Meiosis Labeling

 

 

Meiosis

 

On each of the images, label the phase of meiosis

1. _______________

2. _______________

3. _______________

4. _______________

5. _______________

6. _______________

7. _______________

8._______________

9._______________

10. _______________

 

11. A cell with a diploid number of 20 undergoes meiosis. This will produce ________ daughter cells, each with ________ chromosomes.

12. Synapsis occurs during this phase: _______________________

13 How many different possible combinations are there for a cell that has 10 chromosomes (5 pairs): _____________

14. Tetrads line up along the equator during this phase: ______________

15. At the end of meiosis I, ________ daughter cells are created. These daughter cells are [ diploid | haploid ].

16. Meiosis occurs in what type of cells: ____________________________

 

Now label the photographs.
17. _______________
18. _______________
19. _____________
20. _______________
21. _______________
22. _____________
23. _______________
24. _______________
24. _____________
25. _______________

 

Genetics

 

Mendelian Genetics
All Materials © Cmassengale 

 

 

Mendel 1862 Mendel 1868 Mendel 1880
1862 1868 1880

 

Genetic Terminology:

  • Trait – any characteristic that can be passed from parent to offspring
  • Heredity – passing of traits from parent to offspring
  • Genetics – study of heredity
  • Alleles – two forms of a gene (dominant & recessive)
  • Dominant – stronger of two genes expressed in the hybrid; represented by a capital letter (R)
  • Recessive – gene that shows up less often in a cross; represented by a lowercase letter (r)
  • Genotype – gene combination for a trait (e.g. RR, Rr, rr)
  • Phenotype – the physical feature resulting from a genotype (e.g. tall, short)
  • Homozygous genotype – gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure 
  • Heterozygous genotype – gene combination of one dominant & one recessive allele    (e.g. Rr); also called hybrid
  • Monohybrid cross – cross involving a single trait
  • Dihybrid cross – cross involving two traits
  • Punnett Square – used to solve genetics problems

Blending Concept of Inheritance:

  • Accepted before Mendel’s experiments
  • Theory stated that offspring would have traits intermediate between those of its parents such as red & white flowers producing pink
  • The appearance of red or white flowers again was consider instability in genetic material
  • Blending theory was of no help to Charles Darwin’s theory of evolution 
  • Blending theory did not account for variation and could not explain species diversity
  • Particulate theory of Inheritance, proposed by Mendel, accounted for variation in a population generation after generation
  • Mendel’s work was unrecognized until 1900

Gregor Mendel:

  • Austrian monk
  • Studied science & math at the University of Vienna
  • Formulated the laws of heredity in the early 1860’s
  • Did a statistical study of  traits in garden peas over an eight year period

 

drawing of a flower cross-section showing both male and female sexual structures

 

Why peas, Pisum sativum?

  • Can be grown in a small area
  • Produce lots of offspring
  • Produce pure plants when allowed to self-pollinate several generations
  • Can be artificially cross-pollinate

Picture of Pisum sativum
GARDEN PEA

Mendel’s Experiments:

  • Mendel studied simple traits from 22 varieties of  pea plants (seed color & shape, pod color & shape, etc.)
  • Mendel traced the inheritance of individual traits & kept careful records of numbers of offspring
  • He used his math principles of probability to interpret results
  • Mendel studied pea traits, each of which had a dominant & a recessive form (alleles)
  • The dominant (shows up most often) gene or allele is represented with a capital letter, & the recessive gene with a lower case of that same letter (e.g. B, b)
  • Mendel’s traits included:

         a. Seed shape —  Round (R) or Wrinkled (r)
            b. Seed Color —- Yellow (Y) or  Green (y)
            c. Pod Shape — Smooth (S) or wrinkled (s)
            d. Pod Color —  Green (G) or Yellow (g)
            e. Seed Coat Color —  Gray (G) or White (g)
            f. Flower position — Axial (A) or Terminal (a)
            g. Plant Height — Tall (T) or Short (t)
            h. Flower color — Purple (P) or white (p)


  •  Mendel produced pure strains by allowing the plants to self-pollinate for several generations
  • These strains were called the Parental generation or P1 strain
  • Mendel cross-pollinated two strains and tracked each trait through two
    generations (e.g. TT  x  tt )

     

                  Trait – plant height

                  Alleles – T tall, t short

    P1 cross    TT  x  tt

    genotype      —    Tt
    t t phenotype    —    Tall
    T Tt Tt genotypic ratio –all alike
    T Tt Tt phenotypic ratio- all alike

     

 

  • The offspring of this cross were all hybrids showing only the dominant trait & were called the First Filial or F1 generation
  • Mendel then crossed two of his F1 plants and tracked their traits; known as an F1 cross

 

              Trait – plant height

              Alleles – T tall, t short

F1 cross    Tt  x  Tt

genotype      —    TT, Tt, tt
T t phenotype    —    Tall & short
T TT Tt genotypic ratio —1:2:1
t Tt tt phenotypic ratio- 3:1

 

 

  • When 2 hybrids were crossed, 75% (3/4) of the offspring showed the dominant trait & 25% (1/4) showed the recessive trait; always a 3:1 ratio
  • The offspring of this cross were called the F2 generation
  • Mendel then crossed a pure & a hybrid from his F2 generation; known as an F2 or test cross

 

Trait   –  Plant Height
Alleles – T  tall, t  short

F2 cross       TT  x Tt

F2 cross       tt  x Tt

T t T t
T TT Tt t Tt tt
T TT Tt t Tt tt
          genotype – TT, Tt           genotype – tt, Tt
          phenotype  –  Tall           phenotype  –  Tall & short
          genotypic ratio  – 1:1           genotypic ratio  – 1:1
          phenotypic ratio – all alike           phenotypic ratio – 1:1

 

  • 50% (1/2) of the offspring in a test cross showed the same genotype of one parent & the other 50% showed the genotype of the other parent; always a 1:1 ratio

Problems: Work the P1, F1, and both F2 crosses for all of the other pea plant traits & be sure to include genotypes, phenotypes, genotypic & phenotypic ratios.

  • Mendel also crossed plants that differed in two characteristics (Dihybrid Crosses)
    such as seed shape & seed color
  • In the P1 cross, RRYY  x  rryy, all of the F1 offspring showed only the dominant form for both traits; all hybrids, RrYy

 

Traits:      Seed Shape & Seed Color

Alleles:     R round                Y yellow
r wrinkled             y green

 P1 Cross:     RRYY          x     r r yy  

      

ry Genotype:      RrYy
RY RrYy
Phenotype:      Round yellow seed
Genotypic ratio:      All alike
Phenotypic ratio:      All Alike

 

  • When Mendel crossed 2 hybrid plants (F1 cross), he got the following results

 

 

Traits:       Seed Shape & Seed Color

Alleles:     R round                Y yellow
r wrinkled             y green

     F1 Cross:     RrYy           x     RrYy                   
RY Ry rY ry
RY
RRYY

RRYy

RrYY

RrYy
Ry
RRYy

RRyy

RrYy

Rryy
rY
RrYY

RrYy

r rYY

r rYy
ry
RrYy

Rryy

r rYy

r ryy

 

 

 

Genotypes Genotypic Ratios Phenotypes Phenotypic Ratios
RRYY 1 Round yellow seed
9
RRYy 2
RrYY 2
RrYy 4
RRyy 1 Round green seed
3
Rryy 2
r rYY 1 Wrinkled yellow seed
3
r rYy 2
r ryy 1 Wrinkled green seed
1

 

Problems: Choose two other pea plant traits and work the P1 and F1 dihybrid crosses. Be sure to show the trait, alleles, genotypes, phenotypes, and all ratios. 

Results of Mendel’s Experiments:

  • Inheritable factors or genes are responsible for all heritable characteristics
  • Phenotype is based on Genotype
  • Each trait is based on two genes, one from the mother and the other from the father
  • True-breeding individuals are homozygous ( both alleles) are the same
  • Law of Dominance states that when different alleles for a characteristic are inherited (heterozygous), the trait of only one (the dominant one) will be expressed. The recessive trait’s phenotype only appears in true-breeding (homozygous) individuals

 

Trait: Pod Color
Genotypes: Phenotype:
GG Green Pod
Gg Green Pod
gg Yellow Pod

 

  • Law of Segregation states that each genetic trait is produced by a pair of alleles which separate (segregate) during reproduction

 

Rr
R r

 

  • Law of Independent Assortment states that each factor (gene) is distributed (assorted) randomly and independently of one another in the formation of gametes

 

RrYy

RY Ry rY ry

 

 

Other Patterns of Inheritance:

  • Incomplete dominance occurs in the heterozygous or hybrid genotype where the 2 alleles blend to give a different phenotype
  • Flower color in snapdragons shows incomplete dominance whenever a red flower is crossed with a white flower to produce pink flowers

  • In some populations, multiple alleles (3 or more) may determine a trait such as in ABO Blood type
  • Alleles A & B are dominant, while O is recessive

 

Genotype Phenotype
IOIO Type O
IAIO Type A
IAIA Type A
IBIO Type B
IBIB Type B
IAIB Type AB

 

  • Polygenic inheritance occurs whenever many variations in the resulting phenotypes such as in hair, skin, & eye color
  • The expression of a gene is also influenced by environmental factors (example: seasonal change in fur color)

 

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Karyotype Lab

 

 

Karyotype lab

 

Introduction:

We can learn a lot by looking at chromosomes! They can tell us everything from the likelihood that an unborn baby will have a genetic disorder to whether a person will be male or female. Scientists often analyze chromosomes in prenatal testing and in diagnosing specific diseases. Fetal cells from an unborn child are contained in the amniotic fluid and can be tested for hereditary disorders such as Tay-Sachs or Phenylketonuria. Chromosomes are compact spools of DNA. If you were to stretch out all the DNA from one of your cells, it would be over 3 feet (1 meter) long from end to end! You can think of chromosomes as “DNA packages” that enable all this DNA to fit in the nucleus of each cell. Normally, we have 46 of these packages in each cell; we received 23 from our mother and 23 from our father. A karyotype is an organized profile of a person’s chromosomes. In a karyotype, chromosomes are arranged and numbered by size, from largest to smallest. This arrangement helps scientists quickly identify chromosomal alterations that may result in a genetic disorder.

To make a karyotype, scientists take a picture of someone’s chromosomes, cut them out and match them up using size, banding pattern and centromere position as guides. Homologous pairs are arranged by size in descending order (largest to smallest) with the sex chromosomes (XX for female or XY for male) as the last or 23 pair. Homologous chromosomes have genes for the same trait at the same location.

Since humans have 46 chromosomes in their somatic or body cells, they have 23 pairs of chromosomes in their karyotype. If chromosomes fail to separate in meiosis, a condition called nondisjunction, a person may have more or less than the normal 46 chromosomes on their karyotype. A disorder called Down Syndrome would be a example of this. A person with Down Syndrome will have 3 chromosomes in their 21st pair. The image below shows chromosomes as they are seen on the slide (left panel) and after arrangement (right panel).

Materials:

karyotype background (run on colored paper), 1-3 sheets of numbered chromosomes, stick glue, scissors, envelope, black ink pen or fine-point marker

Procedure:

  1. Use your assigned sex and chromosome condition to determine how many of each chromosome you will need for your karyotype. (Assigned conditions include Normal male, Normal female, Female with Turner Syndrome, Male with Klinefelter’s Syndrome, Female with Down Syndrome, Male with Down Syndrome, Female with three X chromosomes, Male with no X chromosome, female with Cri-du-chat, Male with Cri-du-chat)
  2. Cut out this number of chromosomes keeping the homologous pairs together. (Do not cut off the chromosome numbers until you are ready to glue the chromosomes to your karyotype sheet.)
  3. Start arranging the chromosome pairs on the construction paper karyotype sheet in descending order by their size. Do not glue the chromosomes until  all of them are arranged correctly.
  4. Evenly space out 4 rows of chromosomes on your karyotype sheet. Row 1 should contain pairs 1-6, row 2 has pairs 7-12, row 3 has pairs 13-18, and row 4 pairs 19 through the sex chromosomes.
  5. If any additional chromosomes are needed to complete your karyotype, cut these out from additional chromosome sheets.
  6. Make sure ALL PAIRS are in the same direction with their SHORTER END TOWARDS THE TOP OF THE CONSTRUCTION PAPER. 
  7. Cut off the numbers from one homologous pair of chromosomes at a time and glue that pair to your construction paper karyotype sheet.
  8. With your ink pen or marker, neatly number each pair 1-23 below the glued pair.
  9. In the lower left corner of your karyotype, write the sex of your individual and their genetic condition (normal, Cri-du-chat, Down’s…).
  10. In the lower right corner, write the total number of chromosomes for this person.

Karyotype Template: (Click here for additional templates)

Questions & Observations:

  1. What is a karyotype? 

 

2. How can a karyotype be useful to a couple wanting to have children?

 

3. What makes up chromosomes?

4. How is a karyotype of an unborn infant obtained?

 

5. What was the sex of the individual you were assigned?

6. What is this person’s GENOTYPE for sex?

7. What is a mutation?

 

8. What mutation, if any, occurred in this person’s karyotype?

9. How many chromosomes are in a somatic or body cell of this individual?

10. How many chromosomes are in a gamete or sex cell of this individual?

11. How many chromosomes are in a normal person’s somatic cells?

12. How many chromosomes are in a normal person’s gametes?

13. How many UNPAIRED chromosomes are their in this organism’s somatic cells?

14. What is the sex of an individual with 23 MATCHED pairs of chromosomes?

15. What is the diploid number for this organism?

16. Explain nondisjunction.

 

17. Name and explain 3 disorders due to nondisjunction of chromosomes.

 

 

 

 

AP Lab 2 Report 2001

 

Enzyme Catalysis

 

Introduction
Enzymes are proteins produced by living cells that act as catalysts, which affect the rate of a biochemical reaction. They allow these complex biochemical reactions to occur at a relatively low temperature and with less energy usage.

In enzyme-catalyzed reactions, a substrate, the substance to be acted upon, binds to the active site on an enzyme to form the desired product. Each active site on the enzyme is unique to the substrate it will bind with causing each to have an individual three-dimensional structure. This reaction is reversible and is shown as following:

E + S—-ES—- E + P

Enzymes are recyclable and unchanged during the reaction. The active site is the only part of the enzyme that reacts with the substrate. However, its unique protein structure under certain circumstances can easily be denatured. Some of the factors that affect enzyme reactions are salt concentration, pH, temperature, substrate and product concentration, and activators and inhibitors.

Enzymes require a very specific environment to be affective. Salt concentration must be in an intermediate concentration. If the salt concentration is too low, the enzyme side chains will attract each other and form an inactive precipitate. Likewise, if the salt concentration is too high, the enzyme reaction is blocked by the salt ions. The optimum pH for an enzyme-catalyzed reaction is neutral (7 on the pH scale). If the pH rises and becomes basic, the enzyme begins losing its H+ ions, and if it becomes too acidic, the enzyme gains H+ ions. Both of these conditions denature the enzyme and cause its active site to change shape.

Enzymes also have a temperature optimum, which is obtained when the enzyme is working at its fastest, and if raised any further, the enzyme would denature. For substrate and product concentrations, enzymes follow the law of mass action, which says that the direction of a reaction is directly dependent on the concentration. Activators make active sites better fit a substrate causing the reaction rate to increase. Inhibitors bind with the enzymes’ active site and block the substrate from bonding causing the reaction to subside.

The enzyme in this lab is catalase, which produced by living organisms to prevent the accumulation of toxic hydrogen peroxide. Hydrogen peroxide decomposes to form water and oxygen as in the following equation:

2H2O2 ® 2H2O + O2

This reaction occurs spontaneously without catalase, but the enzyme speeds the reaction considerably. This lab’s purpose is to prove that catalase does speed the decomposition of hydrogen peroxide and to determine the rate of this reaction.

 

Hypothesis
The enzyme catalase, under optimum conditions, effectively speeds the decomposition of hydrogen peroxide.

 

Materials
Exercise 2A: Test of Catalase Activity

In Part 1, the materials used were 10mL of 1.5% H2O2, 50-mL glass beaker, 1 mL catalase, and 2 10-mL pipettes and pipette pumps. In Part 2, the materials used were 5 mL of catalase, a boiling water bath, 1 test tube, a test tube rack, 10 mL of 1.5% H2O2, 50-mL beaker, and 2 10-mL pipettes and pipette pumps. In Part 3, the materials used were 10 mL of 1.5% H2O2, 50-mL beaker, liver, and a syringe.

Exercise 2B: The Baseline Assay

This part of the lab required 10 mL of 1.5% H2O2, 1 mL distilled H2O, 10 mL of H2SO4, 2 50-mL beakers, a sheet of white paper, 5 mL KMnO4, 2 5-mL syringes, and 2 10-mL pipettes and pumps.

Exercise 2C: The Uncatalyzed Rate of H2O2 Decomposition

The materials used for this section were 15 mL of 1.5% H2O2, 1 mL distilled H2O, 10 mL H2SO4, 2 50-mL beakers, a sheet of white paper, 5 mL KMnO4, 2 5-mL syringes, and 2 10-mL pipettes and pumps.

Exercise 2D: An Enzyme-Catalyzed Rate of H2O2 Decomposition

The materials required for Exercise 2D were 70 mL of 1.5% H2O2, 70 mL of H2SO4, 6 mL of catalase solution, 13 plastic, labeled cups, 3 100-mL beakers, 1 50-mL beaker, 1 10-mL syringe, 1 5-mL syringe, 1 60-mL syringe, a sheet of white paper, a timer, and 30 mL of KMnO4.

 

Method
Exercise 2A: Test of Catalase Activity

In Part 1, 10 mL of 1.5% H2O2 were transferred into a 50-mL beaker. Then, 1 mL of fresh catalase solution was added and the reaction was observed and recorded. In Part 2, 5 mL of catalase was placed in a test tube and put in a boiling water bath for five minutes. 10 mL of 1.5% H2O2 were transferred to a 50-mL beaker and 1 mL of the boiled catalase was added. The reaction was observed and recorded. In Part 3, 10mL of 1.5% H2O2 were transferred to a 50 mL beaker. 1 cm3 of liver was added to the beaker and the reaction was observed and recorded.

Exercise 2B: The Baseline Assay

10 mL of 1.5% H2O2 were transferred to a 50-mL beaker. 1 mL of H2O was added instead of catalase, and then, 10 mL of H2SO4 were added. After mixing well, a 5 mL sample was removed and placed over a white sheet of paper. A 5-mL syringe was used to add KMnO4, 1 drop at a time until a persistent brown or pink color was obtained. The solution was swirled after every drop, and the results were observed and recorded. The baseline assay was calculated.

Exercise 2C: The Uncatalyzed Rate of H2O2 Decomposition

A small quantity of H2O2 was placed in a beaker and stored uncovered for approximately 24 hours. To determine the amount of H2O2 remaining, 10 mL of 1.5% H2O2 were transferred to a 50-mL beaker. 1 mL of H2O was added instead of catalase, and then, 10 mL of H2SO4 were added. After mixing well, a 5 mL sample was removed and placed over a white sheet of paper. A 5-mL syringe was used to add KMnO4, 1 drop at a time until a persistent brown or pink color was obtained. The solution was swirled after every drop, and the results were observed and recorded. The percent of the spontaneously decomposed H2O2 was calculated.

Exercise 2D: An Enzyme-Catalyzed Rate of H2O2 Decomposition

 

The baseline assay was reestablished following the directions of Exercise 2B. Before starting the actual experiment a lot of preparation was required. Six labeled cups were set out according to their times and 10 mL of H2O2 were added to each cup. 6 mL of catalase were placed in a 10-mL syringe, and 60 mL of H2SO4 were placed in a 60-mL syringe. To start the actual lab, 1 mL of catalase was added to each of the cups, while simultaneously, the timer was started. Each of the cups were swirled. At 10 seconds, 10 mL of H2SO4 were added to stop the reaction. The same steps were repeated for the 30, 60, 120, 180, and 360 second cups, respectively.

Afterwards, a five 5 mL sample of each of the larger cups were moved to the corresponding labeled smaller cups. Each sample was assayed separately by placing each over a white sheet of paper. A 5-mL syringe was used to add KMnO4, 1 drop at a time until a persistent brown or pink color was obtained. The solution was swirled after every drop, and the results were observed and recorded.

 

Results

Table 1
Enzyme Activity

 

 

 

Activity

 

Observations

Enzyme activity The solution only bubbled slightly and slowly.
Effect of Extreme temperature

 

 

The catalase had no reaction with the H2O2; there were no bubbles
Presence of catalase The solution foamed up immediately

 

 

Table 2
Establishing a Baseline

 

 

 

Volume

 

Initial reading

 

 

5.0 mL

 

Final reading

 

 

0.8 mL

 

Baseline ( final volume – initial volume)

 

 

4.2 mL

 

 

Table 3
Rate of Hydrogen Peroxide Spontaneous Decomposition

 

 

 

Volume

 

Initial KMnO4

 

 

5.0 mL

 

Final KMnO4

 

 

1.2 mL

 

Amount of KMnO4 used after 24 hours

 

 

3.8 mL

 

Amount of H2O2 spontaneously decomposed
( ml baseline – ml after 24 hours)

 

0.4 mL

 

Percent of H2O2 spontaneously decomposed
( ml baseline – ml after 24 hours/ baseline)

 

9.52%

 

 

Table 4
Rate of Hydrogen Peroxide Decomposition by Catalase

 

Time ( Seconds)
10 30 60 120 180 360
 

Baseline KMnO4

 

 

4.0 mL

 

4.0 mL

 

4.0 mL

 

4.0 mL

 

4.0 mL

 

4.0 mL

 

Initial volume KMnO4

 

 

5.0 mL

 

5.0 mL

 

5.0 mL

 

5.0 mL

 

5.0 mL

 

5.0 mL

 

Final volume KMnO4

 

 

2.2 mL

 

1.4 mL

 

2.0 mL

 

1.7 mL

 

2.4 mL

 

2.3 mL

 

Amount KMnO4 used
(baseline – final)

 

2.8 mL

 

3.6 mL

 

3.0 mL

 

3.3 mL

 

2.6 mL

 

2.7 mL

 

Amount H2O2 used
(KMnO4 – initial)

 

1.2 mL

 

0.4 mL

 

1.0 mL

 

0.7 mL

 

1.4 mL

 

1.3 mL

 

Amount of Hydrogen Peroxide Decomposed by Catalase

Exercise 2A: Test of Catalase Activity

1. Observing the reaction of catalase on hydrogen peroxide:

a. What is the enzyme in this reaction?  catalase

b. What is the substrate in this reaction? Hydrogen peroxide

c. What is the product in this reaction? Oxygen & water

d. How could you show that the gas evolved is O2? The gas could be shown to be O2 if the gas were collected in a tube, and a glowing splint was held in the tube. If the splint glowed, it would prove the gas was oxygen.

2. Demonstrating the effect of boiling on enzyme action:

a. How does the reaction compare to the one using the unboiled catalase? Explain the reason for this difference. While the unboiled catalase caused bubbles to form in the solution, the boiled catalase did not react at all because boiling an enzyme causes the protein to unfold and therefore denatures it.

3. Demonstrating the presence of catalase in living tissue:

a. What do you think would happen if the potato or liver was boiled before being added to the H2O2? The catalase in the liver would have been denatured by the boiling and would not have reacted with the H2O2.

Analysis of Results

1. Determine the initial rate of the reaction and the rates between each of the time points.

 

 

Time Intervals (Seconds)

 

Initial 0 to 10

 

10 to 30

 

30 to 60

 

60 to 120

 

120 to 180

 

180 to 360

 

Rates

 

0.12 mL/sec

 

-0.04 mL/sec

 

0.02 mL/sec

 

-0.005 mL/sec

 

0.01167 mL/sec

 

-0.00083

mL/sec

 

 

2. When is the rate the highest? Explain why.

 

The rate is the highest in the initial ten seconds because the concentration of catalase is at its highest. As more of the product is formed, it blocks the reaction between the catalase and the hydrogen peroxide.

3. When is the rate the lowest? For what reasons is the rate low?

The rate is lowest during the 180-360 seconds time period because of the law of mass action. This law says that when there is a high concentration of product as in this period, the enzymes will be blocked by the product (water) from reaching and reacting with the substrate (H2O2).

 

4. Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to enzyme structure and chemistry

 

Sulfuric acid has an inhibiting effect on catalase function because it causes the pH level in the solution to lower considerably. Acidic solutions cause the protein structure of the enzyme to gain H+ ions causing it to denature.

 

5. Predict the effect lowering the temperature would have on the rate of enzyme activity. Explain your prediction.

 

Lowering the temperature of the catalase would slow the rate of reaction until it finally caused the enzyme to denature, and it would no longer react with the substrate. Most enzymes are only affective in a temperature range between 40° – 50° C.

6. Design a controlled experiment to test the effect of varying pH, temperature, or enzyme concentration.

Part 1: Enzyme Activity at Room Temperature

Add 10 mL of 1.5% H2O2 to a 50-mL beaker, and add 1 mL of room temperature catalase. Mix well and add 10 mL of H2SO4. Watch the reaction and record the results.

Part 2: The Effect of Excessive Heat on Enzyme Activity

Put 5 mL of catalase into a test tube and heat thoroughly over a Bunsen burner. Add 1 mL of the heated catalase to 10 mL of 1.5% H2O2 in a 50-mL beaker. Add 10 mL of H2SO4. Watch the reaction and record the results.

Part 3: The Effect of Excessive Cooling on Enzyme Activity

Put 5 mL of catalase in a freezer until completely frozen. Add 1 mL of the frozen catalase to 10 mL of 1.5% H2O2 in a 50-mL beaker. Add 10 mL of H2SO4. Watch the reaction and record the results.

 

Error Analysis
Any number of factors in this lab could have affected the results of this experiment. To get the desired results all of the measurements had to be precisely accurate and fully planned before hand. In Exercise D especially, the factor of planning became increasingly essential. The first attempt at 2D was unsuccessful due to several reasons. First of all, the measurements, which were taken, could have possibly been inaccurate and the 60-mL syringe containing H2SO4 also dripped into one of the cups early which did not allow the reaction to fully take place. There was also some confusion on the operation of the timer and precise planning in its use. The second attempt at 2D contained errors as well. The measurements were still not as accurate as they should have been, and the solution did not appear entirely uniform. In one cup, for example, the first drop of KMnO4 left a persistent pink color, and then after over a minute, it returned back to being clear. It then took several milliliters more to get it back to a pink color.

 

Discussion and Conclusion
This lab showed how catalase increased the rate of decomposition of hydrogen peroxide. In 2A, it was shown that catalase causes a visual reaction with H2O2, that when boiled catalase is no longer reactive, and that catalase is present in living tissue. Lab 2C shows that the natural decomposition of H2O2 is much slower than the enzymatic reaction. Lab 2D showed the decomposition of H2O2 over just a period of six minutes, and it had already decomposed more than the uncatalyzed H2O2 had done in 24 hours.

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