mslinscience.weebly.commslinscience.weebly.com/.../2/3/2/...and_the_hardy_weinb… · Web viewEvolution is any change in the frequency of alleles in a population. Evolution is the

Name _____________________________________

Date ________________

Lab: Population Genetics, The Hardy-Weinberg Principle, and Evolution

Introduction: Evolution is any change in the frequency of alleles in a population. Evolution is the result of changes in the gene pool. Two men, G. H. Hardy and W. Weinberg, proposed a mathematical model for detecting changes in the gene pool.

The Hardy-Weinberg Principle states: “In the absence of mutation, migration, and natural selection, and in a population that is sufficiently large, the frequencies of alleles will remain the same.” No population is free of these agents of change. The Hardy-Weinberg equation is used to detect changes in the population from one generation to the next.

A population is said to be in Hardy-Weinberg equilibrium if 5 conditions are met: The population is large. (The effect of “chance” is greatly reduced in a large population.) Mating is random. (Individuals show no preference for a particular phenotype when choosing a

mate.) There is no mutation. There is no immigration or emigration. (No individuals move into or out of the population.) There is no natural selection. (All offspring have an equal chance of surviving and reproducing.)

If a population is in Hardy-Weinberg equilibrium, there are no changes in allele frequency and the population is not evolving. If the above five conditions are met, the population’s allele and genotype frequencies will remain constant from one generation to the next.

p = the frequency of the dominant allele in a populationq = the frequency of the recessive allele in a population

p2 = the frequency of the homozygous dominant genotype in a populationq2 = the frequency of the homozygous recessive genotype in a population2pq = the frequency of the heterozygous genotype in a population

Purpose: 1. To simulate how changes in the gene pool might occur by using the class as a breeding population of

individuals.2. To observe how the Hardy-Weinberg equation is used to detect changes in allele frequencies in a

population.

Materials: PTC test papers, Calculator, Allele cards, Coins, Pencil and paper

Safety Precautions: None

The Hardy-Weinberg equation: p2 + 2pq + q2 = 1

1

Procedure Part A: Determining the Frequency of a Trait in a Population

The ability to taste the chemical PTC is due to a dominant allele. A person would have the ability to taste PTC if they were homozygous (TT) for this trait or heterozygous (Tt) for this trait. The homozygous recessive individual (tt) is unable to taste the chemical PTC.

Obtain a piece of PTC paper from your teacher. Press the paper against your tongue. PTC tasters will notice a bitter taste.

1. Are you a taster or a nontaster?

2. What is the genotype of the nontasters?

3. What is the genotype of the tasters?

4. What percent of the class is a nontaster?

5. What part of the Hardy-Weinberg equation represents the individuals who are nontasters?

6. Determine the frequency of p and q in the classroom population.

7. What percent of the class population should be homozygous for tasting?

8. What percent of the class population should be heterozygous for tasting?

Procedure Part B: Testing the Hardy-Weinberg Principle

Now, let’s run a simulation to test the validity of the Hardy-Weinberg Principle. The class will represent a stable, breeding population of individuals. Each individual in the class is heterozygous for a particular trait.

Procedure: Each class member should obtain 4 cards from the instructor. You should receive two cards with “A” written on them, and two cards with “a” written on them.

2

Your genotype is Aa. You will pass one of these alleles to your future offspring in the process of meiosis. Remember that in meiosis the Aa will be replicated and then one allele will be placed into each of four gametes. Thus, you should have four cards. Two cards should have “A” and two cards should have “a”.

Move about the room, and find another student to “mate” with. (I will refer to Student 1 and Student 2 in the remainder of the instructions.)

Place your four cards face down and shuffle them. Student 1 should draw a card from student 2, and student 2 should draw a card from student 1. These two drawn cards represent the alleles of the first offspring.

Student 1 should record the genotype of this offspring in the data table below (question #10). Repeat the “mating” process. Student 1 should draw a card from student 2, and student 2 should

draw a card from student 1. These two drawn cards represent the alleles of the second offspring. Student 2 should record the genotype of this offspring in the data table below (question #10). The first round of mating is over! Student 1 and Student 2 should now assume the identity of the

two offspring. Student 1 assumes the genotype of the first offspring. Student 2 assumes the identity of the second offspring.

If necessary, each student should obtain new cards to represent the alleles of the first and second offspring. If the offspring was AA, then obtain 4 cards with “A”. If the offspring was “a”, then obtain four cards with “a”. If the offspring was “Aa”, then obtain two cards with “A” and two cards with “a”.

Randomly move about the room, and choose another student with whom to mate for the second round.

Repeat this mating process until you have completed five rounds. Be sure to fill out the data table at the end of each round.

9. If every individual in the class is heterozygous for this trait at the beginning of the simulation, what is the initial frequency of p and q in the class population? Explain your answer.

10. Record your genotype at the end of each mating cycle:

First generation: _______________________

Second generation: ____________________

Third generation: ______________________

Fourth generation: ____________________

Fifth generation: _______________________

11. Determine the total number of each genotype for the class population at the end of five generations.

3

The instructor will ask for a show of hands. You will raise your hand to indicate your genotype at

the end of the fifth generation.

Class TotalsAAAaaa

Total number of students in class

12. At the end of five mating generations, what does the Hardy-Weinberg Principle predict as the value of the new p and q? Explain your answer.

13. Determine the value of p and q after five generations:

How many ”A” alleles are present in the population after five generations?

Number of students with “AA” genotype _____________ x 2 = ______________ “A” alleles

Number of students with “Aa” genotype _____________ x 1 = ______________ “A” alleles

Total number of “A” alleles = ______________

Total number of “A” allelesp = ____________________________________________________________ = _________________

Total number of alleles in the class population(# students x 2)

How many ”a” alleles are present in the population after five generations?

Number of students with “aa” genotype _____________ x 2 = ______________ “a” alleles

Number of students with “Aa” genotype _____________ x 1 = ______________ “a” alleles

Total number of “a” alleles = ______________

Total number of “a” allelesq = ____________________________________________________________ = _________________


14. Do the results of this simulation agree with the Hardy-Weinberg prediction? If not, why?

4

15. Determine if each major assumption of the Hardy-Weinberg Principle was supported or violated by this simulation.

a) Was mating completely random? __________b) Were there any mutations? __________c) Was there any migration into or out of the population? __________d) Did all offspring have an equal chance (no natural selection) of surviving and reproducing?

__________e) Was the population size sufficiently large? __________

16. Which part of the Hardy-Weinberg Principle was violated by this simulation? What effect did this have on the predicted values of p and q?

Procedure Part C: Testing the Hardy-Weinberg Principle for Selection Pressure

Now let’s make the simulation a bit more realistic. In nature, not every offspring has an equal chance of surviving and reaching reproductive age. Natural selection occurs in a population when some offspring are “favored” by the environment. Some of the offspring will be better suited to survive in a particular environment, and are more likely to reach reproductive age to pass their alleles on to future offspring. We will use the disease of sickle cell anemia as an example. Sickle cell anemia is caused by a recessive allele, so that individuals that are homozygous recessive have the disease. It is rare that these individuals reach reproductive age. There is selection pressure against the “aa” genotype.

Procedure: In this simulation, 100% of the homozygous recessive individuals will die, and 100% of the

normal (AA or Aa) individuals will survive. The procedure is similar to that of the first simulation. Each student will begin the simulation with the heterozygous (Aa) genotype. Randomly select other students with whom you will mate. However, if you produce an offspring

that is “aa”, this offspring will not survive. You must try again until two surviving offspring are produced.

At the end of each generation, assume the identity of the offspring, and proceed to the next generation. Continue the process through five generations.

At the end of each generation, record your genotype in the table below (question 18).

17. If every individual in the class is heterozygous for this trait at the beginning of the simulation, what is the initial frequency of p and q in the class population?


5







The instructor will ask for a show of hands. You will raise your hand to indicate your genotype at the end of the fifth generation.

Class TotalsAAAaaa










6






21. How does your calculated value of p and q after five generations compare to the initial frequency of p and q?


a) Was mating completely random? __________b) Were their any mutations? __________c) Was there any migration into or out of the population? __________d) Did all offspring have an equal chance (no natural selection) of surviving and reproducing?

__________e) Was the population size sufficiently large? __________

23. Which part(s)of the Hardy-Weinberg Principle was violated by this simulation? What effect did this have on the predicted values of p and q?

24. If the simulation was allowed to continue for an additional five generations, what would happen to the value of p and q?

25. Will the presence of the recessive sickle cell allele ever completely disappear from the population? Why or why not?

Procedure Part D: The Heterozygote Advantage

7

In some instances, individuals who are heterozygous for a particular trait have an advantage over individuals who are homozygous dominant and homozygous recessive for the trait. Such is the case with the sickle cell anemia allele. There is intense selection pressure against individuals who are homozygous recessive for sickle cell. These individuals rarely survive to reproductive age. However, individuals who are heterozygous for the sickle cell trait are actually “favored” by the environment and have an advantage. These individuals are more resistant to malaria than individuals who are AA. This protection is very important in tropical regions where malaria causes the death of many individuals. Thus, the heterozygous individual (Aa) will be favored by the environment. This “Aa” individual does not have sickle cell anemia, and they are more resistant to malaria than the “AA” individual.

Procedure: The procedure is similar to that of the first and second simulation. Each student will begin the simulation with the heterozygous (Aa) genotype. Randomly select other students with whom you will mate. If you produce an offspring that is “aa”,

this offspring will not survive. The new twist: If the offspring is AA, they will not have sickle cell anemia, but they may die from

malaria, since they are not as resistant as the “Aa” individuals. If your offspring is “AA”, you must flip a coin. If the coin is “tails”, the individual dies of malaria. If the coin is “heads”, the individual survives.

You must continue until two surviving offspring are produced. At the end of each generation, assume the identity of the offspring, and proceed to the next

generation. Continue the process through five generations. At the end of each generation, record your genotype in the table below (question 27).









8

The instructor will ask for a show of hands. You will raise your hand to indicate your genotype at the end of the fifth generation.

Class TotalsAAAaaa















30. How does the new value of p and q in this simulation compare with the value of p and q from Part C? Explain why this happened.

9

31. Which individuals were favored in this simulation? How will this affect the frequency of the harmful allele, “a”?

32. Why are individuals who are heterozygous for different traits so important to a population?

Final Analysis Questions:

33. Define evolution.

34. Write the Hardy-Weinberg equation. What is indicated by each component of the equation?

35. What five conditions must be met in order for a population to be in Hardy-Weinberg equilibrium?

a)

b)

c)

d)

e)

Complete the following Hardy-Weinberg Problems:

10

36. The allele for dimples (D) is dominant. In a random sample of 2000 individuals, 1745 individuals had no dimples. Determine the frequency of the dominant and recessive allele in this population. How many individuals would be expected to be homozygous dominant for dimples? How many individuals would be expected to be heterozygous for dimples?

37. The allele for brown eye color is dominant over the allele for blue eye color. In a population of 15,000 individuals, 25% were found to have blue eyes. How many individuals in this population are homozygous and heterozygous for brown eyes?

38. The ability to taste the chemical PTC is controlled by a dominant allele. In a random sample of 5000 individuals, it was determined that 3500 of the individuals were tasters. Determine the number of individuals that are TT, Tt, and tt for this trait.

39. In a particular population, it has been determined that the dominant phenotype is present in 84% of the population. What is the frequency of the dominant and recessive allele in this population?

40. A particular recessive allele is known to be lethal when present in the homozygous condition (aa). The number of offspring born with this lethal disease is 1 in every 6000 births. In order for children

11

to be born with this recessive combination (aa) a certain percentage of the population must be heterozygous for the trait. What percentage of this population is heterozygous for this trait?

41. If a population has the following genotype frequencies:

AA = .32 Aa = .46 aa = .22

What is the frequency of the dominant allele (A)? What is the frequency of the recessive allele (a)?(Note: Do not use the Hardy-Weinberg equation to solve this. Solve this problem using the method learned in this lab.)

42. A small group of 30 people decide to isolate themselves from the world and move to a small and remote deserted island. 10 of the individuals have albino skin, which is due to the homozygous recessive condition. The remaining 20 individuals have normal skin but are heterozygous for this condition. What is the frequency of the dominant allele? What is the frequency of the recessive allele?

43. When the island population reaches 750 individuals, how many of the inhabitants are predicted to have albino skin?

Copyright © April 2013 Amy Brown (aka Science Stuff)

Teacher Preparation and Answer Key

12

1. The materials list is short and simple. The only item you will have to order is the PTC test papers. If PTC paper is not available, other traits can easily be observed in the classroom population:

a) The presence of dimples is dominant over the absence of dimples.b) Free earlobes are dominant over attached earlobes.c) The ability to roll the tongue is dominant over the inability to roll the tongue.

Note: The above examples are not always so clear-cut, and there is evidence that more than one pair of alleles governs these traits. For the purpose of this lab exercise, let’s assume that each trait is controlled by a single pair of alleles and that there is a clear-cut difference between the phenotypes.

2. This is most definitely a guided activity. The class must work together to generate a large amount of data. Students will not be able to complete this unless the teacher serves as a moderator during the course of the activity.

3. The handouts for the student are numerous. I usually have 4 to 5 classes of students doing this lab. I usually run off only a class set of handouts, and ask the students to record their work on their own notebook paper. Running off a set of handouts for each student will require a lot of paper.

4. For consistency, I have my students carry out all answers to three places past the decimal. Final answers can be greatly affected by too much rounding.

5. Prior to the lab, print out multiple copies of the allele cards. I highly recommend that you print these on card stock paper. Otherwise, the students can see through the card from the backside. After printing on card stock, laminate the pages, then cut out the cards into individual cards. These laminated cards should last for years.

You will need to have two “A” cards and two “a” cards for each student in the class. Print extra cards since the genotype of the student will change during the simulation. The student will need to exchange their cards from the pool of extra cards during the exercise.

Answers to Questions and Problems:

Part A: Determining the Frequency of a Trait in a Population

1. Are you a taster or a nontaster? Student answers will vary.

2. What is the genotype of the nontasters? tt

3. What is the genotype of the tasters? TT or Tt

4. What percent of the class is a nontaster?

Sample Data: Consider a class size of 30 students. It is determined that 22 of the students are tasters and 8 of the students are nontasters.

8/30 = 26.67%5. What part of the Hardy-Weinberg equation represents the individuals who are nontasters? q2

6. Determine the frequency of p and q in the classroom population.13

Since q2 = 26.67%

q2 = .2667q = .516

Since all of the alleles in the population are either dominant or recessive, p + q = 1.

p + q = 1 p + .516 = 1 p = .484

7. What percent of the class population should be homozygous for tasting?

The homozygous dominant portion of the population is represented by p2 in the Hardy-Weinberg equation.

p2 (TT) = (.484)2

p2 = .2343 or 23.43%

8. What percent of the class population should be heterozygous for tasting?

The heterozygous portion of the population is represented by 2pq in the Hardy-Weinberg equation.

2pq (Tt) = (2) (.484) (.516)

2pq = .4995 or 49.95%

Part B: Testing the Hardy-Weinberg Principle

9. If every individual in the class is heterozygous for this trait at the beginning of the simulation, what is the initial frequency of p and q in the class population? Explain your answer.

If all students begin the simulation with the genotype of “Aa”, then 50% of the alleles in the class population are “A” and 50% of the alleles are “a”.

p = .5 and q = .5

10. Record your genotype at the end of each mating cycle: Student answers will vary.


14

The instructor will ask for a show of hands. You will raise your hand to indicate your genotype at the fifth generation.

Class TotalsAA 8

Aa 15

aa 7

Total number of students in class 30

12. At the end of five mating generations, what does the Hardy-Weinberg Principle predict as the value of the new p and q? Explain your answer.

At the end of five generations, the Hardy-Weinberg Principle predicts that p will remain .5 and that q will remain .5. (The assumption is that the population is in Hardy-Weinberg equilibrium and there will be no changes in allele frequencies. )



Number of students with “AA” genotype: 8 x 2 = 16 “A” allelesNumber of students with “Aa” genotype: 15 x 1 = 15 “A” alleles

Total number of “A” alleles = 31

p = Total number of “A” alleles / Total number of alleles in the class population

p = 31 / 60 = .517


Number of students with “aa” genotype: 7 x 2 = 14 “a” allelesNumber of students with “Aa” genotype 15 x 1 = 15 “a” alleles

Total number of “a” alleles = 29

q = Total number of “a” alleles / Total number of alleles in the class population

q = 29 / 60 = .483

14. Do the results of this simulation agree with the Hardy-Weinberg prediction? If not, why?

We predicted that both p and q would remain at the frequency of .5. Our actual values (p = .517 and q = .483) are very close to the expected values.


Sample Data:

15

a) Was mating completely random? Yesb) Were there any mutations? Noc) Was there any migration into or out of the population? Nod) Did all offspring have an equal chance (no natural selection) of surviving and reproducing?

Yese) Was the population size sufficiently large? No

16. Which part of the Hardy-Weinberg Principle was violated by this simulation? What effect did this have on the predicted values of p and q?

Our classroom population is not sufficiently large. The smaller the population, the greater the effect of “chance” on the outcome. Our calculated value for p and q were slightly different than the expected outcome.

Part C: Testing the Hardy-Weinberg Principle for Selection Pressure



p = .5 and q = .5




Class TotalsAA 17

Aa 13

aa 0




Sample Data:

16




p = 47 / 60 = .783





q = 13 / 60 = .217

21. How does your calculated value of p and q after five generations compare to the initial frequency of p and q?

The initial frequency: p = .5 and q = .5

After five generations of applied “selection pressure”, the value of p increased and the value of q decreased.

There was selection pressure against the homozygous recessive (aa) and the value of q decreased as a result.


a) Was mating completely random? Yesb) Were their any mutations? Noc) Was there any migration into or out of the population? Nod) Did all offspring have an equal chance (no natural selection) of surviving and reproducing?

Noe) Was the population size sufficiently large? No

23. Which part(s)of the Hardy-Weinberg Principle was violated by this simulation? What effect did this have on the predicted values of p and q?

Violated:

17

a) The population size is too small. Chance plays a greater role in affecting the outcome of the simulation.

b) All offspring do NOT have an equal chance of surviving and reproducing. The homozygous recessive offspring never survived, and this created selection pressure against the “a” allele.

Effect: This caused the frequency of the dominant allele (A) to increase and the frequency of the recessive allele (a) to decrease.

24. If the simulation was allowed to continue for an additional five generations, what would happen to the value of p and q?

The value of p would continue to increase and the value of q would continue to decrease. However, the value of q will never reach zero because the recessive allele (a) is maintained in the heterozygous individuals.

25. Will the presence of the recessive sickle cell allele ever completely disappear from the population? Why or why not?

No. The recessive allele continues on in the heterozygous individuals.

Part D: The Heterozygote Advantage



p = .5 and q = .5




Class TotalsAA 12

Aa 18

aa 0




18




p = 42 / 60 = .7





q = 18 / 60 = .3

30. How does the new value of p and q in this simulation compare with the value of p and q from Part C? Explain why this happened.

As expected, the value of p increased from the original frequency of .5. The value of q decreased from the original frequency of .5.

Compared to the results in Part C, however, the value of p increased less and the value of q decreased less than the results of Part C.

In Part C, the “aa” individual was selected against and the “Aa” individual was not favored. In Part D, the heterozygous individual was favored. Therefore, the value of q will not decrease as rapidly since “a” is favored in the heterozygote.

31. Which individuals were favored in this simulation? How will this affect the frequency of the harmful allele, “a”?

The heterozygous individual is favored. This will prevent the frequency of the harmful recessive allele from decreasing at a rapid rate.

32. Why are individuals who are heterozygous for different traits so important to a population?

The heterozygous individual has two different alleles that might be passed to their offspring. This maintains the genetic variations that are possible amongst the offspring. There is greater potential for variation in the offspring. Without heterozygous individuals in a population, genes would become “fixed” and variation would be limited.

Final Analysis Questions:

33. Define evolution.

19

Evolution is any change in the frequency of alleles in a population. Evolution is the result of changes in the gene pool.

34. Write the Hardy-Weinberg equation. What is indicated by each component of the equation?

The Hardy-Weinberg equation: p2 + 2pq + q2 = 1

p = the frequency of the dominant allele in a populationq = the frequency of the recessive allele in a population

p2 = the frequency of the homozygous dominant genotype in a populationq2 = the frequency of the homozygous recessive genotype in a population2pq = the frequency of the heterozygous genotype in a population

35. What five conditions must be met in order for a population to be in Hardy-Weinberg equilibrium?

A population is said to be in Hardy-Weinberg equilibrium if 5 conditions are met: The population is large. (The effect of “chance” is greatly reduced in a large population.) Mating is random. (Individuals show no preference for a particular phenotype when

choosing a mate.) There is no mutation. There is no immigration or emigration. (No individuals move into or out of the population.) There is no natural selection. (All offspring have an equal chance of surviving and

reproducing.)

Complete the following Hardy-Weinberg Problems:

36. The allele for dimples (D) is dominant. In a random sample of 2000 individuals, 1745 individuals had no dimples. Determine the frequency of the dominant and recessive allele in this population. How many individuals would be expected to be homozygous dominant for dimples? How many individuals would be expected to be heterozygous for dimples?

q2 = 1745 / 2000q2 = .873q = .934

If q = .934, then p must equal .066.

Homozygous dominant (p2) = (.066)2 = .00436 or .436%.00436 x 2000 = 9 individuals that are homozygous dominant

Heterozygous (2pq) = (2) (.066) (.934) = .1233 or 12.33%.1233 x 2000 = 246 individuals that are heterozygous

37. The allele for brown eye color is dominant over the allele for blue eye color. In a population of 15,000 individuals, 25% were found to have blue eyes. How many individuals in this population are homozygous and heterozygous for brown eyes?

q2 = 25%

20

q2 = .25q = .5p = .5

Homozygous Brown eyes (p2) = (.5)2 = .25 or 25%.25 x 15,000 = 3750 individuals that are homozygous dominant for brown eyes

Heterozygous Brown eyes (2pq) = (2) (.5) (.5) = .50 or 50%.5 x 15,000 = 7500 individuals that are heterozygous for brown eyes

38. The ability to taste the chemical PTC is controlled by a dominant allele. In a random sample of 5000 individuals, it was determined that 3500 of the individuals were tasters. Determine the number of individuals that are TT, Tt, and tt for this trait.

3500 are tasters (TT or Tt)1500 are nontasters (tt)

q2 = 1500 / 5000q2 = .3q = .548p = .452

TT (p2) = (.452)2 = .2043 or 20.43%.2043 x 5,000 = 1022 individuals that are homozygous dominant

Tt (2pq) = (2) (.452) (.548) = .4954 or 49.54%.4954 x 5,000 = 2477 individuals that are heterozygous

tt = 1500 individuals that are homozygous recessive

39. In a particular population, it has been determined that the dominant phenotype is present in 84% of the population. What is the frequency of the dominant and recessive allele in this population?

If the dominant phenotype is present in 84% of the population, then the recessive phenotype is present is 16% of the population.

q2 = 16%q2 = .16q = .4p = .6

40. A particular recessive allele is known to be lethal when present in the homozygous condition (aa). The number of offspring born with this lethal disease is 1 in every 6000 births. In order for children to be born with this recessive combination (aa) a certain percentage of the population must be heterozygous for the trait. What percentage of this population is heterozygous for this trait?

q2 = 1 / 6000

21

q2 = .000167q = .0129p = .9871

Heterozygous individuals (2pq) = (2) (.9871) (.0129) = .0254 or 2.54% of the population

41. If a population has the following genotype frequencies:

AA = .32 Aa = .46 aa = .22

What is the frequency of the dominant allele (A)? What is the frequency of the recessive allele (a)?(Note: Do not use the Hardy-Weinberg equation to solve this. Solve this problem using the method learned in this lab.)

Since each portion of the population is known, this problem should be solved in this manner:

How many ”A” alleles are present in the population?

Number with “AA” genotype: 32 x 2 = 64 “A” allelesNumber with “Aa” genotype: 46 x 1 = 46 “A” alleles



p = 110 / 200 = .55


Number with “aa” genotype: 22 x 2 = 44 “a” allelesNumber with “Aa” genotype 46 x 1 = 46 “a” alleles



q = 90 / 200 = .45

42. A small group of 30 people decide to isolate themselves from the world and move to a small and remote deserted island. 10 of the individuals have albino skin, which is due to the homozygous recessive condition. The remaining 20 individuals have normal skin but are heterozygous for this condition. What is the frequency of the dominant allele? What is the frequency of the recessive allele?

How many ”A” alleles are present in the population?

22

Number with “AA” genotype: 0 x 2 = 0 “A” allelesNumber with “Aa” genotype: 20 x 1 = 20 “A” alleles



p = 20 / 60 = .333


Number with “aa” genotype: 10 x 2 = 20 “a” allelesNumber with “Aa” genotype 20 x 1 = 20 “a” alleles



q = 40 / 60 = .667

43. When the island population reaches 750 individuals, how many of the inhabitants are predicted to have albino skin?

Assuming that the population is in Hardy-Weinberg equilibrium:

q = .667q2 = (.667) 2 = .4449 or 44.49% of the population will be expected to have albino skin.

44.49% of 750 individuals = 334 individuals with albino skin

Created by Amy Brown – Science StuffCopyright © April 2013 Amy Brown (aka Science Stuff)

All rights reserved by author.This document is for your classroom use only.

This document may not be electronically distributed or posted to a web site.http://www.teacherspayteachers.com/Store/Science-Stuff

23

Documents

mslinscience.weebly.commslinscience.weebly.com/.../2/3/2/...and_the_hardy_weinb… · Web viewEvolution is any change in the frequency of alleles in a population. Evolution is the