EDVO-Kit · Advanced Placement Biology ... Biology AP Lab # 3 EVT 001188AM EDVO-Kit # 288...

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Advanced Placement Biology

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AP Biology

Lab 8

Population Geneticsand Evolution

EXPERIMENT OBJECTIVE

The objective of this experiment module is to use theHardy-Weinberg theorem to calculate allele andgenotype frequencies. Students will predict the effectof selection processes on allelic frequencies, anddiscuss the effect of changes in allelic frequencies onevolution.

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EVT 001188AM

EDVO-Kit # 288 Population Genetics and Evolution

8Table of Contents

Page

Experiment Components 3

Experiment Requirements 3

Background Information 4

Experiment Procedures

Experiment Overview 6

Part A 7

Part B 9

Part C 13

Part D 17

Part E 19

Instructor's Guidelines

Notes to the Instructor 21

Pre-Lab Preparations 22

Expected (SAMPLE CASE) Results and Selected Answers 23

Material Safety Data Sheets 26

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8• PTC taste paper• Control taste paper

Experiment Components

• Calculator with square root function• 100 index cards - 3 x 5”

Requirements

Storage:Store entire

experiment at roomtemperature.

This experimentis designed for10 lab groups

All components areintended foreducational researchonly. They are not tobe used fordiagnostic or drugpurposes, noradministered to orconsumed byhumans or animals.

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Population Genetics and Evolution

Population genetics deals with analysis of gene frequencies in apopulation over many generations. The concept of describingfrequencies of inherited traits owes its origin to scientific works pub-lished at the beginning of the 20th century. A 1908 paper, “Mende-lian Proportions in a Mixed Population” published in Science 28 (49-50)by British mathematician G.H. Hardy, and a separate independentstudy also published in 1908 by the German physician W. Weinberg,both suggested that gene frequencies were not dependent upondominance or recessiveness but may remain unchanged from onegeneration to the next under a set of “idealized conditions”. Theseclassic papers describe an equation which has come to be called theHardy-Weinberg theorem of genetic equilibrium. This theorem hasbecome the basis for population genetics.

The Hardy-Weinberg theorem is used to determine the frequencies ofindividual alleles of a pair of genes, and the frequency of heterozy-gotes and homozygotes in the population. The theorem states that inthe absence of outside forces such as mutation, selection, randomgenetic drift, and migration, gene frequencies remain constant overmany generations in a large population. It is important to rememberthat in natural populations, events such as gene mutation, selectionof genotypes which confer enhanced viability, presence of lethalhomozygous recessive genes, nonrandom mate selection, andimmigration and emigration of individuals of a population, are eventsthat do occur. Nevertheless, the Hardy-Weinberg theorem is usefulsince unexpected deviations can point to the occurrence of evolu-tionary significant events such as speciation.

Distribution frequencies of two alleles for a given gene at a singlelocus, one being dominant, the other recessive, will follow a binomialdistribution in the population. Consider the case of two alleles for agene, one dominant and the other recessive.

Let p = the frequency of one allele and q = the frequency of theother. If gene frequencies are expressed as decimals, the followingmust be true,

Equation # 1: p + q = 1and,

Equation # 1a: p = 1 - q therefore,Equation # 2: (p + q)2 = 1.

Expanding equation #2 generates,

Equation #3: p2 + 2pq + q2 = 1.



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Population Genetics and Evolution

When equation 3 is applied to an ideal population, it follows that thefrequency of homozygous dominant individuals is p2, the frequency of theheterozygotes is 2pq, and the frequency of homozygous recessives is q2.

As an example, consider the following hypothetical situation. The famousEuropean geneticist Professor Ed V. Otek, tested his rather large geneticsclass for the ability to taste the chemical phenylthiocarbamide, PTC. Heknew that the gene for this ability to taste PTC had two alleles, the domi-nant allele for tasting called T, and the recessive allele called t. He foundthat out of 1000 students, there were 700 students with the ability to tastePTC and 300 who lacked the ability to taste PTC. He used the Hardy-Weinberg equation to determine the gene frequencies for the T and talleles of the gene for the ability to taste PTC. His notes show the followinganalysis:

A Converted raw data to decimals.

• Frequency of two genotypes for tasting, TT and Tt, was 700/1000 = 0.7.• Frequency of genotype for inability to taste PTC, t t, was 300/1000 = 0.3.

B Determined gene frequency of the unique allele.

• From the Hardy-Weinberg equation # 3, (p2 + 2pq + q2 = 1), the frequencyof non-tasters, tt = 0.3 = q2.

• Taking the square root of 0.3, q = 0.5477, and 0.5477 is the frequency ofthe t allele in Dr. Otek’s student population.

C Determined gene frequency of other allele, p:From equation # 1a, (p = 1-q), the frequency of p is 0 .4523.

D. Determined frequency of homozygous TT and heterozygous Tt individu-als in the population. Using equation #3:

p2 + 2pq + q2 = 1,

(0.4523)2 + 2(0.4523 x 0.5477) + (0.5477)2 = 1

• The frequency of homozygous tasters is, TT = p2 = 0.45232 = 0.2046.• The frequency of heterozygous tasters is Tt = 2pq = 2 (0.4523 x 0.5477) = 0.4954.

During this experiment students will utilize the Hardy-Weinberg equation toanalyze population data from the class.



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Experiment Overview

EXPERIMENT OBJECTIVE

• To use the Hardy-Weinberg theorem to calculate allele andgenotype frequencies.

• To predict the effect of selection processes on allelic frequenciesand discuss the effect of changes in allelic frequencies on evolu-tion.

WORKING HYPOTHESIS

If there is no selection for any allele in a large randomly-matingpopulation, then the gene frequencies will remain constant overmany generations. However, if there are outside forces such asselection for an allele, heterozygote advantage, and genetic driftworking in a population, then the gene frequencies will change overtime.

LABORATORY SAFETY

Remember to wear gloves and safety goggles during all hands-onlaboratory activities. Always wash hands thoroughly with soap andwater after handling reagents or biological materials in the laboratory.

MATERIALS FOR THE EXPERIMENT

Each Lab Group should have the following materials:

• PTC taste paper• Control taste paper• Index cards - 3 x 5”



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Part A: Student Experimental Procedures

ESTIMATION OF GENE FREQUENCIES FOR THE TRAIT TO TASTEPTC WITHIN A SMALL SAMPLE POPULATION.

This experiment deals with the determination of the gene frequency of ahuman trait amongst students with no known selective advantage. Theability to taste the chemical phenylthiocarbamide, PTC, is one suchhuman trait.

The ability to taste PTC is due to the presence of a dominant allele, T.Therefore, all tasters will either be homozygous, TT, or heterozygous, Tt.Non-tasters will be homozygous for the recessive gene, tt. Using theHardy-Weinberg equation #3, first determine the frequency, q2, of thehomozygous recessive gene t, which is directly determined from your classdata. The frequency of the non-taster phenotype in your class equals thefrequency of the genotype tt. q is determined as the square root of thefrequency of non-tasters. Using equation # 1a, the value of p is deter-mined by subtraction. We begin this analysis using the frequency of thehomozygous recessive, since for this experiment it is impossible to deter-mine whether the true genotype of the tasters is either TT or Tt.

1. Students groups should obtain a PTC taste strip and a control strip.

2. Every member of the group should first taste the control strip of paper.

3. Every person should taste the PTC impregnated strip of paper. Com-pare the taste of the control and the PTC paper.

If you are a taster, the PTC paper strip will be bitter. Non-tasters will notnotice a difference between either strip of paper.

4. For the class, record the total number of tasters and the total numberof non-tasters on the blackboard.

5. Determine decimal value by division for tasters (p2 + 2pq), andlikewise the decimal value for non-tasters (q2).

• For example, there are 100 people in your class. 25 are non-tasters and 75 are tasters.

• Then 25/100, or 0.25, is the frequency of non-tasters, and 75/100,or 0.75, is the frequency of tasters.



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Answer the following study questions in your laboratory notebook or ona separate worksheet.

1. What is the frequency of homozygous tasters, p2, in your class-room?

2. What is the frequency of heterozygous tasters, 2pq, in your class-room?

3. What is the frequency of homozygous non-tasters in your class-room?

4. Determine the percentage of the three genotypes TT, Tt, and tt inyour classroom. Hint: multiply the values obtained for question 1, 2,and 3 by 100.

Part A: Study Questions

Part A: Student Experimental Procedures

6. Record your values in Table 1. Use Hardy-Weinberg as describedabove to determine the value of p and q for your class.

CLASSPOPULATION

NORTH AMERICANPOPULATION

CLASS PHENOTYPES

0.55 0.45

p2 + 2pq q2

ALLELE FREQUENCYCALCULATED BY THE

HARDY WEINBERG EQUATION

p% TASTERS % NON-TASTERS

q

TABLE 1: Phenotypes and GeneFrequencies for Trait to Taste PTC



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Part B: Student Experimental Procedures

TESTING AN IDEALIZED HARDY-WEINBERG POPULATIONWITHOUT SELECTION

For this example, we will consider the class to be a natural randomlymated group with no selection and each individual being heterozygousfor a particular phenotype having the genotype A. The initial genefrequency in this population is 0.5 for “A”, the dominant allele, and 0.5 for“a”, the recessive allele. Therefore p and q both equal 0.5. Every pair ofindividuals should keep track of their child’s genotype after each mating.In addition, the class as a group must keep track of totals for each matingduring each of five (5) generations.

1. Work in pairs. For each generation, each pair will produce twooffspring.

2. For this example, each person in the class should obtain 4 index cards.Label two “A” and two “a”. These cards represent the only twounique gametes which a heterozygous individual with a genotype ofAa can produce. You have 4 cards since there are two cell divisionsduring meiosis.

3. Each person in a group should place the index cards face down onlab bench. Shuffle cards so that they are randomly mixed. Eachperson selects only one card from their pile of 4 cards, and places ittogether with one selected by their partner.

4. Flip the cards over and record the genotype of this “mating” inTable 2.

AA Aa aaAA, Aa, or aa

GENERATION NUMBER

OFFSPRING GENOTYPE

CLASS TOTALSFOR EACH GENOTYPE

CLASS TOTALS FOR EACH GENOTYPEADDED OVER 5 GENERATIONS

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TABLE 2: Hardy-Weinberg Population



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5. Recover your index card so again you have two of each allele“A”, and “a”. Reshuffle the cards between selections.

6. With your partner, produce a second offspring and record thegenotype. Be sure to collect the class totals on the blackboard atthe end of this generation and for the next 4 generations.

7. Now you will assume the genotype of the first child and yourpartner will assume the genotype of the second child.

8. Re-label your four index cards to reflect the gametes which youwould produce with the new genotype. For example if your childwas AA, your cards would be labeled A,A,A,A. Similarly, if yourchild was Aa, your cards would be labeled A,A,a,a.

9. Now randomly select a different partner in your class.

10. Using the same procedure for the first generation, produce twonew offsprings with your new “mate”. Again record the genotypeof each child in the table. Remember, the class as a group shouldalso keep track of each genotype after each generation. Recordthe class totals.

11. As before, one student of the pair will assume the genotype of thefirst child and the other student should assume the genotype ofthe second child produced during this mating. If necessary, re-label your index cards to reflect the four gametes which youwould produce.

12. Again, each person should randomly seek out another mate andrepeat the procedure generating two new children. Record thenew genotypes as before. Keep track of the class totals aftereach mating.

13. This process of “random mating” without selection should continueuntil you have produced 5 generations of children.

14. Be sure to place your individual results and the class totals for eachmating into Table 2.

Part B: Student Experimental Procedures



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STUDY QUESTIONS

Answer the following study questions in your laboratory notebook or on aseparate worksheet.

1. Using the total of all class data over the 5 generations considered asone large population, compute the genotypic frequency for each ofthe three possible genotypes. The class totals for each genotypeshould be summed over all of the 5 generations as indicated in Table2. These three numbers, which are the bottom 3 cells in Table 2,should be added together to determine the grand total number of allindividuals with genotypes, AA + Aa + aa in your class population. Thisnumber is used as the divisor for determination of the frequencies ofthe AA, Aa, and aa genotypes in the population after 5 generations.

a. AA genotypic frequency = Total number of AA individuals

Total number of all individualsin the population

b. Aa genotypic frequency = Total number of Aa individuals


c. aa genotypic frequency = Total number of aa individuals


2. Record your results of the actual class genotypic frequencies deter-mined in question 1 above.

a. AA (p2) genotypic frequency ______

b. Aa (2pq) genotypic frequency ______

c. aa (q2) genotypic frequency ______

3. Determine the theoretical Hardy-Weinberg frequencies of the begin-ning population when p = 0.5, and q =0.5.

p2 (AA) =

2pq (Aa) =

q2 (aa) =

Part B: Study Questions



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4. From the data in question #3, determine q from q2, anddetermine p.

5. Calculate the values of p and q in your class population for the 5thgeneration only. This represents the frequencies of the two allelesafter 5 generations of totally random mating for a gene whichdoes not confer any selective advantage. Use these p and qvalues to determine the gene frequencies with the Hardy-Weinberg equation #3, p2 + 2pq + q2 = 1.

From Table 2, the total number of individuals in the 5th generationis obtained by summation across the last 3 cells in the row for the5th generation, that is AA + Aa + aa (this should also be the samevalue as the number of students in your class!).

a. AA genotypic frequency of the 5th generation =

Total number of AA individuals in the 5th generation

Total number of all individuals in the 5th generation

b. Aa genotypic frequency of the 5th generation =

Total number of Aa individuals in the 5th generation


c. aa genotypic frequency of the 5th generation =

Total number of aa individuals in the 5th generation


d. Remember that you determine q as the square root of thevalue determined in 5c. Also p = 1 - q. Determine the follow-ing values for the 5th generation only;

p2 (AA) =

2pq (Aa) =

q2 (aa) =

6. How have the values of p and q changed?

Part B: Study Questions



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Part C: Student Experimental Procedures

TESTING AN IDEALIZED HARDY-WEINBERG POPULATION WITHSELECTION

A number of recessive alleles are lethal when they occur in homozygousaa individuals. These individuals usually do not survive until they are oldenough to mate, or are unable to mate. Such a selection serves as abiological mechanism to remove the recessive allele “a” from the popula-tion. Only the AA, and Aa individuals survive to mate. Therefore, overmany generations there is a gradual decline in frequency of the recessive“a” allele since the aa individuals do not service to mate. Consider thefollowing example:

Where p (A)= 0.5, and q (a) = 0.5, and genotypes AA, Aa, and aa arepossible,

p2 + 2pq + q2 = 0.25 + 0.50 + 0.25.

If aa individuals are unable to mate, the new genotypic frequency wouldbe:

AA = 0.25/ (0.25 + 0.5) = 0.33Aa = 0.5/ (0.25 + 0.5) = 0.67

The aa allele is eliminated from the calculation since it is unable toproduce further offspring for the next generation.

Therefore, the only possible matings are shown in Table 3.

Genotypic Frequencies

Matings Frequencies AA Aa aa

AA x AA (0.33) 2 = 0.11

AA x Aa 2 (0.33 X 0.67) = 0.22 0.22

Aa x Aa (0.67)2 = 0.45 0.11 0.23 0.11

0.44 0.45 0.11

TABLE 3: Hardy-Weinberg Population with Selection

Total PopulationGenetic Frequencies



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Part C: Student Experimental Procedures

The frequency of the “a” allele has decreased to 0.33 (the square rootof 0.11) after only 1 generation. With selection against the doublerecessive, the “a” allele is gradually eliminated from the population.The following equation will predict gene frequency of the recessiveallele, q (“a”) after n, number of generations, with complete selectionagainst the double recessive.

qn = qo/(1 + nqo)

qn = the frequency of the recessive allele after n generationsqo = the initial frequency of the recessive allele.

Each pair of students should keep track of their child’s genotype aftereach mating. In addition, the class as a group must record totals foreach mating during each of the 5 generations.

1. Each student should begin with 4 index cards. Two labeled “A”,and two labeled “a”. Proceed in a manner similar to that of Part Bwith one slight difference. If you produce an “aa” offspring youmust do the following:• Record it in Table 4 for the class total.• In this exercise you must mate again with the same partner

until you produce either an AA or Aa offspring.• Continue randomly mating for 5 generations and be sure to

record your data.• Reshuffle and reselect cards until you obtain either an AA or

Aa child, keeping the population size constant. Record yourindividual results and class totals in Table 4.

2. Continue the procedure for 5 generations.


OFFSPRING GENOTYPE



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TABLE 4: Hardy-Weinberg Population with Selection



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Part C: Study Questions

Answer the following questions in your laboratory notebook or on aseparate worksheet.

1. Calculate, from your experimental data, gene frequency values ofthe A, and a alleles, as well as p and q in your class population for the5th generation only. \This represents frequencies of two alleles after 5generations of totally random mating for a gene in which recessivegenotype aa is lethal.

From Table 3, the total population in the 5th generation (which is usedas the divisor below) is obtained by summation across the last 3 cellsin the row for the 5th generation, that is AA + Aa + aa.

a. AA genotypic frequency of the 5th generation =

Total number of AA individuals in the 5th generation


b. Aa genotypic frequency of the 5th generation =

Total number of Aa individuals in the 5th generation


c. aa genotypic frequency of the 5th generation =

Total number of aa individuals in the 5th generation


d. Remember that you determine q as the square root of thevalue determined in 1c and p = 1- q. Determine the follow-ing values for the 5th generation only;

p2 (AA) =

2pq (Aa) =

q2 (aa) =

2. How have values of p and q changed since the first generation?



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Part C: Study Questions

3. As discussed above, the following equation can be used topredict the frequency of the recessive allele in the populationwhen there is selection against the double recessive individuals.

qn = qo/ (1 + nqo)qn = frequency of the recessive allele after n generationsqo = the initial frequency of the recessive allele.

a. Using this equation, predict frequency of the recessive alleleafter 1, 5, 10, and 100 generations. Assume an initial p = 0.5,and q = 0.5.

b. How does the calculated value for the 5th generation com-pare to your experimental value for the 5th generation?

4. Compare the values of p and q determined after 5 generationshere with selection against the double recessive, to the value of pand q which you have already determined in Part B, after 5generations without selection.



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Part D: Student Experimental Procedures


OFFSPRING GENOTYPE



GENERATION NUMBER

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TABLE 5: Hardy-Weinberg Population with Heterozygotic Advantage

TESTING AN IDEALIZED HARDY-WEINBERG POPULATION, WITHHETEROZYGOTIC ADVANTAGE

It was easy to see in Part C that selection against the homozygous recessivelethal genotype results in a decrease in the original “q” value. However, it hasbeen shown this is not always the case as in certain recessive alleles remain ahigh level. An example is the case of Sickle-cell anemia, common amongindividuals of African origin. In this disease, the heterozygote has an advan-tage over the homozygous dominant allele. Individuals who have heterozy-gotic genotype have an increased resistance to malaria which historically wasdeadly in Africa. This type of advantage is incorporated in the followingsimulation.

1. Keep everything the same as in Part C, except if the offspring is AA, flip acoin. The individual survives if the coin lands tails up; the individual doesnot survive if it lands heads up.

2. Simulate five generations, starting again with the initial genotype from PartB. The genotype aa will never survive. The homozygous dominantgenotypes survive only if the coin toss comes up tails. In order to keep aconstant population size, the same two parents must try again until theyproduce two surviving offspring. Obtain new “allele” cards from the poolas needed. Collect and record class genotypes in Table 5. Calculate the“p” and “q” frequencies.

3. If time permits, start with the F5 genotype, go through five more genera-tions, and total the genotypes. Calculate the frequencies of “p” and “q”.



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Part D: Study Questions


1. Compare the changes in “p” and “q” frequencies in Part C withPart B and Part D. Explain.

2. Will the recessive allele be completely eliminated in either Part Cor Part D? Explain.

3. Why is heterozygote advantage important in maintaining geneticvariation in a population?



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Part E: Student Experimental Procedures

TESTING AN IDEALIZED HARDY-WEINBERG POPULATION, WITHGENETIC DRIFT

Changes in a small gene pool is known as genetic drift. The simulationthat follows incorporates this fact into the following simulation.

1. Divide the class into three isolated populations. Individuals from oneisolated population will not mate with individuals from anotherpopulation.

2. Go through five generations. Record the new genotypic frequencies.Calculate the new frequencies of “p” and “q” for each population.


OFFSPRING GENOTYPE



GENERATION NUMBER

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TABLE 6: Hardy-Weinberg Population with Genetic Drift



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1. How do the initial genotypic frequencies of the populationscompare?

2. Show a difference between small and large populations in allelicfrequency? Explain.

Part E: Study Questions



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scientists in laboratories conducting biotechnology research. Some ofthe experimental procedures may have been modified or adapted tominimize equipment requirements and to emphasize safety in theclassroom, but do not compromise the educational experience forthe student. The experiments have been tested repeatedly tomaximize a successful transition from the laboratory to the classroomsetting. Furthermore, the experiments allow teachers and studentsthe flexibility to further modify and adapt procedures for laboratoryextensions or alternative inquiry-based investigations.

ORGANIZING AND IMPLEMENTING THE EXPERIMENT

Class size, length of laboratory sessions, and availability of equipmentare factors which must be considered in the planning and the imple-

Notes to the Instructor




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Pre-Lab Preparations

1. Each student group should receive the following:

• PTC tasting strips for each student

• Control taste strips for each student

• 8 - 12 index cards

2. The experiment can be expanded to include several other“Mendelian” phenotypes, which also appear to lack any selec-tive advantage. For example, students can measure the classfrequency of free versus attached ear lobes. Free ear lobes aredominant, attached ear lobes are recessive.

3. Students may also use the Chi-square statistic to compare theirclassroom data with the average for North America.


Please refer to the kit insert for the Answers to

Study Questions

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Specific Gravity (H 0 = 1) 2

PTC taste paper

CAS # 103-85-5 Phenylthiourea, 1- Phenyl - 2 thiourea, Phenylthiocarbamide

08-19-05

NO data

NO data

NO data

NO data

NO data

NO data

NO data NO data NO data

use media suitable to extinguish surrounding fire

Firefighters should wear full protective equipment and NIOSH breathing apparatus

Thermal decomposition or contact with acids or acid fumes produces toxic fumes.

Soluble

paper strip

Stability

Section V - Reactivity DataUnstable

Section VI - Health Hazard Data

Incompatibility

Conditions to Avoid

Route(s) of Entry: Inhalation? Ingestion?Skin?

Other

Stable

Hazardous Polymerization

May Occur Conditions to Avoid

Will Not Occur

Health Hazards (Acute and Chronic)

Carcinogenicity: NTP? OSHA Regulation?IARC Monographs?

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Medical Conditions Generally Aggravated by Exposure

Emergency First Aid Procedures

Section VII - Precautions for Safe Handling and UseSteps to be Taken in case Material is Released for Spilled

Waste Disposal Method

Precautions to be Taken in Handling and Storing

Other Precautions

Section VIII - Control Measures

Ventilation Local Exhaust Special

Mechanical (General)

Respiratory Protection (Specify Type)

Protective Gloves

Other Protective Clothing or Equipment

Work/Hygienic Practices

Eye Protection

Hazardous Decomposition or Byproducts

X Heat, acid, or acid fumes

Acids

NOx, SOx

X Heat, acid, or acid fumes

No Yes Yes

Ingesting copious amounts of the chemical can be harmful or fatal.

None No data No data No data

Skin irritation. Ingesting copious amounts of the chemical will cause gastrointestinal discomfort.

None noted

Skin: Wash exposed area for 15 minutes. Seek medical attention if irritation persists.If copious amounts of the chemical are ingested, immediately call poison control center. Induce vomiting.

NA

Observe all federal, state, and local regulations.

Avoid skin contact. Avoid heat, acid or acid fumes. Keep in a cool, dry place.

Do not ingest more than the one taste strip provided by your instructor.

None needed under normal conditions with adequate ventilation.Yes

Yes

None

None

Splash-proof gogglesYes

Impervious clothing to prevent skin contact

EDVO-Kit · Advanced Placement Biology ... Biology AP Lab # 3 EVT 001188AM EDVO-Kit # 288...

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