AP Sample 5 Lab 5 Cellular Respiration

 

 

Lab 5     Cellular Respiration

 

 

Introduction:

 

Cellular respiration is an ATP-producing catabolic process in which the ultimate electron acceptor is an inorganic molecule, such as oxygen. It is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Carbohydrates, proteins, and fats can all be metabolized as fuel, but cellular respiration is most often described as the oxidation of glucose, as follows:

C6H12O6 + 6O2 → 6CO2 + 6H2O + 686 kilocalories of energy/mole of glucose oxidized

Cellular respiration involves glycolysis, the Krebs cycle, and the electron transport chain. Glycolysis is a catabolic pathway that occurs in the cytosol and partially oxidizes glucose into two pyruvate (3-C). The Krebs cycle is also a catabolic pathway that occurs in the mitochondrial matrix and completes glucose oxidation by breaking down a pyruvate derivative (Acetyl-CoA) into carbon dioxide. These two cycles both produce a small amount of ATP by substrate-level phosphorylation and NADH by transferring electrons from substrate to NAD+ (Krebs cycle also produces FADH2 by transferring electrons to FAD). The electron transport chain is located at the inner membrane of the mitochondrion, accepts energized electrons from reduced coenzymes that are harvested during glycolysis and Krebs cycle, and couples this exergonic slide of electrons to ATP synthesis or oxidative phosphorylation. This process produces 90% of the ATP.

Cells respond to changing metabolic needs by controlling reaction rates. Anabolic pathways are switched off when their products are in ample supply. The most common mechanism of control is feedback inhibition. Catabolic pathways, such as glycolysis and the Krebs cycle, are controlled by regulating enzyme activity at strategic points. A key control point of catabolism is the third step of glycolysis, which is catalyzed by an allosteric enzyme, phosphofructokinase. The ratio of ATP to ADP and AMP reflects the energy status of the cell, and phosphofructokinase is sensitive to changes in this ratio. Citrate and ATP are allosteric inhibitors of phosphofructokinase, so when their concentration rise, the enzyme slows glycolysis. As the rate of glycolysis slows, the Krebs cycle also slows since the supply of Acetyl-CoA is reduced. This synchronizes the rates of glycolysis and the Krebs cycle. ADP and AMP are allosteric activators for phosphofructokinase, so when their concentrations relative to ATP rise, the enzyme speeds up glycolysis, which speeds of the Krebs cycle.

Cellular respiration is measure in three manners: the consumption of O2 (how many moles of O2 are consumed in cellular respiration?), production of CO2 (how many moles of CO2 are produced in cellular respiration?), and the release of energy during cellular respiration.

PV = nRT is the formula for the inert gas law, where P is the pressure of the gas, V is the volume of the gas, n is the number of molecules of gas, R is the gas constant, and T is the temperature of the gas in degrees K. This law implies several important things about gases. If temperature and pressure are kept constant then the volume of the gas is directly proportional to the number of molecules of the gas. If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume. If the temperature changes and the number of gas molecules is kept constant, then either pressure or volume or both will change in direct proportion to the temperature.

Hypothesis:

 

The respirometer with only germinating peas will consume the largest amount of oxygen and will convert the largest amount of CO2 into K2CO3 than the respirometers with beads and dry peas and with beads alone. The temperature of the water baths directly effects the rate of oxygen consumption by the contents in the respirometers (the higher the temperature, the higher the rate of consumption).

 

Materials:

The following materials are necessary for the lab: 2 thermometers, 2 shallow baths, tap water, ice, paper towels, masking tape, germinating peas, non-germinating (dry) peas, glass beads, 100 mL graduated cylinder, 6 vials, 6 rubber stoppers, absorbent and non- absorbent cotton, KOH, a 5-mL pipette, silicon glue, paper, pencil, a timer, and 6 washers.

 

Methods:
Prepare a room temperature and a 10oC water bath. Time to adjust the temperature of each bath will be necessary. Add ice cubes to one bath until the desired temperature of 10oC is obtained.

Fill a 100 mL graduated cylinder with 50 mL of water. Add 25 germinating peas and determine the amount of water that is displaced. Record this volume of the 25 germinating peas, then remove the peas and place them on a paper towel. They will be used for respirometer 1. Next, refill the graduated cylinder with 50 mL of water and add 25 non-germinating peas to it. Add glass beads to the graduated cylinder until the volume is equivalent to that of the expanded germinating peas. Remove the beads and peas and place on a paper towel. They will be used in respirometer 2. Now, refill the graduated cylinder with 50 mL of water. Determine how many glass beads would be required to attain a volume that is equivalent to that of the germinating peas. Remove the beads. They will be used in respirometer 3. Then repeat the procedures used above to prepare a second set of germinating peas, dry peas and beads, and beads to be used in respirometers 4,5,and 6.

Assemble the six respirometers by obtaining 6 vials, each with an attached stopper and pipette. Then place a small wad of absorbent cotton in the bottom of each vial and, using the pipette or syringe, saturate the cotton with 15 % KOH. Be sure not to get the KOH on the sides of the respirometer. Then place a small wad of non-absorbent cotton on top of the KOH-soaked absorbent cotton. Repeat these steps to make the other five respirometers. It is important to use about the same amount of cotton and KOH in each vial.

Next, place the first set of germinating peas, dry peas and beads and beads alone in vials 1,2, and 3. Place the second set of germinating peas, dry peas and beads, and glass beads in vials 4,5, and 6. Insert the stoppers in each vial with the proper pipette. Place a washer on each of the pipettes to be used as a weight.

 

Respirometer Temperature Contents
1 Room Germinating Peas
2 Room Dry Seeds + Beads
3 Room Beads
4 10oC Germinating Peas
5 10oC Dry Seeds + Beads
6 10oC Beads

 

Make a sling using masking tape and attach it to each side of the water baths to hold the pipettes out of the water during the equilibration period of 10 minutes. Vials 1,2, and 3 should be in the bath containing water at room temperature. Vials 4, 5, and 6 should be in the bath containing water that is 10oC. After the equilibration period, immerse all six respirometers into the water completely. Water will enter the pipette for a short distance and stop. If the water does not stop, there is a leak. Make sure the pipettes are facing a direction from where you can read them. The vials should not be shifted during the experiment and your hands should not be placed in the water during the experiment.

Allow the respirometers to equilibrate for three more minutes and then record the initial water reading in each pipette at time 0. Check the temperature in both baths and record the data. Every five minutes for 20 minutes take readings of the water’s position in each pipette, and record.

 

Results:

Table 1: Measurement of O2 Consumption by Soaked and Dry Pea Seeds at Room Temperature and 10˚C Using Volumetric Methods

 

 

Beads Alone

Germinating Peas

Dry Peas and Beads

 

Reading at time X

 

Diff.

 

Reading at time X

 

Diff.

 

Corrected Diff.∆

 

Reading at time X

 

Diff.

 

Corrected Diff.∆

 

Initial-0

1.38 1.35 1.47
 

0-5

1.38 0 1.16 .19 .19 1.46 .01 .01
 

5-10

1.38 0 1.04 .31 .31 1.44 .03 .03
 

10-15

1.38 0 0.93 .42 .42 1.43 .04 .04
 

15-20

1.38 0 0.57 .78 .78 1.42 .05 .05
 

Initial-0

1.40 1.32 1.40
 

0-5

1.39 .01 1.20 .12 .11 1.40 0 .01
 

5-10

1.38 .02 1.11 .21 .19 1.40 0 .02
 

10-15

1.38 .02 1.00 .32 .30 1.39 .01 .01
 

15-20

1.38 .02 0.95 .37 .93 1.38 .02 0

 

 

In this activity, you are investigating both the effects of germination versus non-germination and warm temperature versus cold temperature on respiration rate. Identify the hypothesis being tested on this activity. The rate of cellular respiration is higher in the germinating peas in cold than in the beads or non-germinating peas; the cooler temperature in the cold water baths slows the process of cellular respiration in the both germinating and non-germinating peas.

This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each. The constant temperature in the water baths yielding stable readings, the unvarying volume of KOH from vial to vial leading to equal amounts of carbon dioxide consumption, identical equilibration periods for all the respirometers, precise time intervals between measurements, and glass beads acting as a control for barometric pressure all served as controls.

 

Describe and explain the relationship between the amount of oxygen consumed and time. There was a constant, gradual incline in the amount of oxygen consumed over precise passage of time.

 

 

Condition

 

Calculations

 

Rate in mL O2/ minute

 

Germinating Peas/ 10 oC

 

(1.40-1.38)

20 min.

.001
 

Germinating Peas/ 20 oC

 

(1.35-.57)

20 min.

.040
 

Dry Peas/ 10 oC

 

(1.40-1.38)

20 min.

.001
 

Dry Peas/ 20 oC

(1.47-1.42)

20 min.

.003

 

 

Why is it necessary to correct the readings from the peas with the readings from the beads? The beads served as a control variable, therefore, the beads experienced no change in gas volume.

 

Explain the effects of germination (versus non-germination) on pea seed respiration. The germinating seeds have a higher metabolic rate and needed more oxygen for growth and survival. The non-germinating peas, though alive, needed to consume far less oxygen for continued subsistence.

Above is a sample graph of possible data obtained for oxygen consumption by germinating peas up to about 8 oC. Draw in predicted results through 45 oC. Explain your prediction. Once the temperature reached a certain point, the enzymes necessary for cellular respiration denatured and germination (and large amounts of oxygen consumption) was inhibited.

 

What is the purpose of KOH in this experiment? The KOH drops absorbed the carbon dioxide and caused it to precipitate at the bottom of the vial and no longer able to effect the readings.

 

Why did the vial have to be completely sealed under the stopper? The stopper at the top of the vial had to be completely sealed so that no gas could leak out of the vial and no water would be allowed into the vial.

 

If you used the same experimental design to compare the rates of respiration of a 35g mammal at 10 oC, what results would you expect? Explain your reasoning. Respiration would be higher in the mammal since they are warm-blooded and endothermic.

 

If respiration in a small mammal were studied at both room temperature (21 oC) and 10 oC, what results would you predict? Explain your reasoning. Respiration would be higher at 21 degrees because it would be necessary for the animal to maintain a higher body temperature. The results would proliferate at 10 degrees because the mammal would be required to retain its body temperature at an even lower temperature in comparison to room temperature.

 

Explain why water moved into the respirometer pipettes. While the peas underwent cellular respiration, they consumed oxygen and released carbon dioxide, which reacted with the KOH in the vial, resulting in a decrease of gas in the pipette. The water moved into the pipette because the vial and pipette were completely submerged into the bath.

 

Design an experiment to examine the rates of cellular respiration in peas that have been germinating for 0, 24, 48, and 72 hours. What results would you expect? Why? Respirometers could be set up with respirometer 1 containing non-germinating peas, respirometer 2 holding peas that have been germinating 24 hours, 3 would contain the peas that germinated 48 hours, and 4 would hold the peas that germinated 72 hours. All the respirometers should have the KOH added to the bottom in the same manner as in lab described earlier. The respirometers should be placed in baths with the same temperature for all the respirometers. The seeds that have not begun germination would consume very little oxygen. The peas that have been germinating for 72 hours will have the greatest amount of oxygen consumption, while the other two samples will consume a medium (in comparison to respirometers 1 and 4 results) amount of oxygen.

 

Error Analysis:

 

Numerous errors could have occurred throughout the lab. The temperature of the baths may have been allowed to fluctuate, the amounts of peas, beads, KOH, and cotton may have varied from vial to vial damaging the results, and these problems would have occurred only during set up. Air may have been allowed to creep into the vial via a leaky stopper or poorly sealed pipette. Timing for the equilibration of the respirometers and the five-minute time intervals may have been erroneous. It was somewhat difficult to read the markings on the pipettes and so errors are always likely. Mathematical inaccuracies may have taken place when filling out the table and finding the corrected difference by using the formula provided.

 

Discussion and Conclusion:

 

The lab and the results gained from this lab demonstrated many important things relating to cellular respiration. It showed that the rates of cellular respiration are greater in germinating peas than in non-germinating peas. It also showed that temperature and respiration rates are directly proportional; as temperature increases, respiration rates increase as well. Because of this fact, the peas contained by the respirometers placed in the water at 10 oC carried on cellular respiration at a lower rate than the peas in respirometers placed in the room temperature water. The non-germinating peas consumed far less oxygen than the germinating peas. This is because, though germinating and non-germinating peas are both alive, germinating peas require a larger amount of oxygen to be consumed so that the seed will continue to grow and survive.

In the lab, CO2 made during cellular respiration was removed by the potassium hydroxide (KOH) and created potassium carbonate (K2CO3). It was necessary that the carbon dioxide be removed so that the change in the volume of gas in the respirometer was directly proportional to the amount of oxygen that was consumed. In the experiment water will moved toward the region of lower pressure. During respiration, oxygen will be consumed and its volume will be reduced to a solid. The result was a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The respirometer with just the glass beads served as a control, allowing changes in volume due to changes in atmospheric pressure and/or temperature.

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AP Genetics Problems

 

Genetics Problems

1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.

  • What is the simplest explanation for the inheritance of these colors in chickens?
  • What offspring would you predict from the mating of a gray rooster and a black hen?

2. In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) —> Rr (pink). If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white? What will be the ratios in the F2 generation?

3. Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

 

Character Dominant Recessive
Flower position Axial (A ) Terminal (a )
Stem length Tall (T ) Dwarf (t )
Seed shape Round (R ) Wrinkled (r)

 

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)

  1. homozygous for the three dominant traits
  2. homozygous for the three recessive traits
  3. heterozygous
  4. homozygous for axial and tall, heterozygous for seed shape

4. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second one, 7 blacks and 5 albinos were obtained.

  • What is the best explanation for this genetic situation?
  • Write genotypes for the parents, gametes, and offspring.

5. In sesame plants, the one-pod condition (P ) is dominant to the three-pod condition (p ), and normal leaf (L ) is dominant to wrinkled leaf (l) . Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring:

  1. 318 one-pod normal, 98 one-pod wrinkled
  2. 323 three-pod normal, 106 three-pod wrinkled
  3. 401 one-pod normal
  4. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled
  5. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled

6. A man with group A blood marries a woman with group B blood. Their child has group O blood.

  • What are the genotypes of these individuals?
  • What other genotypes and in what frequencies, would you expect in offspring from this marriage?

7. Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant, and allele i is recessive to both. How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

8. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

  1. all three of their children will be of normal phenotype
  2. one or more of the three children will have the disease
  3. all three children will have the disease
  4. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

9. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

  1. aabbccdd
  2. AaBbCcDd
  3. AABBCCDD
  4. AaBBccDd
  5. AaBBCCdd

10. In 1981, a stray black cat with unusual rounded curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the “curl” cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true breeding variety.

  • How would you determine whether the curl allele is dominant or recessive?
  • How would you select for true-breeding cats?
  • How would you know they are true-breeding?

11. What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs?

  1. AABbCc x aabbcc —-> AaBbCc
  2. AABbCc x AaBbCc —–> AAbbCC
  3. AaBbCc x AaBbCc —–> AaBbCc
  4. aaBbCC x AABbcc —-> AaBbCc

12. Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have another child, the child will have sickle-cell anemia.

13. Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume both parents are heterozygous for the “disease” gene.

14. In tigers, a recessive allele causes an absence of fur pigmentation (a “white tiger”) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

15. In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation?

16. The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the filled-in circles and squares, are unable to break down a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant or recessive allele? Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?

 
17. A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?

18. Imagine you are a genetic counselor, and a couple planning to start a family came to you for information. Charles was married once before, and he and his first wife had a child who has cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)

19. In mice, black color (B ) is dominant to white (b ). At a different locus, a dominant allele (A ) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a ) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring?

20. The pedigree below traces the inheritance of a vary rare biochemical disorder in humans. Affected individuals are indicated by filled-in circles and squares. Is the allele for this disorder dominant or recessive? What genotypes are possible for the individuals marked 1, 2, and 3.

 

 

Solutions

Amino Acids

Listed below are the common names and structural formulas of the twenty amino acids found in proteins. They are arranged alphabetically.

Structure of alanine. [str5ala.jpg]
alanine
Structure of arginine. [str5arg.jpg]
arginine
Structure of asparagine. [str5asn.jpg]
asparagine
Structure of aspartic acid. [str5asp.jpg]
aspartic acid
Structure of cysteine. [str5cys.jpg]
cysteine
Structure of glutamic acid. [str5glu.jpg]
glutamic acid
Structure of glutamine. [str5gln.jpg]
glutamine
Structure of glycine. [str5gly.jpg]
glycine
Structure of histidine. [str5his.jpg]
histidine
Structure of isoleucine. [str5ile.jpg]
isoleucine
Structure of leucine. [str5leu.jpg]
leucine
Structure of lysine. [str5lys.jpg]
lysine
Structure of methionine. [str5met.jpg]
methionine
Structure of phenylalanine. [str5phe.jpg]
phenylalanine
Structure of proline. [str5pro.jpg]
proline
Structure of serine. [str5ser.jpg]
serine
Structure of threonine. [str5thr.jpg]
threonine
Structure of tryptophan. [str5trp.jpg]
tryptophan
Structure of tyrosine. [str5tyr.jpg]
tyrosine
Structure of valine. [str5val.jpg]
valine

Amylase on Starch Lab

 

Enzyme Amylase Action on Starch

INTRODUCTION:

In this experiment you will observe the action of the enzyme amylase on starch. Amylase changes starch into a simpler form: the sugar maltose, which is soluble in water. Amylase is present in our saliva, and begins to act on the starch in our food while still in the mouth.
Exposure to heat or extreme pH (acid or base) will denature proteins. Enzymes, including amylase, are proteins. If denatured, an enzyme can no longer act as a catalyst for the reaction.
Benedict’s solution is a test reagent that reacts positively with simple reducing sugars like maltose, but will not react with starch. A positive test is observed as the formation of a brownish-red cuprous oxide precipitate. A weaker positive test will be yellow to orange.

MATERIALS:

Cornstarch
Distilled water
Saliva
Vinegar
Benedict’s qualitative solution
3 graduated cylinders (10mL)
250-ml beaker
Stirring rod
3 test tubes (16 x 125mm)
Test tube rack
Wax pencil
Water Bath

PRE-LAB:

Add 1g of cornstarch to a beaker containing 100ml of cold distilled water. While stirring frequently, heat the mixture just until it begins to boil. Allow to cool.

PROCEDURE:

1. Fill the 250-mL beaker about 3/4 full of water and place on the hot plate for a boiling water bath. Keep the water JUST AT BOILING.

2. Mark 3 test tubes A, B and C. “Spit” between 1 and 2 mL of saliva into each test tube.

3. Into tube A, add 2 mL of vinegar. Into tubes B and C, add 2 mL of distilled water. Thump the tubes to mix.

4. Place tube B into the boiling water bath for 5 minutes. After the five minutes, remove from the bath, and place back into the test tube rack.

5. Add 5 mL of the starch solution to each tube and thump to mix. Allow the tubes to sit for 10 minutes, occasionally thumping the tubes to mix.

6. Add 5 mL of Benedict’s solution to each tube and thump to mix. Place the tubes in the hot water bath. The reaction takes several minutes to begin.

OBSERVATIONS:

Tube A: Starch + saliva treated with vinegar (acid)

    • Was the test positive or negative? _______________________

What does this indicate?__________________________________________________

____________________________________________________________________

____________________________________________________________________

Tube B: Starch + saliva and water, treated in a boiling water bath

    • Was the test positive or negative? _______________________

What does this indicate?__________________________________________________

____________________________________________________________________

____________________________________________________________________

Tube C: Starch + saliva

    • Was the test positive or negative? _______________________

What does this indicate?__________________________________________________

____________________________________________________________________

____________________________________________________________________

QUESTIONS:

1. What is the function of an enzyme?

 

2. Where does a substrate attach to an enzyme?

3. If an enzyme is present in a reaction, less ________________ _________________ will be needed to get the reaction started.

4. What is a common suffix found at the end of most biological enzymes?

5. Most enzymes are macromolecules called ________________.

6. Define denaturation of proteins.

 

 

7. Name 3 things that can denature or unfold an enzyme.

 

8. In this lab, what weak acid denatured the protein?

9. What was the purpose of placing one test tube in a hot water bath?

 

10. What happens to enzymes in your body whenever you run fever?

 

 

 

Amylase Writeup

What to Include in Your Lab Write–Up
Lab: Enzyme Amylase Action on Starch

Introduction:

  • What is an enzyme
  • Describe an enzyme’s structure
  • Explain how an enzyme works (substrate, active site)
  • What’s amylase
  • Where is amylase found
  • What denatures amylase (proteins)
  • Describe the Benedict’s test

Hypothesis:
Exposure to heat or extreme pH will …

Materials:
The materials used include…

 Methods:
Type the procedure in paragraph form.

Results:

Complete this table

Test Tube Contents of Tube Color of Tube After Heating + or – Benedict’s test Enzyme Denatured

Yes or no

A Starch + Saliva + Vinegar
B (Starch + Saliva) Heated
C Starch + Saliva

 Conclusion:

  • Restate the hypothesis
  • Explain the results of the Benedict’s test on Tube C (color change, contents of tube, NOT heated)
  • Explain the results of the Benedict’s test on Tube A ( color change, contents on tube, what was added before heating)
  • Explain the results of the Benedict’s test on Tube B ( color change, contents on tube, heating tube)
  • Tell what denaturing proteins is & how did the Benedict’s test show this.

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