Population Genetics and Evolution Within a Gene Pool

INTRODUCTION: The Hardy-Weinberg scheme is a way of viewing evolution as changes in the frequency of alleles in a population of organisms. If A and a are alleles for a particular gene and each individual has two alleles then p is the frequency of the A allele and q is the frequency of a alleles. The frequency of the possible diploid combinations is expressed in the equation p2+2pq+q2=1. In order for the Hardy-Weinberg equation to work five conditions must be met:
  1. The breeding population must be large.
  2. Mating must be random.

  3. There must not be mutations of the alleles.
  4. No differential migration may occur.
  5. All allele combinations must survive equally.
If the conditions are met then the allele genotype frequencies in a population remain the same. HYPOTHESIS:
  • A: If we use PTC paper to determine the allele frequencies for the dominant and recessive alleles in our classroom population, then we can use the Hardy-Weinberg equation to determine the number of homozygous dominant, heterozygous, and homozygous recessive individuals.
  • B: If the requirements for the Hardy-Weinberg equation are met, then the predictions of the allele frequencies by the Hardy-Weinberg equations will be similar to the actual allele frequencies.

MATERIALS:
  • 8A: ~PTC paper
  • 8B: ~Note cards
PROCEDURE: 8A:
  1. Using the PTC paper, rip off a piece and press it to the tip of your tongue. PTC tasters will sense a bitter taste. These individuals are considered tasters.
  2. A number representing the frequency of tasters (p2+2pq) should be calculated by dividing the number of tasters by the total number of students.

    A number representing the frequency of nontasters (q2) should be obtained by dividing the number of nontasters by the total number of students.

  3. Use the Hardy-Weinberg to determine the frequencies of the dominant (p) and recesive (q) alleles. The frequency of q can be calculated by taking the square root of q2. once q is determined, p can be found by the equation 1-q=p.
8B:

Case I

  1. Turn the cards over so that the letters do not show, shuffle them, and take the card on top. The partner should do the same thing.

    Put the cards together, they represent the alleles of the first offspring. Record the genotype. Reshuffle until each partner has one offspring.

  2. Both partners should record their genotype.

    They are now the next generation.

  3. Each partner should obtain cards that correspond with the genotype. Randomly seek other people and mate with them to produce offspring of the next generation. Record genotype after each generation.

  4. Calculate the allele frequencies of the population after five generations of random mating.

Case II

1. Start again with the same initial genotype and produce offspring the same way as case
  • I. Everytime the offspring is aa it will not survive. Retry mating until AA or Aa phenotype occurs.
  • II.

    Proceed through five generations selecting against homozygous recessive offspring. Calculate the new allele frequencies.

Case III

  1. Keep everything the same as Case II. If the offspring is AA flip a coin, if it is heads the individual does not survive.
  2. Simulate five generations, keeping the same initial genotype.

    The genotype aa does not survive. The same parents must retry until an offspring that survives is produced. Calculate the new frequencies.

  3. Starting at the F5 generation go through five more generations and calculate the new frequencies.

Case IV

  1. Divide the lab into small populations.
  2. Go through five generations inside the new populations. Record the new frequencies.
RAW DATA: See lab manual DATA:8A: | Phenotypes| Allele Frequency Based on the H-W Equation| | Tasters (p2 + 2pg)| Nontasters (p2)| p| q| Class Population| #| %| #| %| .

46| . 54| | 10| 71| 4| 29| | | North American Population| . 55| . 33| .

33| . 67|8B: Case 1 Hardy-Weinberg Equilibrium Initial Class Frequencies AA: . 25 Aa: . 5 aa: .

25 My initial Genotype: Aa F1 Genotype: AA F2 Genotype: aa F3 Genotype: Aa F4 Genotype: aa F5 Genotype: Aa Final Class Frequencies: AA: . 43 Aa: . 36 aa: . 21 p: . 54 q: .

46 Case 2 SelectionInitial Class Frequencies: AA: . 25 Aa: . 5 aa: . 25 My Initial Genotype: Aa F1 Genotype: Aa F2 Genotype: AA F3 Genotype: Aa F4 Genotype: Aa F5 Genotype: AA Final Class Frequencies: AA: . 29 Aa: .

71 aa:0 p: . 65 q:. 35 Case 3 Heterozygote Advantage Initial Class Frequencies: AA: . 25 Aa: . 5 aa: .

25 My Initial Genotype: Aa F1 Genotype: Aa F2 Genotype: aa F3 Genotype: Aa F4 Genotype: AA F5 Genotype: aa F6 Genotype: Aa F7 Genotype: Aa F8 Genotype: AA F9 Genotype: aa F10 Genotype: Aa Final Class Frequencies: (after five generations) AA: . 29 Aa: . 36 aa: . 36 p: . 4 q: . 6 Final Class Frequencies: (after ten generations)AA: .

21 Aa: . 71 aa: . 07 p: . 74 q: . 26 Case 4 Genetic Drift Initial class Frequencies: AA: .

25 Aa: . 5 aa: . 25 p: . 5 q: . 5 My Initial Genotype: Aa F1 Genotype: Aa F2 Genotype: AA F3 Genotype: AA F4 Genotype: Aa F5 Genotype: Aa Final class Frequencies AA: .

21 Aa: . 71 aa: . 07 p: . 74 q: . 26 8B: Class Results Case 1| | AA| Aa| aa| F1| 5| 4| 5| F2| 3| 5| 6| F3| 7| 4| 3| F4| 4| 6| 4| F5| 6| 5| 3| Case 2| | AA| Aa| aa| F1| 9| 5| 0| F2| 9| 5| 0| F3| 6| 8| 0| F4| 6| 8| 0| F5| 4| 10| 0| Case 3| | AA| Aa| aa| F1| 3| 9| 2| F2| 4| 7| 3| F3| 1| 9| 4| F4| 2| 9| 3| F5| 4| 5| 5| F6| 4| 9| 1| F7| 4| 7| 3| F8| 7| 5| 2| F9| 2| 11| 1| F10| 3| 10| 1| Case 4| | AA| Aa| aa| F1| 4| 7| 3| F2| 4| 8| 2| F3| 3| 9| 2| F4| 3| 9| 2| F5| 3| 10| 1|RESULTS: 8A:

  1. The percentage of heterozygous tasters in our class is 49. 7%.

  2. The percentage of the North American population that is heterozygous for the taster trait is 44. 2 %
8B:

Case I

  1. The Hardy-Weinberg equation predicts that the new p and q is 54% for the dominant allele and 46% for the recessive allele.
  2. The results of our simulation do not agree with the predicted outcome because the Hardy-Weinberg is just a prediction and mating is random so outcomes may vary.
  3. The major assumption not strictly followed was a large breeding population.

Case II

  1. In the new allele frequencies the recessive alleles frequency dropped.

  2. The major assumption not strictly followed was the fact that all genotypes must survive.
  3. I predict that if five more generations were simulated the recessive allele would drop futher in frequency.
  4. In a large population, it would most likely not be possible to completely eliminate a deleterious recessive allele because with such a large population the chances of the allele completely disappearing are very slim.

Case III

  1. The changes of p and q frequencies in case II compared with I and III is severely different because the Hardy-Weinberg conditions are not met because homozygous recessive do not survive.

  2. I do not think that the recessive allele will be completely eliminated because the population is too large to eliminate all the recessive alleles.
  3. The importance of heterozygotes in maintaining variation in populations is that they ensure the survival of recessive alleles.

Case IV

  1. I would expect 160 individuals to be homozygous dominant and 480 individuals to be heterozygous. 60/1000=. 36 sqr(.

    36= . 6 1-. 6=. 4 .

    4sqrd= . 16 . 16*1000=160 . 36*1000=360 360+160=520 1000-520=480

  2. I would expect 125 individuals to be homozygous dominant and 250 individuals to be heterozygous.
  3. I would expect 90 individuals to be AA, 420 individuals to be Aa, 490 individuals to be aa.
  4. I would expect 720 individuals to be AA, 960 to be Aa, and 320 aa.
  5. I would expect 640 individuals to be AA, 320 individuals to be Aa, and 40 individuals to be aa.
  6. The frequency of the dominant allele is.

Conclusion:

  • 8A: By the results of the PTC paper testing, we were successfully able to determine the allele frequencies for the dominant and recessive tasting alleles in our classroom population and use the Hardy-Weinberg equation to determine the number of homozygous dominant, heterozygous, and homozygous recessive individuals.
  • 8B: The results of the experiment confirm the observer's hypothesis that if the Hardy-Weinberg equation are met, then the Hardy-Weinberg equation's prediction will be similar to the actual allele frequencies within the population's gene pool.