it left a space between the two; and you could actually observe that.1075

Then, you added distilled water, so if you add distilled water, that is a hypotonic solution.1083

It would, then, enter the cell via osmosis and return the cell to its initial state.1096

So, plasmolysis is just further investigation of these concepts that we talked1101

about with osmosis and water potential and the direction of movement of water.1105

In the second recommended AP Biology lab, you studied enzyme catalysis.1111

Recall that enzymes increase the rate of the reaction by lowering the activation energy required for the reaction, so enzymes are catalysts.1116

This was all reviewed in the lecture on cellular energetics, so make sure you understand the concepts behind how enzymes work.1128

In this lab on catalysis, what you studied was the following reaction: hydrogen peroxide breaking down to form water plus oxygen in its gas form,1136

so breakdown of hydrogen peroxide to form water and oxygen gas.1156

Recall that hydrogen peroxide is a by-product of aerobic respiration, and that hydrogen peroxide is toxic to cells.1161

Therefore, it is important for cells to somehow get rid of this by-product, and they can do it by converting it to water and oxygen gas.1173

So, in the first part of this lab, what you did is added catalase to a beaker of hydrogen peroxide, and catalase catalyzes this reaction right here.1180

So, you add catalase to hydrogen peroxide, and what you are going to do is observe. What you should see is bubbling.1193

And this bubbling is the result of the production of oxygen gas.1205

Your next step was to boil catalase and then, add that catalase to hydrogen peroxide, and what you are going to see is no bubbling. Well, why not?1212

Well, the reason is if you recall most enzymes are proteins, and proteins can be denatured by high temperature because they lose their tertiary structure.1231

Other causes of denaturation of enzymes are applying a strongly basic or acidic condition to them or a high salt concentration.1241

And as it says up here, enzyme activity is affected by factors such as substrate concentration,1251

salts, temperature, pH and the presence of activators and inhibitors.1257

So, what you did in this lab is you investigated the effect of temperature on enzyme activity.1261

In the next part of the lab, what you did, instead of just observing and saying "OK, the reaction occurred or did not occur",1268

is you actually made a more quantitative measurement of what is going on.1275

So, during the second part of the lab - so, lab 2B - what you did is ran a baseline1279

assay to determine the initial concentration of hydrogen peroxide in the solution.1288

So, first determine hydrogen peroxide concentration, and how did you do that?1294

Well, you used an assay that required potassium permanganate, and you ran a titration of potassium permanganate with hydrogen peroxide.1302

And it is based on the following reaction, so I will put this right here.1319

Hydrogen peroxide plus potassium permanganate plus sulfuric acid is going to yield potassium sulfate plus manganese sulfate plus water plus oxygen.1326

And you certainly do not need to learn this whole reaction or anything, but I just wanted to show you what it is based on.1351

So, if you have hydrogen peroxide in the presence of sulfuric acid, and you add potassium permanganate,1356

once the hydrogen peroxide has been used up, if you add any more potassium permanganate, the solution is going to turn a light pink.1365

So, let me explain this again. What you did to determine the concentration of hydrogen peroxide is you have your hydrogen peroxide in your beaker.1374

You add sulfuric acid, so you added the sulfuric acid to the hydrogen peroxide.1385

Then, one drop at a time, you add the potassium permanganate, so you titrate it.1404

You have got the hydrogen peroxide. You have the sulfuric acid one drop at a time, potassium permanganate.1419

And once the hydrogen peroxide is used up, that next drop you add after the hydrogen peroxide being used up,1425

the solution turns light pink when the hydrogen peroxide is used up.1436

Therefore, the amount of potassium permanganate that needs to be added to get this1450

pink color is proportional to the amount of hydrogen peroxide that you started out with.1455

This allows you to determine the amount of hydrogen peroxide that was present in the initial solution, and this is called a baseline assay.1461

So, we have not gotten into the whole experiment yet.1470

But, this is the baseline, your starting point, the initial concentration of hydrogen peroxide that you had.1473

So, what you did in the lab, then, after determining that baseline amount was to explore the conversion of1480

hydrogen peroxide to water and oxygen in the uncatalyzed reaction and in the catalyzed reaction.1489

So, first what you did after determining the baseline, after doing your baseline assay and determining how much1499

hydrogen peroxide you had in, is you got a beaker of hydrogen peroxide and simply left it uncovered for 24 hours.1506

Then, you came back, and you performed a second titration with potassium1514

permanganate to determine the final amount of hydrogen peroxide in the beaker.1518

So, first you are going to determine initial and final hydrogen peroxide concentration in the uncatalyzed reaction.1524

Then, you repeated the process or repeated the basic experiment but this time with the use of catalase.1543

So, you started out with your hydrogen peroxide, and added catalase.1552

And you stopped that reaction after 10 seconds 30, 60 90, and various increments.1558

So, how do you stop the reaction? Well, by using sulfuric acid.1565

This time you had your hydrogen peroxide. You had your baseline amount.1568

You knew what you had. You added catalase and stopped reaction after, let's say, 10 seconds.1572

So you stopped the reaction by adding sulfuric acid.1583

And then, you ran another titration to determine the final amount of hydrogen peroxide in the beaker.1586

How does sulfuric acid stop the reaction? Well, recall, as I said, pH affects enzymes, and strongly acidic or strongly basic conditions can denature enzyme.1593

So, you are denaturing the catalase and stopping the reaction.1606

OK, then, you repeated that procedure. You added your hydrogen peroxide.1609

You added catalase, and this time after 30 seconds, you added sulfuric acid to stop the reaction.1613

And you measured your final amount of hydrogen peroxide.1620

You did that again, except stopping the reaction after 60 seconds, and you went on and did that until the maximum of 360 seconds.1623

And this allows you to determine the rate of the reaction in this catalyzed reaction.1631

And you did this, looked at the data in graph form by plotting time on the X axis and the amount1639

of hydrogen peroxide consumed on the Y axis to determine the rate of the catalyzed reaction.1648

And you can compare that two with the results that you saw from the uncatalyzed reaction after that whole 24 hours.1653

In the third lab, you explored mitosis and meiosis. We discussed both of these in detail during the lecture on cell reproduction.1660

So, make sure that you are familiar with the stages of both mitosis and meiosis and that you understand the concept of crossing over.1670

In the first part of lab three, what you did was observed either the apical meristem of an onion root tip.1679

or you may - in you lab - have observed a whitefish blastula.1687

Now, as far as the plant, recall that the meristems of a plant, the apical or1692

the meristem regions of plants in general are the regions of active growth.1700

And that is why they are a great place to observe if you want to study the process of mitosis.1704

So, what you probably did in you lab, every school has a little bit different set up.1711

But, typically, what you do is you would observe previously prepared slides.1715

And you would look at these cells, these apical meristem cells or whitefish blastula cells.1719

And you would identify cells in various stages of mitosis and sketch those cells.1726

So, here is just a figure reviewing the stages of mitosis:1733

So, prophase, metaphase characterized by the chromosomes lining up single file along the metaphase plate,1739

anaphase in which sister chromatids separate, and telophase followed, then, by cytokinesis and the separation into the two identical daughter cells.1751

As part of this lab, in addition to just identifying the different stages of mitosis in the cell,1762

you also may have had to determine the relative length of time that a cell spends in each stage of mitosis.1768

Well, it takes about 24 hours for a cell in an onion root tip to complete the cell cycle.1776

About 24 hours for the cell cycle to be completed in cells in the root tip of an onion.1784

So, what you would do is you would look under the microscope and determine the phase of each cell for a total of 100 or even better 200 cells.1790

So, you observe 200 cells and determine the stage of mitosis that they are in.1801

And you will tally those up and also determine some of the cells maybe in interface,1815

so interface, as well, where the chromosomes are not yet in their condensed forms.1821

So, all five of these stages including interface determine the percentage of cells that were in each phase.1825

And then, use this formula to calculate how many minutes of the cell cycle were spent in a particular stage.1834

So, the formula is the percentage of cells in the stage times 1440 minutes, and that equals the number of minutes spent in a stage.1840

So, if you determine 10% of the cells were in interface, then, you times that by 1440 minutes.1867

And that will give you the number of minutes spent in a particular stage.1872

This was the first part of the lab, and then, in the second part of lab 3, what you likely did was studied meiosis using models of chromosomes.1876

The models usually consist of strands of beads, and you should especially have focused on the process of crossing over.1887

So, it is important that you understand the process of crossing over. Again, this was explained in detail in the lecture on cell reproduction.1894

You also should make sure that you are familiar with the differences between mitosis and meiosis.1902

In the next part of lab 3, so this is still lab 3, you used what are called sac fungi, and we discussed these types of fungi back in the lecture on fungi.1911

The fungi that you used are a type of Ascomycota, but in particular, you used Sordaria fimicola.1930

I just want to review their life cycle and talk about what you did in this part of the laboratory.1943

So, first, let's review the life cycle. Recall that hyphae of opposite mating types plus and minus undergo plasmogamy.1950

The hyphae that result from this process are what we call dikaryotic or dikaryons, so they have unfused nuclei.1960

They are not truly 2n. They are n + n.1969

An ascus forms on these hyphae, and numerous asci are contained within a fruiting body called a perithecium.1973

Let me correct that. So, peritheciums are a fruiting body that contain asci- singular is ascus.2001

The nuclei inside the ascus, then, undergo karyogamy, and they fuse- fusion of nuclei.2026

So, at first the nuclei were dikaryotic inside an ascus, and then, these nuclei fuse to form a diploid cell.2040

These diploid cells, then, we have a zygote, but they, then, undergo meiosis and produce four haploid nuclei; so, we end up with four haploid nuclei.2048

This is followed by a round of mitosis to result in eight haploid nuclei, so we end up with eight haploid nuclei.2064

And cell walls form around these nuclei with the result that each ascus is going to contain eight ascospores.2075

So, we end up with eight haploid spores per ascus.2091

So, I wrote this off to the side, but this is all actually happening right in here.2097

We are going from meiosis to give us four haploid nuclei to mitosis to give us eight haploid nuclei.2101

Cell walls form around those eight haploid nuclei to form eight haploid ascospores per cell.2108

These spores, then, are released from the ascocarp, and they germinate to form a haploid fungus.2115

And the fungus actually spends most of its life cycle in this haploid state.2123

OK, so this is the background of the life cycle of ascomycota. Now, what did you do in the lab?2127

Well, the purpose of this lab was to observe and study the process of crossing over, and you did this by studying wild type and mutant strains of Sordaria.2133

The wild type has black ascospores, so if you open up an ascus, what you would see would be 1, 2, 3, 4, 5, 6, 7, 8 black ascospores.2148

The mutant has 10 ascospores, so you are going to see - we will use red instead of tan here -1, 2, 3, 4, 5, 6, 7, 8 tan ascospores.2172

If you mate a wild type mycelium in a mutant mycelium, mutant for ascospore color,2191

you are going to end up with meiosis followed by mitosis, which is going to produce a hybrid.2200

So, this is the hybrid, and this is a cross between wild type and mutant for ascospore color, mutant with the tan color.2209

What you are going to end up with is 1, 2, 3, 4 black ascospores and 1, 2, 3, 4 tan ascospores.2219

Now, the arrangement of the ascospores in the ascus reflects whether or not crossing over has occurred during meiosis.2234

What is shown here is no crossing over.2244

You are going to see this four to four pattern, and it does not matter if the tan ones were on this side and the black one is here or vice versa.2251

There is no crossing over. What you are going to want to look for are these versus the patterns that you will see from crossing over.2258

If crossing over has occurred, you will either see a 2:2:2:2 pattern or a 2:4:2 pattern. Let me show you what I mean.2268

So, there are four possibilities. We could have this 2:2:2:2 pattern with black, tan, black, tan.2277

Oops, that is not tan. This is tan, and we can even make this bigger.2293

So, we have a 2:2:2:2 pattern, or they could be the other way.2300

We could end up with tan first on this end then, black, tan, black. That is from crossing over.2306

The second pattern that you could see is a 2:4:2 pattern. There are two possibilities there.2320

We could have 2 of the black ascospores on the end, then, 4 tan, 2 black, or the other way, 2 tan, 4 black, 2 tan.2329

So, what you did in this lab was under a microscope, you observed the asci and counted 100 of these hybrid asci.2346

And when you count this, you are going to see some that do not have crossing over and others that did have crossing over.2360

And you are going to calculate the results, you are going to write down your results.2367

And then, you are going to use that to calculate the distance between the centromere and the low side for ascospore color.2371

This gives you distance in what we call map units, and recall that map units are equal to 1% recombination.2383

So, to understand this lab, you needed to know some basics of the life cycle of Sordaria.2395

You also needed to understand the concepts of crossing over and that in this organism, we can visualize the results of crossing over.2399

You easily identify which organisms had crossing over by the pattern of the spores inside the ascus.2407

You counted 100 hybrid asci, and then, you determined which percentage had resulted from crossing over2413

and used that to map or calculate the distance between the allele for ascus color and the centromere, OK?2424

So, that was lab 3, which is the study of meiosis and mitosis.2431

In lab 4, you studied plant pigments and photosynthesis.2437

So, in this lab, what you did first was to separate out plant pigments using paper chromatography.2442

And that is what shown here- this set of paper chromatography.2449

Chromatography is the method that is used to separate substances in a mixture.2458

What you do is you take chromatography paper, and you place it in a solvent; and the solvent will move up the paper via capillary action.2465

And here you can see the solvent front right there.2477

The solutes inside the solvent have differences in their solubility. Some are more soluble than the solvent.2480

Some are less soluble, and they also have differences in their attraction to the paper, into the chromatography paper.2486

Those pigments that...we are going to be studying plant pigments.2493

So, the pigments that dissolve better in the solvent will move up the paper more rapidly.2498

The result is that substances move up the paper at different rates.2506

And you are going to end up with a series of lines where the substances have separated out.2510

In this lab, what you used is an extract of photosynthetic pigments that you extracted from spinach leaves.2519

So, you take the solvent. You place a little bit in the bottom of a graduated cylinder.2527

And then, take a coin or some other item that you are instructed to use and crash up the spinach leaves.2532

This will extract the pigments. You place a little bit of the pigments on to the paper.2538

And then, you put the paper in your graduated cylinder with the bottom of the paper just touching the solvent.2542

This shows the paper more submerged, but the really the bottom of the paper should barely be just touching the solvent.2548

And then, the pigments should be outside of the solvent. Then, you allow the solvent to move up the paper.2554

And what you are going to do is measure the distance of each band of pigment from the bottom of the paper, and they are going to be different distances.2561

In this lab, you would be looking at four pigments, and those pigments are keratin, xanthophyll, chlorophyll A and chlorophyll B.2570

And each of these is going to form a band along the chromatography paper.2591

Then, you are going to calculate what is called the resistance front ratio for each of the pigments, so R_f is the resistance front ratio.2596

And it is given as the distance a particular pigment migrated in millimeters divided by the distance the solvent migrated- again, in millimeters.2611

So, that was the first part of this lab studying photosynthesis and plant pigments, so right here, you are focusing on the plant pigments.2636

In the next part of the lab, what you did was you measured the rate of photosynthesis for various plant pigments.2643

Recall that different pigments absorb different wavelengths of light, and also, that different wave lengths of light are different colors.2651

So, we talked earlier on this course, for example, that violet light has a shorter wavelength than red light.2659

Remember that we talked about chloroplast containing light-harvesting complexes.2668

And these light-harvesting complexes contain various pigment molecules.2679

So, they contain various pigment molecules such as the ones I have mentioned, chlorophyll A, chlorophyll B, carotenoids.2684

And because they contain multiple types of pigments,2694

the light-harvesting complexes can obtain light from a broader portion of the electromagnetic spectrum.2696

What the light-harvesting complexes do is they absorb photons.2703

And the result is that the electrons in the pigment are excited and moved to a higher energy state.2707

So, they absorb photons, and the result is that electrons move to a higher energy state.2714

Recall, then, that this energy is used to transfer electrons along an electron transport chain.2728

And this ends up reducing NADP to NAPH and producing a proton gradient.2735

Energy from that proton gradient is used. It is harnessed to produce ATP.2744

And then, in the dark reactions, the ATP and NADPH are used to fix carbon into organic molecules.2750

So, that was a really fast review of the complex subject of photosynthesis.2757

But definitely, watch the lecture on photosynthesis if you need more clarification, but it is important to have that background as you go into this lab.2762

Because what you did in this lab is you used DPIP, which is a dye, and this is substituted for NADP in this experiment.2769

So, in this experiment, DPIP is used as an electron acceptor, and the reason that this is used is because DPIP is blue.2780

But, when it is reduced, so when it accepts an electron, it turns colorless, so it allows us to actually visually know when it has been reduced.2795

Now, you also used a device called a spectrophotometer or sometimes you can just call it a spec.2808

And what a spec does is it measures the transmitted light, so it measures the amount of light transmitted.2822

In this experiment, you, again, used your spinach leaves, and you extracted the photosynthetic pigments from them; and then, you had a beaker.2834

One of these contained fresh photosynthetic pigment extract from spinach leaves, and the other contained boiled extract.2843

You took, then, a tube of the boiled extract and placed one of them, so we have another fresh one and another boiled one here.2861

And you placed one of the fresh ones in the light, one of the fresh ones in the dark, one of the boiled ones in the light and the other boiled one in the dark.2871

So, you have fresh extract and boiled extract, one tube in the light of each and one in the darker of each.2882

Also, for a control, you are going to have a tube with no extract that you place in the light as a control.2888

You would set that up and then you, at time intervals of 5 minutes, take some extract,2894

place it in a cavat, and then, you put the cavat into your spec to measure the transmittance of light.2901

Now, remember that if photosynthesis is occurring, the result is that this electron is going to be transferred to the DPIP.2912

And the DPIP is going to be reduced.2922

So, what you would expect is that there will be a greater transmittance of light in the tube where there was2924

more photosynthesis because the DPIP would have turned colorless so light can pass through more easily.2930

And that is your expected outcome, is that the tube containing fresh extract in the light2935

- and that was placed in the light - had an increase in the transmittance of light.2950

So, as you took these samples from each of these tubes at 5 minute intervals and then, graphed them out,2957

you would see that over time, there is a good increase in the transmittance of light in the tube with fresh extract placed in the light.2963

Whereas, the one that was in the dark did increase but not as well and so on for each of these substances.2975

So, what you can conclude from this is that both the chloroplast - intact not boiled, intact working chloroplast - and light are necessary for photosynthesis.2984

In the 5th lab, you studied cell respiration, and you studied it in germinating and dormant pea seeds.2998

So, here we have the germinating seeds, and here, we have the dry or non-germinating dormant seeds. These grey substances are just glass beads.3007

Let's first just briefly review that during aerobic respiration, oxygen and glucose are used to produce ATP.3027

CO₂ is produced as a waste product, and in this lab, what you did is you measured oxygen consumption to determine the rate of cell respiration.3039

Recall that the overall equation for cell respiration is C₆H₁₂O₆.3063

This is glucose plus 6 molecules of oxygen yields 6 molecules of CO₂ plus 6 molecules of water plus energy in the form of ATP.3069

Therefore, the energy from glucose is used to generate ATP, which is the energy currency of the cell.3085

In this lab, what you did is you measured the oxygen consumption in both dry and germinating pea seeds at different temperatures.3092

And then, the glass beads alone were controlled.3100

Now, glass beads were added to the device containing the dry seeds because germinating seeds have absorbed water so they take up a greater volume.3102

And we need to have the same starting volume in all of these setups so we added glass beads just to make sure that we have the same starting volume.3113

And what this device is called is a respirometer, so let's look at the setup.3123

What we have here is a respirometer, and in the bottom, we have potassium hydroxide placed in the bottom.3135

And the reason that we have potassium hydroxide is that it is going to react with the CO₂ to form a solid.3143

And the reason that we want to do that is CO₂ is being produced during respiration.3153

But, I just want to focus on the change in volume of gas in the respirometer caused by the changed in oxygen.3157

I wanted to measure oxygen consumption, but if I have another gas being produced, CO₂,3165

I am not going to be able to just focus on what happened with the oxygen if it is being, then, replaced by CO₂.3170

Therefore, the KOH at the bottom is going to go ahead and react with the CO₂ to form a solid.3176

I should not put "to form CO₂". It should be to form a solid.3185

And that will leave me free to just focus on the consumption of oxygen.3188

And you were supposed to measure the change in the volume of gas in the respirometer at different temperatures,3194

so measure the change in the volume of gas at different temperatures such as 25°C and the cooler temperature of 10°C.3202

So, what you did is you first submerged the respirometers in a water bath.3216

As oxygen is consumed and the volume of gas inside the respirometer decreases, water will flow into the respirometer via the pipette.3221

So, what you did is you recorded the position of the water in the pipette every 5 minutes for 20 minutes.3231

What you should have found from your results is that the germinating seeds at the higher temperature had the greatest rate of O₂ consumption.3239

And this translates to the greatest respiratory rate because what we are using is oxygen consumption as a measure of respiration- respiratory rate.3266

The germinating seeds at lower temperature consumed oxygen but at a lower rate.3283

And the dormant seeds had a very low rate same with the beads, the control.3287

So, this was the lab in cell respiration, and again, you measured oxygen consumption with germinating and dormant seeds at different temperatures.3294

In laboratory 6, you focused on molecular biology.3307

Recall that when we talked in the bio technology lecture earlier in the course, I talked about transformation.3312

And transformation refers to the uptake of DNA by bacterial cells from their surroundings, and this can occur naturally.3319

However, in the lab, we can also induce the ability of cells to be able to transform or take up DNA.3328

And cells that have the ability to take up DNA like this are called competent cells, so we will make the cells competent.3337

In this lab, you may have used the heat shock method to make these cells competent.3345

And you have probably worked with E. coli, a very common bacteria used in the lab.3354

And the heat shock method involves heating and cooling, first cooling then, heating the bacterial cells.3360

And also adding calcium chloride in order to disrupt the cell membrane and cause the cells to be competent,.3367

So, what you started out with is the tube that contained E. coli and another tube with E. coli that you added plasmid to.3374

And the plasmid that you were using...remember plasmid is a ring of DNA found in3388

bacterial cells outside the main genome that can contain usually just a few genes.3394

And the gene that was on this plasmid is ampicillin-resistant meaning that this bacteria will not be stopped - its growth - or destroyed by ampicillin.3398

It will be unaffected by ampicillin. It can grow in the presence of ampicillin.3413

So, you had your tube with E. coli in. Then, you had your other tube with E. coli.3418

You added plasmid to it, and then, you kept these tubes on ice for 15 minutes.3423

Right after that, you placed them in a heated water bath. This is the heat shock portion.3432

So, you placed them on a heated water bath for 90 seconds, so just a burst of heat. Then, you put these cells back into ice.3437

The tubes were returned to the ice, and then, something called LB broth was added.3447

LB broth is just a nutrient-rich broth medium. It is a growth medium.3454

And the bacterial cells were then plated on plates, incubated overnight at 37°.3460

And you came back the next day and looked at your results, and what you would have used is 4 different plates.3472

All of these had the regular growth medium. There will be all growth medium on it, but two of the plates also had ampicillin.3478

And what you did is you plated, from the E. coli a lawn tube on a regular plate and on an ampicillin-containing plate.3490

And then, your other tube had E. coli plus plasmid on a plate without ampicillin and a plate with ampicillin.3501

And then, you observed what happened, and what you ended up with or should have ended up with is that3515

the E. coli that was on a plate without ampicillin there was just a solid lawn of bacteria.3525

The whole plate was just covered with bacteria. It did not matter if there was a plasmid or not.3535

It is on regular growth medium. It is going to grow well.3540

Then, you looked on the E. coli that was not transformed with the plasmid on the plate containing ampicillin.3545

There, you would have expected no growth because the growth of E. coli would be inhibited by this antibiotic.3551

Now, the E. coli with the plasmid, so here I would have had no growth.3560

It is plated, but it did not actually grow; so we will just write E. coli right here, but actually, there is nothing on the plate.3569

Now, here, a plated E. coli that was transformed with the plasmid.3577

When you make cells competent, some of those E. coli will take up the plasmid, some will not.3588

So, I will not see a lawn here. Instead, what I will see are individual colonies.3594

Those E. coli cells that contain the antibiotic-resistant plasmid can survive and grow in the presence of ampicillin.3598

And their offspring, they would multiply. The cells would divide and grow, and there would be discreet colonies; and then, I will count the colonies.3605

Maybe there is 15. Maybe there is 5.3615

So, what I ended up with, then, is seeing that, as expected, I had good growth on a regular plate.3620

Here is my control, that there is not growth on a plate with ampicillin where the bacteria were not transformed.3629

And then, I have E. coli where the growth was in colonies.3636

So, this also points to something that we see in terms of medicine and disease, and that is antibiotic resistance.3643

And that as people take antibiotics and bacteria become exposed to these antibiotics, the antibiotic resistant strains are, then, selected for.3652

And they can even pass this antibiotic resistance along to other organisms through these plasmids-3665

with antibiotic-resistant genes that formed on the plasmids.3672

Recall that in gel electrophoresis, what you do is take digested fragments and separate them out by size.3679

So, I am going to start out by reviewing the concept of restriction digest.3690

In the second part of the molecular biology lab, what you did is you performed a restriction digest on DNA,3693

and then, separated out the DNA using gel electrophoresis.3699

Recall that restriction enzymes are found naturally in bacteria. However, we use them frequently in the lab.3707

And they actually serve as a defense for bacteria by chopping up DNA from invading organisms such as bacteriophage.3718

These enzymes recognize specific DNA sequences called restriction sites.3725

So, they cut DNA at the restriction sites, and these sites are usually about 4 to 8 nucleotides long.3732

A very common one is EcoRI, and it actually cleaves DNA after the sequence CTTAAG.3741

It cuts after that sequence, and it cleaves both strands.3752

So, what you end up with, for example, is 5-prime. We will have GAATTC and then, 3-prime end and here, the complimentary strand.3757

We are going to have CTCAAG, and the enzyme EcoRI is going to cleave both strands.3769

So, we are going to end up with on the 5-prime end...actually, it cleaves right here.3782

Actually, so what you are going to end up with is, on the 5-prime strand, G.3790

And then, over here, because it is going to cleave right here before the G, you are going to end up with AATTC on this other strand.3797

Then, on the complimentary strand, you also got cleavage between the A and the G, which will give you on the 3-prime end,3810

you are going to have the CTTAA and then, over here, G, and then, that is the 5-prime end.3819

So, these are what is called "sticky ends", these overhanging ends like these, and you can ligate these fragments to other pieces of DNA.3827

And we talked about that with cloning earlier on in the lecture series.3836

Now, you are not going to do cloning in that molecular biology lab, but what you did do is from the bacteriophage lambda,3841

which is frequently used in laboratory studies, you digested phage lambda with restriction enzymes and then, separated out those fragments on a gel.3851

And you used actually two different restriction enzymes. Most likely in your lab, you used EcoRI.3863

And another possibility that a lot of labs use is HindIII.3870

So, take phage lambda DNA and digest with these two restriction enzymes. Then, you used an agarose gel to separate out the fragments.3874

So, again, recall from the biotechnology lecture that you can separate out DNA fragments according to size by running them through a gel.3885

And the gels are made of a polymer called agarose.3895

At the top of the gel are some wells, and you placed different DNA samples of each of the wells; and the lecture current is, then, applied.3900

So, there is the positively charged end, the anode and then, the negatively charged end, the cathode.3913

And because the phosphate groups on DNA are negatively charged, they are going to move towards positive charge.3923

So, if I placed DNA in these wells, and then, I immersed the gel in water, or not water actually, but in a solution,3931

and then, I applied an electric current, the DNA is going to be attracted to this positive end.3939

After the current is turned off, a dye is added, and the dye binds to DNA.3947

And when you put the gel under fluorescent light, you can see a banding pattern.3953

The DNA moves down, and it moves at different rates depending on the size. Then, you get these different banding patterns.3957

In the lab, again, you used EcoRI and HindIII, and the DNA digested by the EcoRI would be placed in one well and by HindIII in another.3970

And HindIII, those fragments would be of known size. The fragment is created by digestion with EcoRI are going to be of unknown size.3981

And after running your gel, adding the dye, observing under ultraviolet light3998

- you should observe under ultraviolet light - then, you are going to see these fluorescent bands.4006

And you can determine the distance that these different bands migrated along the gel.4011

And you could plot those on a graph along with the distances that the fragments of known size from the HindIII digests travelled.4017

By having your known sizes of fragments for HindIII and measuring their distances,4026

you can use that to determine the fragment sizes in units of base pairs for the EcoRI fragments.4033

So, in this lab, what you did is you performed a restriction digest.4040

And then, you ran the fragments on an agarose gel and determined the size of the EcoRI digested DNA fragments.4044

In laboratory 7, you studied the genetics of organisms using drosophila melanogaster, which is a fruit fly that is very frequently-used in genetic studies.4053

Drosophila are well-suited to genetic studies, and this is because they can be easily bred; and they produce many offspring.4072

It only takes about 12 days to produce a new generation. They have distinct easily-observable traits.4081

Also, what is important in breeding studies is that the males can easily, well, fairly easily be distinguished from the females.4092

And finally, drosophila have only 4 pairs of chromosomes.4099

So, just to start you out with a little bit of the life cycle of drosophila because when you are doing these crosses, you need to understand the life cycle.4104

And the life cycle begins when the female lays the eggs. After about a day, these eggs hatch into larva.4112

These are also called instars, and there are three instar stages.4127

After the first two, molting occurs. Then, in the third instar stage, the larva develops into the pupa.4133

The pupa becomes encased in a hard covering or cocoon.4147

And over a period of about 4 days, it undergoes metamorphosis and then, emerges in its adult form.4156

Mating can occur as soon as 12 hours or so after maturity, and the lifespan of the adult is about 1 month.4174

And in this lab, there were several types of crosses that you studied.4187

Recall, when we talked about Mendelian genetics, I mentioned a monohybrid cross.4191

And in a monohybrid cross, it is a cross between individuals who are heterozygous for one trait.4198

Monohybrid cross is a cross between individuals who are heterozygous for one trait.4210

In a dihybrid cross, you study individuals who are heterozygous for two traits, so a cross between individuals who are heterozygous.4216

And finally, in sex-linked crosses, you were studying traits that are carried or determined by alleles that are located on the X chromosome.4236

Although, a sex-linked trait could be on a Y chromosome, there are not many.4248

And what you are studying here is the sex-linked crosses with traits carrying the X chromosome.4251

And you may have studies various different traits in your lab.4257

Some typical ones you study in drosophila are eye color, and in drosophila, red eye color is...you might have studied red and white eye color.4260

And white is a sex-linked recessive trait, whereas, red is dominant.4271

Wing type: you might have studied flies with wild type wing type,4279

which is autosomal dominant and versus vestigial wings, which are autosomal recessive.4286

This lab procedure required you to be able to first, identify male versus female flies and to handle4298

the flies because in the lab, you needed to be able to immobilize the flies using various methods.4305

One method is chilling them.4312

And just recall, when you are working with genetic crosses, the initial generation that you start out with was the parental generation P.4315

The offspring of that generation, the first filial generation are the F1 generation.4324

And then, from the F1 crosses, you ended up with the F2 or second filial generation.4331

So, you are observing particular traits and doing crosses for those traits.4338

And after each cross, you should have observed and recorded the trait that you were studying.4342

Then, you compiled that data and used it to determine the mode of inheritance.4350

So, you were crossing wild type flies with flies that were mutant for a particular trait.4356

For each generation, you should have observed, collected and recorded data for about 100 to 200 flies.4362

Once you did this, you also then, used your data to perform a statistical analysis using what is called the Chi square analysis or Chi square test.4372

So, the Chi square analysis tests what is called the null hypothesis.4384

The null hypothesis states that there is not a difference between the expected data and the observed data.4392

In other words, what we are trying to determine are deviations in expected versus actual findings due to chance,4401

or if there are deviations from expected versus actual outcome.4416

Is it is because there is a significant finding that my hypothesis in the experiment was not correct?4422

Or was there a flaw in the way I designed or implemented the experiment?4428

To help you understand this, let's go back to talking about pea plants.4435

Recall that one of the traits that Mendel studied in pea plants was height, and in pea plants, tall is dominant.4438

Big T is tall, and little t is dwarf or short pea plants; and that was recessive.4448

So, let's say I do a monohybrid cross, and also in this lab, it is important to recall your skills in using Punnett squares.4455

So, I do my cross, and what I would expect if I performed the experiment rigorously and have a4466

large number of plants that what I am going to end up with is 1, 2, 3 tall plants for every short plant.4473

Let's say I bred 100 plants. What I expect is that I am going to have 75 tall and 25 short.4485

Let's say I breed them, and I end up with 73 that are tall and 27 that are short, so this is my expected: 75 tall and 25 short.4495

But, what if my actual outcome is 73 tall and 27 short?4509

Now, does this mean I messed up in how I did my experiment? Does it mean that my hypothesis that the tall allele is dominant to the short?4519

Or does it just mean this was a reasonable deviation from the expectation due to chance?4530

Well, that is what is a Chi square test can tell us.4537

It can tell us if deviations from the expected outcome are due to chance or due to something4539

else such as experimental data or a hypothesis that was not then proven.4545

So, the Chi square formula is shown here.4553

And this is the sum of the square difference between the observed data and the expected data that is divided by the expected data.4558

In this lab, you did a Chi square analysis on your data from these fruit fly crosses.4577

This chart here, this table, shows degrees of freedom.4584

And the degrees of freedom or DF are the number of phenotypes, the number of classes of phenotypes minus 1.4587

Now, here, I had two phenotypes: tall and short. For flies, I might have red eyes and white eyes for the studies that you did.4603

So, looking here at my pea plants, I will say "OK, I had two possible phenotypes, tall and short, minus 1 equals 1 degree of freedom".4611

Now, let's say I did a study on flies or pea plants, and my degree of freedom was 1.4625

And I did my Chi square analysis, and I got Chi square equals 4.51.4634

What do I do? Well, I look at this table that gives critical values, and what you should do is look at the probability row of 0.05.4643

And then, look for the correct degree of freedom.4658

My degree of freedom is 1, so I am going to go ahead and look here, and my Chi square value is 4.51.4661

If the Chi square value is greater than the critical values...what this is, is a chart of critical values.4671

If the Chi square value that you calculated is greater than or equal to the critical value, you reject the null hypothesis.4691

OK, so, critical value, I have 1 degree of freedom, and I want to use the probability of 0.05, and what I find is the number 3.84.4715

So, I say "OK, is 4.51 greater than or equal to 3.84?". Yes, it is, so I am going to reject the null.4728

If I reject the null, what I am saying is the null says there is not a difference between the expected data and the observed data.4738

And by rejecting the null, what I am saying is, either my hypothesis was wrong, that my hypothesis that the pea plants,4748

that tall is dominant, and short is recessive, autosomal dominant, autosomal recessive is wrong.4758

Or I messed up in the experiment because I have rejected the null.4763

If I accepted the null, if my critical value was less, or if my Chi square was less than the critical value,4765

and I accepted the null, then, I am saying that the data fit my expected findings.4773

I am confirming my initial hypothesis.4779

Now, since we are using the rho P = 0.5, what we are saying is that our results are significant to the probability of 0.05,4782

which is pretty standard for what you want to use in science.4792

So, what this is saying is that if I rejected the null based on my 4.51 value with the probability of 0.05,4799

then, I am saying that 5% of the time, this data would have occurred even though the null is correct.4807

Only 5% of the time would this data occur even though the null is correct.4815

So, this is a somewhat complicated concept.4821

But, what you really need to know is just to understand what this formula is telling you, understand what the null hypothesis states and then,4824

how to determine your degrees of freedom and interpret the Chi squared value and determine whether or not to reject or accept the null hypothesis.4831

In lab 8, you studied population genetics and evolution, and remember that the Hardy-Weinberg equation4843

can be used to predict the frequencies of alleles and genotypes in populations that meet certain conditions.4850

And recall what these certain conditions are: mating is random; no mutations occur; there is no natural selection occurring,4856

so it is a non-evolving population; you have a very large population, and that there is no immigration or emigration, in other words, the population is isolated.4872

And in the lab, we can get pretty close to these conditions. We can never control that there is no mutations or anything.4889

But, working with pea plants or fruit flies, we can get pretty close to these conditions.4896

Here is the Hardy-Weinberg equation, and recall that p² is the frequency of the homozygous - these are all frequencies here - recessive genotype.4903

Actually recessive is q². Recessive is conventionally q².4922

p² is the homozygous frequency. These are all frequency of the homozygous dominant genotype.4927

2pq is the frequency of the heterozygous genotype, and these all have to add up to 1.4939

Those are the only three possibilities: homozygous dominant, heterozygous, homozygous recessive add up to 1.4950

The frequency of the dominant allele is p plus the frequency of the recessive allele is q, and those equal 1.4956

So, it is the frequency of the dominant allele and the frequency of the recessive allele.4966

In this lab, what you did is you used your class as the population being studied.4982

The trait that you studied was the ability to taste PTC, and PTC is a chemical that has a bitter taste for some people and no taste for other people.4988

The ability to taste PTC is dominant. Not tasting it, the inability to taste it is recessive.5001

So, members of your class were given papers with PTC on them, and that is to taste them.5015

And then, what you did is recorded the data and used the Hardy-Weinberg equation to determine the5020

frequency of the allele for the ability to taste PTC in the sample population, which is your class.5027

And you compared these to the frequency of the allele in the population in North America.5035

Lab 9: in this lab, you studied transpiration, and transpiration refers to the loss of water from plants via evaporation.5045

Again, this is reviewed in detail in the plants lecture.5056

But, you should recall that as water will exit the leaf through the open stomata as it moves towards the drier air outside.5059

And that is an area of lower water potential compared to inside the plant.5069

The water evaporates, and it creates negative pressure at the top of the plant.5073

Because of the properties of water, a column is created that allows water molecules to be pulled5079

up all the way from the root through the xylem up and out the leaf, the stoma in the leaf.5088

You should also review the properties of water. Remember that water is cohesive, so water molecules stick together.5099

Because it is highly cohesive, this column of water, the water molecules, sort of, pull each other up the plant one after the other.5109

As one molecule of water is lost from a leaf, that is transmitted all the way down, and one molecule is pulled up from the root.5119

You can think of one molecule of water pulling the other up in the chain.5127

Water is also very adhesive. Adhesive refers to the property that one type of molecule sticks to other substances.5133

So, water sticks to the side of a glass.5140

Water also sticks to the walls of the xylem, so it sticks to each other and the walls of the xylem that helps the capillary action.5143

And these properties are due to the hydrogen bonding between water molecules.5151

So, properties of water are very important for transpiration.5156

Now, what you studied in this lab is transpiration under various conditions.5161

And you used a potometer to study, to measure transpiration on a type of plant you might have used in patients or you might have used bean seedlings.5167

And you tested transpiration under various conditions.5182

For example, you simulated wind by using a fan, varied temperature, light or under more humid conditions.5188

So, the setup here was that you had a plastic piece of tubing and a pipette, and you submerged the pipette in the plastic tubing in water.5197

They filled up with water, and then, you placed the tip of the pipette into the plastic tubing.5209

So, you have your plastic tubing, and then, you have the pipette in here.5215

And you, then, also placed the plant here in the other end of the tubing, so you have your plant here.5221

Then, this whole setup was clamped to a ring stand to hold it up.5233

The level of the water in the pipette is recorded at 1 minute intervals for 30 minutes under the following conditions.5242

So, the conditions you tested were just ambient conditions in the room, so room conditions,5251

so the temperature, light, etc. in your classroom with the fan blowing on the plant, which simulated wind.5257

After misting the plant and then, covering the plant with the plastic bag to hold that water/moisture in,5269

so this is causing the conditions to be humid, so under humid conditions and finally, under strong light.5277

Now, we actually did not focus on temperature in this lab, although, temperature can affect transpiration.5291

We wanted to focus more on light, so what you needed to do in the lab is to use what is called a heat sink.5299

Because if you are putting strong light in the plant, you do not know if it is the light or the heat that is affecting transpiration.5304

So, what you do is between the light source and the plant...so you have your light source here.5310

You have your light bulb, and you placed a bowl of water.5314

And that is called a heat sink because what it is going to do is it is going to absorb the heat that the light will pass through.5320

The light will pass through it, but the heat will be absorbed.5327

Now, when you measured transpiration under these conditions, here is what you should have found that increased in transpiration.5330

The factors that increased transpiration are the light and the wind or the fan, in this case.5341

Factor that decrease transpiration is humidity, so the factor that decrease transpiration is humidity.5349

After this, you may have determined the surface area of the leaves, so you remove the leaves from the plant.5363

You place those leaves on graph paper and trace them.5369

That will allow you to determine the area of the leaves and therefore, water loss per square meter of leaf surface.5373

So, water loss per square meter of leaf surface, and this is usually put in units of milliliters of water loss per meter squared of leaf surface.5386

And then, you could display those results in graph form showing water loss on the Y axis and time on the X axis.5401

In the second part of this lab, you studied the structure of the stem.5411

So, you can take the stem from the plant in the first part of this exercise and cut that into sections and view it under the microscope.5416

Now, tissue and cell types that you should have been able to identify, stem structure, cell types: parenchyma, collenchyma, sclerenchyma,5426

as well as cell types: xylem, phloem, and epidermis.5447

And if you need review on this, you should review under the plant lectures where there5455

were detailed discussions on the structure of the various parts of the plants in section 9.5459

Lab 10 focused on the physiology of the circulatory system, and in this laboratory, a sphygmomanometer is used to measure blood pressure.5468

So, this is the typical blood pressure cuff. It has a cuff on it that is just part of the instrument that5479

you have probably had when you have gone to the doctor and had your blood pressure taken.5485

And a cuff is wrapped around the upper arm and inflated in order to temporarily block off or occlude the flow of blood through the artery.5489

Now, to understand how this works, you need to understand the sounds that blood makes going through the artery.5500

Usually, actually, blood does not make noise as it flows through arteries. It is silent.5506

However, if you block the flow of the blood, then, release it, the flow of the blood going through these compressed arteries is turbulent.5512

And it makes sounds called Korotkoff sounds, and there are actually five different types of Korotkoff sounds.5521

When blood pressure is being assessed, a stethoscope is used to listen to the sounds of the blood passing through the artery in the inside of the elbow.5528

So, as the pressure on the cuff is decreased and the blood is now allowed to pass through the artery, you can hear these Korotkoff sounds.5537

Here is how it works. You pump air into the cuff to about a pressure of 180.5545

And that will occlude the artery so that blood is going to stop flowing through it temporarily.5553

Then, you are listening with your stethoscope, and you open up the valve; and you let air out of the cuff slowly.5559

When you first hear sounds, that is the systolic pressure. So, blood pressure is given as two numbers: 120/80.5567

And the systolic is the top number, and this is the pressure given in millimeters of Mercury at which the Korotkoff sounds begin.5577

Then, you are going to hear the sounds go through different phases get, kind of, louder and softer.5593

And then, when you heard the last sound...through decrease in the pressure, you are listening, listening, decrease in pressure.5596

When you heard the last sound, the last full heartbeat, that it the diastolic pressure.5602

So, the diastolic pressure is the bottom number. This is diastolic, and that is the pressure at which the Korotkoff sounds end.5606

A typical blood pressure for healthy adult is about 120/80. Many factors affect this so: age, gender, fitness level.5623

And the presence of disease can all affect blood pressure.5639

Now, we talked earlier in the course about hypertension.5643

Hypertension refers to a condition in which the blood pressure is elevated, and hypertension is a risk factor for other problems such as stroke.5651

In the next part of this lab on physiology, you measured your lab partner's blood5670

pressure while reclining and then, when standing up immediately after reclining.5676

So, you had your lab partner recline. You measured your lab partner's blood pressure.5683

Then, your lab partner stood up, and you measured it again.5688

This change in systolic pressure of going from reclining to sitting can be used as a measure of fitness.5691

For example...so, you were given a chart, too, that assigns fitness points.5699

And by doing these different tests on blood pressure and pulse, you would be able to assess your fitness level.5703

So, a rise in systolic blood pressure from reclining to standing of 8 or more points could count as 3 fitness points.5710

No rise would be 1 fitness point. A large fall in blood pressure relatively would be negative points.5718

So, that was the second part of the lab.5727

In the third part of the lab, what you did was looked at pulse rate. You assessed pulse rate.5729

And this is also a measure of fitness because a lower resting pulse rate generally correlates5736

with better fitness because people who are in good shape usually have a higher stroke volume.5743

OK, so what is a stroke volume? Stroke volume, recall when we talked about the cardiovascular system,5750

is the amount of blood pumped by the left ventricle per contraction, so in one contraction.5759

So, the heart is pumping really efficiently. It is pumping more blood with each beat of the heart.5780

The pulse rate can be lower.5786

If the heart is not as efficient, you would need a faster pulse rate to get enough blood through the system.5788

So, people who are in really good shape tend to have a lower pulse rate.5792

What you should have done is you can measure resting pulse and also look that up on the fitness chart.5796

Finally, what you did is studied the baroreceptor reflex.5804

Recall that there are special neurons called baroreceptors located in the aortic arch, in the carotid sinuses and in other areas of the body.5814

These baroreceptors monitor blood pressure, and this is yet another mechanism of homeostasis- maintaining a constant internal environment.5826

And if the baroreceptors recognize that there has been a decrease in blood pressure, for example, the information is conveyed to the brainstem.5837

And through the autonomic nervous system, blood pressure can be returned to normal,5847

for example, through increasing the pulse rate or through constricting blood vessels.5853

So, what you did in this part of the lab to measure or observe the baroreceptor reflex is to first measure your partner's pulse when they are reclining.5860

Then, you measured your partner's pulse immediately when they go from reclining to standing up.5873

And you would expect an increase in pulse rate with this change in position.5879

So, this, again, is initiated by the baroreceptor because standing up quickly can cause a drop in blood pressure, which is sensed by the baroreceptors.5885

And the heart rate is increased and also vasoconstriction can occur, so these all help to maintain the blood pressure.5895

In the last part of this lab, you studied temperature and metabolism in ectotherms.5907

You studied the organism Daphnia magna. These are also known as water flees.5917

They are actually a crustacean, and they are an ectotherm.5924

In ectotherms, as the temperature in the environment increases, so does the rate of many of the physiological processes.5931

And the Q₁₀ effect refers to the fact that in some animals, for every 10°C increase in temperature,5941

within a certain range, there is a doubling of the metabolic rate.5950

And you have probably observed that a lot of reptiles like lizards are much more active in warm weather.5955

They are, kind of, scurrying around, and then, in cold weather, they are just inactive. They all, sort of, sit there.5960

So, Daphnia is an ectotherm, and what you will do is observe it under the microscope.5966

And you can take its heart rate because the covering on Daphnia is transparent.5973

And so, you can see the heart, the movement caused by the heartbeat.5978

So, you will place the Daphnia into a depression slide, a slide that has a little depression in it that contain...5983

Place water in here and the Daphnia, and then, initially, you will have water that is 10°C, and you will count the heart rate.5992

So, you count the heartbeat for 10 seconds and multiply that by 6 to get the heart rate in beats per minute of the Daphnia at 10°C.6004

Then, you can repeat this using warmer water. Again, count the heartbeat for 10 seconds.6016

Multiply it by 6, and what you should find is that the heart rate increases at the higher temperatures in water fleas in an ectotherm.6022

Finally, you should have graphed the data with temperature on the X axis and heart rate in beats per minute on the Y axis.6037

In lab 11, you studied animal behavior. Recall that ethology is the study of animal behavior.6053

And a key part of animal behavior is how an animal reacts to stimuli and to their environment.6060

Animals behave in a way that helps them find a hospitable habitat, that helps them find food, avoid predators, mate and raise their young successfully.6069

One of the most basic animal behaviors is movement, and recall that there are a couple types of movement you should be familiar with.6084

One is kinesis. Kinesis is the simplest method that an animal uses to get itself to a better location or desired location.6092

So, the animal simply moves faster when environment or stimulus is less desirable.6104

It just, kind of, moves around, not in any particular direction, moves faster, and then, it moves slower when it is in a desirable location.6123

Taxis is actually directed movement. Taxis is directed movement- movement towards a stimulus.6141

So, it might be towards food or an insect running into a dark place or away from a bright light, a predator running towards prey.6149

It is directed movement. It will be towards a stimulus.6158

In this experiment, or this actually was an observation. It was not a true experiment, but it was an observation.6163

So, you did an observation of pill bugs, and pill bugs are a type of crustacean. They are an isopod.6170

They have a hard exoskeleton, and they are commonly found under rocks and in other damp dark places, so they like damp dark environments.6179

In this lab, you had two containers like petri dishes for example, and you made a small opening between the two.6188

So, they are connected. The pill bugs can move between the two environments.6196

You should have placed one filter paper in each side: a dry piece of filter paper and a wet piece of filter paper.6199

Then, place five bugs on each side. Ten pill bugs, place five on each side.6207

You should have observed that the pill bugs tended to move around more on the dry side.6216

The ones on the wet side, they are, kind of, happy there, so they will hang out by the damp filter paper.6221

After some time, and it can even take hours, but eventually, most or all of the pill bugs will end up on the wet side.6229

Some of the pill bugs, if you have not secured the filter paper, will even try or maybe unsuccessfully crawl under the filter paper.6239

From this, you can conclude a couple of things. One is that pill bugs favor damp locations.6247

Now, light is not supposed to be part of the experiment.6253

But, if you do see your pill bugs trying to get under this paper, you might also conclude that pill bugs favor dark locations.6257

the second thing is, is this movement due to taxis or kinesis.6264

Now, if the pill bugs had a way of sensing the moisture on this wet side even though they were over on the dry side,6270

and if noticed that they had directed movement, they really appeared to just be going that way purposely, then, I might say it is taxis.6278

However, from watching the pill bugs seeing that they tended to move more and just,6286

kind of, move randomly until they got over here where they just moved the less,6291

that is probably what you saw, and you probably, then, concluded that what you are witnessing is kinesis.6295

So, that was the first part of the animal behavior lab. You studied movement.6304

In the second part, you studied mating behavior of the fruit fly.6307

And all you needed to do in this part was carefully observe the courting sequence,6312

which starts with orientation with the male fly approaching the female.6318

The male begins with the courting song that it vibrates its wings and continues on with6323

the male attempting to copulate and the female either accepting or rejecting the male.6329

The final lab is lab 12, and this is a study of dissolved oxygen and aquatic primary productivity.6338

Now, in aquatic environments, dissolved oxygen in the water is a critical abiotic factor for organisms that live in the water.6347

Photosynthetic organisms like plants and phytoplankton generate oxygen during the day.6356

So, photosynthesis during the day increases dissolved oxygen.6362

Oxygen is, then, depleted during decomposition by bacteria, respiration by worms, crustacean, fish and other organisms.6368

So, photosynthesis during the day increases the dissolved oxygen, and respiration reduces the dissolved oxygen.6378

I do want to note that in addition to biological activity, dissolved oxygen levels are dependent on factors like temperature and salinity in the water.6386

Mixing also affects the dissolved oxygen.6396

Water that is better aerated, better mixed that has waves, waterfalls, rapids generally has a higher oxygen level.6400

Now, in the first part of this lab, what you did is measured the amount of dissolved oxygen at various temperatures.6409

And this should be done in water that does not...6424

it should be clean water without organisms in it because the dissolved oxygen will be affected by those biological organisms.6428

Now, saturation is the maximum amount of oxygen that can be dissolved under specific conditions.6435

For example, at a certain temperature, there might be one mark saturation level whereas, another temperature, it is a different level.6464

pH, salinity, these can all, again, affect saturation. The saturation level can be increased by aerating the water.6471

In this part of the lab, when you measure the dissolved oxygen at various temperatures,6480

you will have found that less oxygen can be dissolved at higher temperatures.6485

For example, at 5°C, the saturation is about 12 milligrams per liter, whereas, at 30°C, saturation is more like 7 milligrams per liter.6496

Now, in some cases, it is more important to note the DO level as a percent of the saturation. It is 10% or 50% of the saturation.6512

Other times, you want to know the actual level of dissolved oxygen.6525

One key point to understand is that under certain conditions like high temperature, you can have a higher percent saturation.6529

But, the actual dissolved oxygen is less than it would be in other conditions with a lower percent saturation.6536

In an earlier section of the course under ecology, we talked about the primary productivity of an ecosystem.6550

And primary productivity is the rate at which photosynthetic organisms produce organic compounds, so that is right here.6557

There were two components of primary productivity that we discussed.6569

The first is gross primary productivity, and this is the gross rate or the rate of all organic production by photosynthesis.6572

Net primary productivity is the primary productivity after cellular respiration by the photosynthetic organism uses up a portion of that organic matter.6584

In other words, net primary productivity or NPP equals gross primary productivity minus respiration.6602

Now, there is no way to directly measure this productivity, the mass of organic compounds that are being produced by photosynthetic organisms.6611

So, what we do is we use a surrogate measure when we are trying to study an ecosystem and figure out productivity.6621

In this case, in an aquatic ecosystem, oxygen production during the day is a measure of net primary productivity.6631

So, oxygen production is a measure of NPP.6642

This is because during the day, the plants are producing oxygen, and at the same time, respiration is occurring.6654

At night, photosynthesis is not occurring, so all you have got left at night is respiration.6663

By measuring the net productivity and respiration, you can calculate gross primary productivity because you have two of the unknowns in the equation.6670

There are other ways to measure productivity. Oxygen production is one measure.6684

You could also use CO₂ consumption or sugars produced.6690

Actually producing or actually measuring oxygen produced would take sophisticated apparatus and instruments and things.6698

But, what you can measure reasonably is the change in dissolved oxygen in the water.6708

Change in dissolved oxygen is a good indicator in an aquatic ecosystem of oxygen production.6715

So, in the experiment you did with this lab, what you did is took samples of lake water,6735

preferably water that is very rich in algae and other photosynthetic organisms.6741

You exposed it to light, and you used dissolved oxygen to measure and determine the gross and primary productivity.6746

You also may have simulated the variation in light that occurs at various depths of6757

the lake by varying the amount of light that was allowed to enter each sample.6763

You also used a completely dark sample to determine respiration, so lake water samples, one sample in the dark or exposed to various levels of light.6770

You, then, took the fine measurements, so this, you would, then, have taken the following measurements.6791

One: the initial dissolved oxygen in the various samples at the beginning of the experiment.6797

Two: after 24 hours in full light, you would measure the dissolved oxygen, so that is the final DO/dissolved oxygen.6806

The other measurement that you would take is, again, after 24 hours.6821

OK, this is the sample in full light. After 24 hours in dark, you would measure the dissolved oxygen.6831

Finally, you would also, after 24 hours, have a sample that is exposed to varying levels of light. You would measure the dissolved oxygen.6840

Now, let's go over what these results mean, what these measures and these numbers are going to mean.6855

The change in DO in the dark sample: photosynthesis will not be occurring in there or very, very little.6863

So, the change in dissolved oxygen in a dark bottle is a measure of respiration.6873

In other words, the dissolved oxygen initial minus the dissolved oxygen final in that dark bottle tells us respiration.6884

The change in DO in the light bottle - the bottle exposed to full light - is a measure of net productivity.6898

The DO final minus dissolved oxygen initial gives us NPP- net primary productivity.6910

Now, recall that gross primary productivity, well, if you just rearrange these two equations, what you will see is that gross primary productivity,6922

which is what we are trying to figure out, is the final light DO minus the final dark DO.6930

And so, what you are doing is you are taking the measures that you have.6943

And you are using these equations and your knowledge of the fact that gross primary productivity equals net primary productivity plus respiration.6947

So, you know that gross primary productivity equals net primary productivity plus respiration.6960

So, if you, then, substitute using these equations for net primary productivity and respiration,6965

you can just take final light DO minus final dark DO to get gross primary productivity.6973

Now, remember, we are not directly measuring primary productivity.6979

What we are measuring is dissolved oxygen as an adequate surrogate measure of primary productivity.6985

Now, in this first example, a Chi square analysis is performed on data gathered for a monohybrid cross for the trait of seed shape in pea plants.6992

The hypothesis being tested in the experiment is that round seed shape is inherited in an autosomal dominant manner.7001

And wrinkled seed shape is autosomal recessive.7009

Chi square for the data is determined to be 4.9. Use the table below to explain the significance of this value.7014

Now, we are looking at this chart of critical values, and we first have to determine the degrees of freedom.7027

And recall that the degrees of freedom, DF, is equal to the number of classes of phenotype minus 1.7034

Here, we have two classes. We have seed shape.7046

OK, what are the classes under that? Round and wrinkled.7049

So, that is two classes minus 1 equals 1, so the degree of freedom is 1, so I want to focus on this column.7052

Now, for science, what I wanted... the minimum that I want to see here - this actually should be 0.001 - is 0.05.7060

So, I want to look at the p = 0.05, and I am finding that I need to look at my critical value of 3.84.7072

So, the critical value that I am going to study here or use is 3.84. The Chi square that I have is 4.9.7083

Now, remember that if the Chi square value is greater than or equal to this critical value, then, I will reject the null.7093

And 4.9 is greater than 3.84, so I will reject the null hypothesis.7104

Now, remember that the null hypothesis states that there is not a difference between the expected data and the observed data.7113

So, if I reject the null, what has happened is either my hypothesis was wrong, that the wing type,7125

or excuse me, seed shape, is not inherited as autosomal dominant for round and recessive, for wrinkled.7135

Or my experiment was incorrectly performed.7149

So, by rejecting the null, either I am saying "OK, this initial hypothesis, it is not correct", or I did something wrong in my experiment.7153

I did not have enough plants that I accounted, or I accounted wrong; or something happened.7161

Example two: identify the stage of mitosis that each cell below is in.7169

So, we have two different stages here, and in this stage, I noticed that the chromosomes are lined up in the center along the metaphase plate.7175

And I also see the spindle apparatus is formed. That tells me it is metaphase.7189

Remember prophase, metaphase, anaphase and then, telophase.7194

Next, what I see are the two cells. There is a cleavage furrow here.7205

They are getting ready to split off. I have four chromosomes in each side.7211

The spindle apparatus is broken down, so this is the very end of mitosis. This is a cell in telophase.7219

Example three: an increase in which of the following factors would not be expected to cause an increase in the rate of transpiration in a plant?7229

Notice this says "would not be expected to increase transpiration".7236

Well, wind actually does increase transpiration, so that is not correct.7241

Humidity: well, remember when we misted the plant, the rate of transpiration went down.7248

Humidity actually decreases transpiration, so it would not be expected to increase it. I like that so far.7253

Temperature: increase in temperature increases transpiration, and the increase in light increases transpiration. Therefore, the answer is B.7260

Example four: in a population, 16% of individuals have blue eyes. The rest of the population has brown eyes.7283

Blue eyes are autosomal recessive, and brown eyes are autosomal dominant.7292

Use the Hardy-Weinberg equation to determine the percentage of the population that is heterozygous for eye color.7297

OK, so, the Hardy-Weinberg equation, recall, is p² + 2pq + q² = 1.7308

And what they are telling me here is that 16% of the individuals or 0.16 have blue eyes.7321

Now, blue eyes are autosomal recessive, so if somebody has blue eyes, they are going to have to be homozygous recessive.7328

And q² is the frequency of the homozygous recessive genotype.7337

So, the individuals with blue eyes, I would call them, little b-little b because that would be blue eyes.7348

And little b-little b is homozygous recessive. That is equal to 0.16.7353

Now, what I am being asked to determine...and now, here is what I have. What do I need?7360

I want to figure out the percent of the population that is heterozygous. The frequency of the population that is heterozygous is equal to 2pq.7365

So, here is what I want to determine. I want to figure out 2pq.7384

I know q², so let's figure out q.7387

I can figure out q by just taking the square root of both sides, so the square root of q² and the square root of 0.16.7391

It is going to give me q = 0.4.7399

What I also know is that the frequency of the dominant allele plus the frequency of the recessive allele has to equal 100%, so p + q = 1.7402

The percentages have to equal 100%, so p + q = 1.7412

I know q, so I am going to substitute q in. I am going to substitute 0.4 in here: p + 0.4 = 1.7416

Solving for p, p = 0.6.7427

Now, I have got P. I have got q.7430

I can figure out that 2pq = 2 x p, which is 0.6 x q, which is 0.4, so 2pq = 2 x 0.24 = 0.48 or 48%.7433

Therefore, the percentage of the population that is heterozygous for eye color is 48%.7456

That concludes this laboratory review here at Educator.com.7464

Thanks for visiting.7468