Natural Selection in:

1. Natural Selection in: Teddy Grahams

2. Darwin proposed a mechanism for evolutionary change: natural selection is the differential survival and reproduction of individuals in a population.

3. Allele frequency is a measure of the relative frequency of an allele on a genetic locus in a population. it is expressed as a proportion or a percentage.locus(plural loci) is the specific location of a gene or DNA sequence or position on a chromosome.

4. In population genetics, allele frequencies show the genetic diversity of a species population or the richness of its gene pool.Population genetics studies the different "forces" that might lead to changes in the distribution and frequencies of alleles - or evolution.

5. A gene pool is the total number of genes of every individual in an interbreeding population.

6. A large gene pool indicates high genetic diversity, increased chances of biological fitness, and survival in a changing environment.

7. A small gene pool indicates low genetic diversity, reduced chances of acquiring biological fitness, and increased possibility of extinction.

8. The population is the basic unit of evolution. Populations evolve, but individuals do not evolve.

9. The Hardy–Weinberg principlestates that both allele and genotype frequencies in a population remain constant—that is, they are in equilibrium—from generation to generation unless specific disturbing influences are introduced.

10. Those disturbing influences include: • non-random mating • Mutations • Selection • Limited population size • Random genetic drift • Gene flow

11. The Hardy-Weinberg theorem states that the frequency of alleles in a population will remain the same regardless of the starting frequencies.

12. This theorem is valid only if certain conditions are met:1) The population is very large.2) Matingsare random. 3) There are no net changes in the gene pool due to mutations in the DNA. 4) There is no migration of individuals into and out of the population. 5) All genotypes are equal in reproductive success (no selection).

13. The frequency of the possible combinations of alleles (AA, Aa, and aa) in this population is expressed as: p2 + 2pq + q2 = 1

14. The Hardy Weinberg equilibrium is impossible in nature. Genetic equilibrium is an ideal state that provides a baseline to measure genetic change against.

15. Real Life Application: A patient's child is a carrier of a recessive mutation that causes cystic fibrosis in homozygous recessive children. The parent wants to know the probability of her grandchildren inheriting the disease. In order to answer this question, the genetic counselor must know the chance that the child will reproduce with a carrier of the recessive mutation. This fact may not be known, but disease frequency is known. We know that the disease is caused by the homozygous recessive genotype; we can use the Hardy–Weinberg principle to work backward from disease occurrence to the frequency of heterozygous recessive individuals.

16. Introduction: You are a bear-eating monster. There are two kinds of bears that you like to eat: happy bears and sad bears. You can tell the difference between them by the way they hold their hands. Happy bears hold their hands high in the air, and sad bears hold their hands down low.

17. Happy bears taste sweet and are easy to catch. Sad bears taste bitter, are devious and hard to catch. Because of this you only eat happy bears. The happy trait in bears is caused by the expression of a recessive allele. The homozygous recessive condition is being happy. The sad trait is caused by a dominant allele.

18. New bears are born every year (when they are hibernating in their den: the cardboard box).

19. the birth rate is one new bear for every old bear left from last year.

20. Form a hypothesis about what you expect to happen to the number of Happy and Sad bears over time.

21. Procedure: 1) Obtain a population of 10 bears. Record the number of happy and sad bears, and the total population number. Using the equation for Hardy-Weinberg equilibrium, calculate the frequencies of both the dominant and recessive alleles and the genotypes that are represented in the population.

22. Procedure: p2 + 2pq + q2 = 1 Example: If 5 of the 10 bears are happy. Each bear has 2 alleles, then… * 10 out of 20 alleles would be happy alleles, * or q2(10/20) is 0.5. * To determine the q number, find the square of 0.5.

23. Procedure: 2) Now, go hunting! Eat 3 happy bears. (If you do not have 3 happy bears then eat the difference in sad bears.)

24. Procedure: 3) Once you have consumed the bears, obtain a new generation by removing seven additional bears from the den . Add these new bears to your old ones and chart happy, sad and total populations in your data table. (This is generation 2). You should only have a total of 14 bears – remember birth rate.

25. Procedure: 4) Repeat steps 2and 3 until you have four generations recorded in your data table. Be sure to record the number of each type of bear and the total population.

26. the percentage of Happy and Sad bears Determine the percentage of Happy and Sad bears for each generation and record the percentages in Table 2. To determine the percentage take the number of happy or Sad bears and divide by the total number of bears for that generation and multiply the answer by 100. For example, if there were 12 Sad bears and 4 Happy bears in a generation, then there were 16 bears total. To obtain the percentage of Sad bears you would divide 12 by 16 and then multiply the answer by 100. To obtain the percentage of Happy bears you would divide 4 by 16 and then multiply the answer by 100.

27. Table 1. The number of bears for each generationTable 2. The percentage of bears for each generation

28. the percentage of Happy and Sad bears Graph what happens to the bear population over time. Graph the percentage data for both the Happy and Sad bears on the same graph.

29. Using your actual population data from the above data table and the Hardy Weinberg formulas, complete the following data table to determine the percentage for each genotype in the population:

30. Allele frequencies: Make a second line graph with two lines. One line should compare the “p” vs. generations and the second line should graph the “q” vs. generations.

31. Questions: 1. Describe what is happening to the genotype and allele frequencies in the population of Teddy Grahams? 2. What would you expect to happen if you continued the selection process for additional generations? 3. How would the frequencies change if you were to now select for the sad bears? 4. Why doesn’t the recessive allele disappear from the population? How is it protected?

32. Questions: 5) Is the Hardy-Weinberg theorem valid for the population in this experiment? Which, if any, of the conditions of the theorem did this population violate? 6) Do you think this population was undergoing evolution? Why or why not? 7) In what way(s) is the Teddy Graham ecosystem not like a real world ecosystem? (Hint: read the 5 conditions that must be met by a population)

33. Sample calculations using the HARDY-WEINBERG equation Albinism is a rare genetically inherited trait that is only expressed in the phenotype of homozygous recessive individuals (aa).  The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin.  This condition can occur among any human group as well as among other animal species.  The average human frequencyof albinism in North America is only about 1 in 20,000.

34. The frequency of homozygous recessive individuals (aa) in a population is q².  Therefore, in North America the following must be true for albinism: q² = 1/20,000 = .00005 By taking the square root of both sides of this equation, we get:    (Note: the numbers are rounded off for simplification.) q = .007      

35. The frequency of the recessive albinism allele (a) is .00707 or about 1 in 140 (or 20000 x 0.007 = 140)  Knowing one of the two variables (q) in the Hardy-Weinberg equation, it is easy to solve for the other (p): p = 1 – q p = 1 - 0.007 p = 0.993

36. The frequency of the dominant, normal allele (A) is, therefore, .993 or about 99 in 100. The next step is to plug the frequencies of p and q into the Hardy-Weinberg equation: p² + 2pq + q² = 1 (.993)² + 2 (.993)(.007) + (.007)² = 1 .986 + .014 + .00005 = 1

37. This gives us the frequencies for each of the three genotypes for this trait in the population: p² = predicted frequency of homozygousdominant individuals = .986 = 98.6% 2pq = predicted frequency of heterozygous individuals = .014 = 1.4% q² = predicted frequency of homozygousrecessive individuals (the albinos) = .00005 = .005% Interpret Results: With a frequency of .005% (about 1 in 20,000), albinos are extremely rare.  However, heterozygous carriers for this trait, with a predicted frequency of 1.4% (about 1 in 72), are far more common than most people imagine.  There are roughly 280 times more carriers than albinos (1.4/.005).  Clearly, though, the vast majority of humans (98.6%) probably are homozygous dominant and do not have the albinism allele.

38. EXAMPLE 2: Consider a population of 1000 individuals with a locus and alleles described below. Assume that you have no information on the presence or absence of evolutionary mechanisms in this population. You find that the population consists of: 90 individuals homozygous for the A allele(AA genotype) 490 individuals homozygous for the a allele (aa genotype) 420 heterozygotes (Aa genotype)

39. 90 individuals homozygous for the A allele(AA genotype) 490 individuals homozygous for the a allele (aa genotype) 420heterozygotes(Aa genotype) 1000 Allele Frequencies: The frequency of the A allele will equal: Total number of A alleles in the population / (Total number of alleles in population) (90 x 2) + 420 / (1000 x 2 ) = 0.30 The frequency of the a allele will equal : Total number of a alleles in the population/ (Total number of a alleles in the population) (490 x 2 ) + 420 / (1000 x 2) = 0.7 Or 1 – 0.30 = 0.07

40. Genotype Frequencies: 90 individuals homozygous for the A allele(AA genotype) 490 individuals homozygous for the a allele (aa genotype) 420heterozygotes(Aa genotype) 1000

41. Hardy – Weinberg Prediction: 90 individuals homozygous for the A allele(AA genotype) 490 individuals homozygous for the a allele (aa genotype) 420heterozygotes(Aa genotype) 1000 q² = 490/1000 = 0.49 By taking the square root of both sides of this equation, we get q = .7 p = 1 – q p = 1 - 0.7 p = 0.3 p² + 2pq + q² = 1 (0.3)2 + 2(0.3)(0.7) + (0.7)2 = 1 0.09 + 0.42 + 0.49 = 1

42. Since the observed genotype frequencies equal those predicted by the Hardy – Weinberg Equilibrium Theory, we may (tentatively) conclude that no evolutionary mechanisms operate on this locus in this population. The population meets the assumptions of the Hardy-Weinberg theory: The population is very large. Matingsare random. There are no net changes in the gene pool due to mutations in the DNA. There is no migration of individuals into and out of the population. All genotypes are equal in reproductive success (no selection).