population genetics and evolution - dublin unified school district

Population Genetics andEvolution

74-6540

TEACHER’S MANUAL

World-Class Support for Science & Math

ADVANCED PLACEMENT® BIOLOGY

Laboratory 8

©2004 Carolina Biological Supply Company Printed in USA

This protocol has been adapted from the Advanced Placement® Biology Laboratory Manual with permission from theCollege Entrance Examination Board. The AP designation is a registered trademark of the College Board. Theselaboratory materials have been prepared by Carolina Biological Supply Company, which bears sole responsibility forthe kit contents.

In this laboratory, students will

• learn about the Hardy-Weinberg law of genetic equilibrium

• study the relationship between evolution and changes in allele frequency byusing the class as a sample population

Before beginning this laboratory, students should understand

• the process of meiosis and its relationship to the segregation of alleles

• the basics of Mendelian Genetics

• the Hardy-Weinberg equation and its use in determining the frequency ofalleles in a population

• that natural selection can alter allelic frequencies in a population

• the effects of allelic frequencies of selection against the homozygous recessiveor other genotypes

At the completion this laboratory, students should be able to

• calculate the frequencies of alleles and genotypes in the gene pool of apopulation using the Hardy-Weinberg equations

• discuss natural selection and other causes of microevolution as deviations fromthe conditions required to maintain Hardy-Weinberg equilibrium

Exercise Description Approximate Time Requirement

8A Estimating allele frequencies for a 15 minutesspecific trait within a sample population

8B A Test of Hardy-Weinberg Equilibrium 30 minutes

8C Selection 30 minutes

8D Heterozygote Advantage 30 minutes

8E Genetic Drift (Optional) 30 minutes

Photocopy the Student Guide from this manual for your class.

Mark index cards with the letters A and a (see Setup for Each Student, below, andExercise 8B, Introduction, in the Student Guide).

For Exercise 8A, draw Table 8.2 on the board or overhead for use in collecting classdata.

For Exercises 8B, 8C, and 8D (and optional Exercise 8E), draw Table 8.3 on the boardor overhead. Data collection for the class is best performed by asking for a show ofhands.

TEACHER’S MANUAL LABORATORY 8

3

Time Requirements

Required

Knowledge

Expectations

Objectives

Preparation

LABORATORY 8. POPULATION GENETICS AND EVOLUTION

LABORATORY 8 TEACHER’S MANUAL

4

Following is a list of the materials needed for one student to perform the exercises inthis lab. Prepare as many setups as needed for your class.

*Item not included in kit.

Note: Each student should begin with four cards: two labeled A and two labeled a.Additional cards may be needed for the exercises that come later.

Safety

Some school systems ban the placement of anything in the mouth during a science class.Some school systems ban the use of PTC taste papers. Abide by your system’s state andlocal regulations. If you cannot do Exercise 8A with PTC taste paper, any readilyobservable trait that is controlled by a single gene locus can be substituted. See FurtherInvestigations for some suggested traits.

Exercise 8A: Estimating Allele Frequencies for a Specific Trait Within aSample PopulationNotice that we had to calculate a value for q before we could we could determine p.Why is this true?

1. Use your class data and the Hardy-Weinberg equations to complete Table 8.2.

strip of control paper

strip of PTC paper

3 × 5″ index cards labeled A

3 × 5″ index cards labeled a

extra index cards labeled A

extra index cards labeled a

*calculator with a square root function

Exercise8B

Exercise8C

Exercise8D

Exercise8E

2

2

4

4

1

2

2

4

4

1

2

2

4

4

1

2

2

4

4

1

Setup for Each Student

Student Materialsand Equipment

Exercise8A

1

1

1

Answers toQuestions in theSudent Guide

Phenotypes

Tasters Nontasters Frequency of(p2 + 2pq) (q2) the Alleles (%)

Count % of Total Count % of Total p qClass Data

North AmericanPopulation

Table 8.2*

Phenotypic Proportions of Tasters and Nontasters and Frequency of the

Determining Alleles

SG pg.13


5

Exercise 8B: A Test of Hardy-Weinberg EquilibriumStudents must understand that the four cards used in this exercise represent the allelespresent in the representative gametes produced as a result of the process of meiosis.They should also understand that the values for p and q are estimates of allelefrequencies and are not normally derived from direct counts, as is done in theseexercises. The Hardy-Weinberg equations give the most reliable results when applied tovery large samples. The sample sizes derived in these simulations are too small for theHardy-Weinberg equations to give accurate results; thus the need to use direct counts.

When taking the class data, remember that the totals for all of the genotypes (A/A + A/a+ a/a) should remain constant; that is, population size remains constant. If not, you havemissed counting someone or have counted someone twice. This condition appliesthroughout these simulations.

Exercise 8C: SelectionStudents may expect that the a allele will be rapidly eliminated; it will persist “hidden”in the heterozygote for many generations and the value of p will increase only slightlyafter five or ten generations (for humans, approximately 200 years).

Genotype Totals

Generation A/A A/a a/a

1

2

3

4

5

Table 8.3 Class Totals for

Exercise 8B SG pg. 15

Genotype Totals


1

2

3

4

5


Exercise 8C SG pg. 17

LABORATORY 8 TEACHER’S MANUAL

6

Exercise 8D: Heterozygote Advantage

Genotype Totals


1

2

3

4

5

6

7

8

9

10


Exercise 8D

SG pg. 19


7

Other kinds of forces that affect allele frequencies in a population, e.g., genetic drift,gene flow, changing the value of p, or changing the extent of selection, can also besimulated.

For further reference see “Evolution—More Than a Game,” by A.H. Markart III and P.Hyland, 1977, A.I.B.S. Education Review, Vol.6 (No.3).

The distributions of many other human genetic traits can be investigated. Following aresome suggestions taken from Corriher, Charles M., “Mendelian Inheritance in Humans,”Carolina Tips, Vol. 48, No. 11, November 1, 1985.

Earlobe Attachment: A dominant allele produces a free, or unattached, earlobe (Figure 1). The homozygous recessive condition is a direct attachment of the earlobe tothe head. Other genes affect the size and shape of the ear.

Tongue Rolling: The ability to roll the edges of the tongue upward from the sides toform a tube is due to a dominant allele (Figure 2). Nonrollers are homozygous recessivefor the trait.

Widow’s Peak: The widow’s peak is a distinct downward point of the frontal hairlineand is due to a dominant allele. Homozygous recessive individuals have a straighthairline. Of course, male pattern baldness, a sex-influenced trait, can complicate scoringfor this phenotype.

Middigital Hair: the presence of midphalangeal hair, hair between the second and thirdknuckles (going distally from the hand) is due to a dominant allele (Figure 3). Studentsmay require a hand lens to determine their phenotypes, as the hairs can be quite small.

Students can report on studies of the distribution of human blood groups and othertraits. Population studies of left-handedness are among the most fascinating andcontroversial of all. Many studies indicate that left-handed people are more prone todeath from various diseases and accidents. Since left-handedness seems to be inherited,there should be selection pressure against it, but the percentage of the human populationthat is left-handed seems not to have changed over thousands of years.

Further

Investigations

Figure 3:Middigital hair

Figure 2:Tongue roller

Figure 1:Left, a free earlobe;right, an attachedearlobe

LABORATORY 8 STUDENT GUIDE

8

In this activity, you will

• learn about the Hardy-Weinberg law of genetic equilibrium

• study the relationship between evolution and changes in allele frequency byusing your class as a sample population

Before beginning this laboratory, you should understand

• the process of meiosis and its relationship to the segregation of alleles

• the basics of Mendelian Genetics

• the Hardy-Weinberg equation and its use in determining the frequency ofalleles in a population

• that natural selection can alter allelic frequencies in a population

• the effects of allelic frequencies of selection against the homozygous recessiveor other genotypes

At the completion of this laboratory, you should be able to

• calculate the frequencies of alleles and genotypes in the gene pool of apopulation using the Hardy-Weinberg formula

• discuss natural selection and other causes of microevolution as deviations fromthe conditions required to maintain Hardy-Weinberg equilibrium

The Hardy-Weinberg Theorem states that the frequencies of alleles in a sexuallyreproducing population remain constant (in equilibrium) from generation to generationunless acted upon by outside factors. That is, if we consider two alleles, A and a, in apopulation, the reshuffling of alleles that occurs due to meiosis and recombination doesnot change the numbers of these alleles in the population. Hardy and Weinberg arguedthat a population’s allele and genotype frequencies would remain statistically constantas long as five conditions were met:

1. The breeding population is very large. In a small population, chance events cangreatly alter allele frequency. For example, if only five individuals of a smallpopulation of deer carry allele a and none of the five reproduce, allele a iseliminated from the population.

2. Mating is random. Individuals show no preference for a particular phenotype.

3. There is no mutation of the alleles.

4. No differential migration occurs. No immigration or emigration.

5. There is no selection. All genotypes have an equal chance of surviving andreproducing.

The Hardy-Weinberg Theorem provides a yardstick by which we can measure changesin allele frequency, and therefore, in evolution. If we can determine the frequency of apair of alleles in a population, we can sample that population over several generationsand answer the question, “Is the population evolving with respect to these particularalleles?” The Hardy-Weinberg equations can be applied to estimate the frequencies ofspecified alleles within a population at any given time.

LABORATORY 8. POPULATION GENETICS AND EVOLUTION

Objectives

Required

Knowledge

Background

Expectations

STUDENT GUIDE LABORATORY 8

9

Consider the following data on Rh blood type from a hypothetical human population:

Assume that Rh blood type is inherited through two alleles: D, a dominant allele forRh+ blood type, and a recessive allele d for Rh– blood type.

Using the Hardy-Weinberg equations,

let p = frequency of the dominant allele (D, in this case)

and q = frequency of the recessive allele (d)

Each person is either Rh+ or Rh–

therefore,

p + q = 1 or 100% of the population

Since humans are diploid, individuals may be homozygous dominant (D/D),heterozygous (D/d), or homozygous recessive (d/d). Thus, the basic equation must beexpanded to represent all the genotype frequencies.

Thus, (p + q) (p + q) = 1

or p2 + 2pq + q2 = 1

where p2 = frequency of the homozygous dominant (D/D),

2pq = frequency of the heterozygous condition (D/d),

and q2 = frequency of the homozygous recessive (d/d).

Notice that p2 + 2pq = frequency of the Rh+ phenotype in the population.

From the data in Table 8.1, we can now calculate q2, the frequency of the homozygousrecessive:

q2 = 1320/6000 = 0.22

then q = √ 0.22 = 0.47

p + q = 1

p = 1 – q

p = 1 – 0.47 = 0.53

This tells us that 53% of the population tested has the allele D and 47% has the allele d.

If we repeated our sample over several generations, we could tell if this population isevolving with respect to the Rh factor alleles.

Table 8.1 Results of Testing for Human Rh Blood Type

Blood Number ofType People

Rh+ 4,680

Rh– 1,320

Total 6,000


10

Notice that we had to calculate a value for q before we could we could determine p.Why is this true?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Using the class as a sample population, you will estimate the allele frequency of a genecontrolling the ability to taste the chemical PTC (phenylthiocarbamide). A bitter-tastereaction to PTC is evidence of the presence of a dominant allele in either thehomozygous condition (A/A) or the heterozygous condition (A/a). The inability to tastethe chemical at all depends on the presence of homozygous recessive alleles (a/a).

1. Press a strip of control taste paper to your tongue tip. Wait until your saliva hascompletely saturated the paper, then note the taste of the paper.

2. Now repeat Step 1 using the PTC taste paper. PTC tasters will sense a bitter taste.For the purposes of this exercise, these individuals are considered to be tasters. Ifyou sense little more than the taste of the paper itself, you are a nontaster. Recordthe class data in Table 8.2.

1. Use your class data and the Hardy-Weinberg equations to complete Table 8.2.

2. What percentage of your class are heterozygous tasters? ______________________

Introduction

Exercise 8A: Estimating Allele Frequencies for a Specific Trait Within a Sample Population

Procedure

Questions

Phenotypes

Tasters Nontasters Frequency of(p2 + 2pq) (q2) the Alleles (%)

Count % of Total Count % of Total p qClass Data

North American70 30Population

Table 8.2 Phenotypic Proportions of Tasters and Nontasters and

Frequencies of the Determining Alleles


11

3. What are some reasons that the values for p and q for your class might differ fromthe same frequencies reported for the North American population?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

4. Would you expect the frequencies of the alleles for PTC tasting and nontasting toremain constant for North America over the next 200 years? State your answer interms of the five conditions that Hardy and Weinberg stated must be fulfilled forallele frequencies to remain constant.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

In this activity, your entire class will simulate a population of randomly matingindividuals. Choose another student at random (for this simulation, assume that genderand genotype are irrelevant to mate selection.) The population begins with a frequencyof 0.5 (50%) for the dominant allele A and also 0.5 (50%) for the recessive allele a.Your initial genotype is A/a. Record this on the Data Sheet. You have four cards: eachrepresents a chromosome. Two cards (chromosomes) will have allele A and two cardswill have allele a. The four cards represent the products of meiosis. Each “parent” willcontribute a haploid set of chromosomes to the next generation.

1. Turn the four cards over so the letters are not showing, shuffle them, and take thecard on top to contribute to the production of the first offspring. Your partner shoulddo the same. Put the two cards together. The two cards represent the alleles of thefirst offspring. One of you should record the genotype of this offspring as theGeneration 1 Genotype on his or her Data Sheet.

2. Retrieve your cards and reshuffle them. Repeat Step 1 to produce a secondoffspring. The second partner records the genotype of this offspring on his or herData Sheet. The very short reproductive career of this generation is over.

Exercise 8B: A Test of Hardy-Weinberg Equilibrium

Introduction

Procedure


12

3. You and your partner now become the next generation by assuming the genotypesof the two offspring you have produced. That is, Student 1 assumes the genotype ofthe first offspring and Student 2 assumes the genotype of the second offspring asyou have recorded them on your Data Sheets. Obtain additional cards if necessary.For example, if you now have the genotype a/a, you will need four cards, allmarked a. If you have the genotype A/A, you will need four cards all marked A. Ifyou have the genotype A/a, keep the original four cards.

4. Now, randomly seek out another person with whom to mate in order to produce theoffspring of the next generation. The sex of your mate does not matter, nor does thegenotype. Repeat Steps 1–3, being sure to record your new genotype, after eachgeneration, on your Data Sheet. Repeat this exercise to produce five generations.

5. Your teacher will collect class data for Generation 5 by asking you to raise yourhand to report your genotype. Record the class totals in Table 8.3

Compare your data for Generation 5 in Table 8.3 with the class data for PTC tasting inTable 8.2. What information do you have for Generation 5 that you do not have in Table 8.2?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

From Table 8.3, what is the population size? ___________________________________

Genotype Totals


1

2

3

4

5

Table 8.3 Class Totals for Exercise 8B

Data Analysis


13

The Hardy-Weinberg equations are used to give estimates of allele frequencies for largepopulations. Here, we have a small population and there is no need to estimate allelefrequencies, because we have actual counts. Calculate p and q as follows:

Number of A alleles present at the fifth generation

Number of offspring with genotype A/A _________ × 2 = _________ A alleles

Number of offspring with genotype A/a _________ × 1 = _________ A alleles

Total = _________ A alleles

p = Total number of A alleles = ________Total number of alleles in the population

In this case, the total number of alleles in the population is equal to the number ofstudents in the class × 2.

Number of a alleles present at the fifth generation

Number of offspring with genotype a/a _________ × 2 = _________ a alleles

Number of offspring with genotype A/a _________ × 1 = _________ a alleles

Total = _________ a alleles

q = Total number of a alleles = ________Total number of alleles in the population

1. What are the frequencies of the alleles in Generation 5?

a. the frequency (q) of allele a in Generation 5? _____________________________

b. the frequency (p) of allele A in Generation 5? _____________________________

2. Are the values for p and q in Generation 5 different from the beginning values?____________________________________________________________________

3. Is your answer to Question 1, above, consistent with the alleles being atequilibrium?____________________________________________________________________

If not, why not?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Questions


14

4. Look back at the five conditions that must be met for allele frequencies to remainconstant. Which, if any, of these conditions might not have been met in thissimulation?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

In nature, not all genotypes have the same rate of survival; that is, the environmentmight favor some genotypes while selecting against others. An example is humansickle-cell anemia, a disease caused by a single gene mutation. Individuals who arehomozygous recessive (a/a) often do not survive to reach reproductive maturity. In thissimulation, you will assume that the homozygous recessive individuals never survive(100% selection against), and that heterozygous and homozygous dominant individualssurvive 100% of the time.

Everyone begins with the heterozygous genotype; thus, the initial frequency of each ofthe alleles is again 0.5 (50%). Follow the procedure in Exercise 8B, with the followingmodification: every time your “offspring” is a/a, assume that it does not survive toreproduce. Because you want to maintain a constant population size, the same twoparents must try again until they produce two surviving offspring.

Proceed through five generations, selecting against the homozygous recessive offspring100% of the time. Then, calculate the new p and q frequencies using the method from 8B.


a. the frequency (q) of allele a in Generation 5? _____________________________

b. the frequency (p) of allele A in Generation 5? _____________________________

Exercise 8C: Selection

Introduction

Procedure

Genotype Totals


1

2

3

4

5

Table 8.4 Class Totals for Exercise 8C

Questions


15

2. Are the values for p and q in Generation 5 different from the beginning values?Explain your answer.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

3. Compare the values for p and q calculated for Generation 5, Exercise 8B with thevalues you just calculated for Exercise 8C. How do the new frequencies of p and qcompare to the initial frequencies in Exercise 8B?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

4. Predict what would happen to the frequencies of p and q if you simulated anotherfive generations.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

5. In a large population, would it be possible to completely eliminate a deleteriousrecessive allele? Explain your answer.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

From Exercise 8C, it is easy to see that the lethal recessive allele rapidly decreases inthe population. However, studies show an unexpectedly high frequency of the sickle-cellallele in some human populations. These populations exist in areas where malaria is (oruntil recently was) killing many people. It seems that individuals who are heterozygousfor sickle-cell anemia are slightly more resistant to a deadly form of malaria than arehomozygous dominant individuals. In malaria-ridden areas, there is a slight selectionagainst homozygous dominant individuals as compared to heterozygotes. This fact iseasily incorporated into our simulations.

Exercise 8D: Heterozygote Advantage

Introduction


16

Begin again with the genotype A/a. Follow the procedure in Exercise 8B, with thefollowing modifications: if your offspring is A/A, flip a coin. If heads, the offspring doesnot survive. If tails, the offspring does survive. The genotype a/a never survives. Parentsmust produce two surviving offspring each generation. This time, simulate tengenerations. Total the class genotypes and then calculate the p and q frequencies forGeneration 5 and for Generation 10. If time permits, the results from another fivegenerations would be extremely informative.


a. the frequency (q) of allele a in Generation 10? ____________________________

b. the frequency (p) of allele A in Generation 10? ____________________________

2. Account for the differences in p and q frequencies from Exercise 8C to Exercise 8D.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

3. Do you think the recessive allele will be completely eliminated in either Exercise8C or Exercise 8D? Explain your answer.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Procedure

Questions

Genotype Totals


1

2

3

4

5

6

7

8

9

10

Table 8.5 Class Totals for Exercise 8D


17

4. What is the importance of heterozygotes (the heterozygote advantage) inmaintaining genetic variation in populations?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

5. Suppose you repeated this activity, but you did the coin toss to determine if the A/aindividuals reproduce and all of the A/A individuals reproduced. How would youexpect this to change the allele frequencies for Generation 10?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

It is possible to use our simulation to look at the phenomenon of genetic drift in detail.

Divide the lab into several smaller, isolated populations. For example, a class of 30could be divided into 3 separate populations of 10 individuals each. Individuals fromone population do not interact with individuals from other populations. Follow theprocedure in Exercise 8B through five generations. Record the new genotypicfrequencies and then calculate the new frequencies of p and q for each population.

1. Explain how the initial genotypic frequencies of the populations compare.

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

2. What do your results indicate about the importance of population size as anevolutionary force?

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Exercise 8E: Genetic Drift

Introduction

Procedure

Questions


18

1. In Drosophila, the allele for normal length wings is dominant over the allele forvestigial wings (vestigial wings are stubby, little curls that cannot be used for flight).In a population of 1,000 individuals, 360 show the recessive phenotype. Use theHardy-Weinberg equations to estimate the number of homozygous dominant andheterozygous individuals.

2. The allele for the ability to roll the tongue into a tube is dominant over the allele forthe lack of this ability. In a population of 500 individuals, 25% show the recessivephenotype. Use the Hardy-Weinberg equations to estimate the number ofhomozygous dominant and heterozygous individuals.

3. The allele for the hair pattern called “widow’s peak” is dominant over the allele forno widow’s peak. In a population of 1,000 individuals, 510 show the dominantphenotype. About how many individuals would have each one of the three possiblegenotypes?

4. In the United States, about 16% of the population is Rh–. The allele for Rh– isrecessive to the allele for Rh+. If a high school has a population of 2,000 students,about how many students would have each one of the three possible genotypes?

5. In certain African countries, 4% of the newborn babies have sickle-cell anemia,which is a recessive trait. Out of a random population of 1,000 newborn babies,about how many babies would have each one of the three possible genotypes?

6. In a certain population, the dominant phenotype of a certain trait occurs 91% of thetime. What is the frequency of the dominant allele?

Hardy-WeinbergProblems

19

LABORATORY 8 DATA SHEET

My InitialA/a

Genotype

Generation 1

Genotype

Generation 2

Genotype

Generation 3

Genotype

Generation 4

Genotype

Generation 5

Genotype

Final Class Frequencies

p = q =

Exercise 8B

Hardy-Weinberg Equilibrium

Initial Class Frequencies

p = 0.5 q = 0.5

My InitialA/a

Genotype

Generation 1

Genotype

Generation 2

Genotype

Generation 3

Genotype

Generation 4

Genotype

Generation 5

Genotype


p = q =

Exercise 8C

Selection


p = 0.5 q = 0.5

My InitialA/a

Genotype

Generation 1

Genotype

Generation 2

Genotype

Generation 3

Genotype

Generation 4

Genotype

Generation 5

Genotype


p = q =

Exercise 8E

Genetic Drift


p = 0.5 q = 0.5My Initial

A/aGenotype

Generation 1

Genotype

Generation 2

Genotype

Generation 3

Genotype

Generation 4

Genotype

Generation 5

Genotype

Final Class Frequencies,

Generation 5

p = q =

Exercise 8D

Heterozygote Advantage

Generation 6

Genotype

Generation 7

Genotype

Generation 8

Genotype

Generation 9

Genotype

Generation 10

Genotype

Final Class Frequencies,

Generation 6

p = q =


p = 0.5 q = 0.5

Name ____________________

Date _____________________

20

LABORATORY 8

Carolina Biological Supply Company2700 York Road, Burlington, North Carolina 27215

Phone: 800.334.5551 • Fax: 800.222.7112Technical Support: 800.227.1150 • www.carolina.com

CB251450405

population genetics and evolution - dublin unified school district

Documents