1. AP Biology Heredity PowerPoint presentation text copied directly from NJCTL with corrections made as needed. Graphics may have been substituted with a similar picture. 1 Content from NJCTL

2. Heredity Unit Topics Heredity and Meiosis Law of Independent Assortment What Mendel Didn’t know Probability, Statistics, and Genetics 2 Content from NJCTL

3. Probability, Statistics, and Genetics 3 Content from NJCTL

4. Genetics is Based on Prediction The study of genetics relies on the comparison of expected values and observed values based on actual experimental observation. For example, if we cross 2 individuals with different phenotypes for a trait we expect to get a 3:1 phenotype ratio. This is because we have established a rule based on the observations of Mendel. 4 Content from NJCTL

5. Genetics is Based on Prediction But what if we don’t get exactly 3:1? Did we prove Mendel wrong? Look at the numbers below that were generated by an actual experiment. What are the expected values? Do the observed results adhere to Mendel’s rules? 10,000 offspring produced 6860 purple flowers 3140 white flowers 5 Content from NJCTL

6. Chi Squared Test The Chi-Squared test is designed to help us decide if the difference between the observed results and expected results of an experiment are statistically significant. In other words, it tells us if the difference is due to random chance or it is due to some other factor that effected the results of our experiment. 6 Content from NJCTL

7. Chi Squared Test This is the formula: 2 (O-E) E 2 2 Chi Squared The sum of O = observed frequencies E – Expected frequencies 7 Content from NJCTL

8. Chi Squared Test A simple example: The laws of probability state that if we flip a coin 100 times we should get 50 heads and 50 tails. These are our expected results (e). However, when we actually flip a coin 100 times we get (o) = 56 44 8 Content from NJCTL

9. Chi Squared Test Is this difference because of chance? Or is it because something is effecting the outcome? For example, the way it is being flipped or weight imbalance causing more heads to be flipped. 56 44 9 Content from NJCTL

10. Chi Squared Test This is where Chi Squared is used. The test is assessing the null hypothesis which states: There is no significant difference between the observed and expected frequencies. 56 44 10 Content from NJCTL

11. Chi Squared Test The first step is to get a value for X 2 56 44 X 2 = = 1.44 (56-50) 2 50 + (56-50) 2 50 72 50 = 11 Content from NJCTL

12. Chi Squared Test Now that we have our value for Chi Squared, We can compare it to a statistical analysis chart known as critical values 1.44 Vassar Stats: Critical Values of Chi-Square Level of Significance df0.050.0250.010.005 13.845.026.637.88 25.997.389.2110.6 37.819.3511.3412.84 49.4911.1413.2814.86 511.0712.8315.0916.75 612.5914.4516.8118.55 714.0716.0118.4820.28 815.5117.5320.0921.95 916.9219.0221.6723.59 1018.3120.4823.2125.19 12 Content from NJCTL

13. Chi Squared Test First we must figure out our degrees of freedom (df). This is simply the amount of possible outcomes minus 1. For this example: 2-1 = 1df 1.44 Vassar Stats: Critical Values of Chi-Square Level of Significance df0.050.0250.010.005 13.845.026.637.88 25.997.389.2110.6 37.819.3511.3412.84 49.4911.1413.2814.86 511.0712.8315.0916.75 612.5914.4516.8118.55 714.0716.0118.4820.28 815.5117.5320.0921.95 916.9219.0221.6723.59 1018.3120.4823.2125.19 13 Content from NJCTL

14. Chi Squared Test Next we decide our level of significance. Usually it is 0.05 which means that we are 95% certain of our outcome. If a higher standard is needed another level could be used. 1.44 Vassar Stats: Critical Values of Chi-Square Level of Significance df0.050.0250.010.005 13.845.026.637.88 25.997.389.2110.6 37.819.3511.3412.84 49.4911.1413.2814.86 511.0712.8315.0916.75 612.5914.4516.8118.55 714.0716.0118.4820.28 815.5117.5320.0921.95 916.9219.0221.6723.59 1018.3120.4823.2125.19 14 Content from NJCTL

15. Chi Squared Test 3.84 is our critical value. If X 2 is less than the critical value we accept the null hypothesis. If it is more, than we reject the null hypothesis. 1.44 < 3.84 Vassar Stats: Critical Values of Chi-Square Level of Significance df0.050.0250.010.005 13.845.026.637.88 25.997.389.2110.6 37.819.3511.3412.84 49.4911.1413.2814.86 511.0712.8315.0916.75 612.5914.4516.8118.55 714.0716.0118.4820.28 815.5117.5320.0921.95 916.9219.0221.6723.59 1018.3120.4823.2125.19 Null hypothesis – there is no significant difference between the observed and expected frequencies 15 Content from NJCTL

16. Genetics is Based on Prediction Now that you know Chi Squared, does this observed data meet with the expectations set by Mendel? 10,000 offspring produced 6860 purple flowers 3140 white flowers AP BIO 2015-2016 class LETS FILL THIS OUT! 16 Content from NJCTL

17. Genetics is Based on Prediction What if we look at genotype? Does this observed data coincide with Mendel’s postulates? 10,000 offspring produced 6860 purple flowers 3140 white flowers AP BIO 2015-2016 class LETS FILL THIS OUT! 17 Content from NJCTL

18. Using Probability Rules When Solving Genetics Problems There are two rules of probability which are helpful when solving problems in genetics: The Multiplication Rule The Addition Rule 18 Content from NJCTL

19. The Multiplication Rule The multiplication rule states that the probability that two or more independent events will occur together is the product of their individual probabilities. For example, you would use the multiplication rule if you want to know the probability that a plant will have both yellow seeds and yellow pods. Pea color Yy x Yy = 75% Pod color Gg x Gg=25% 0.75 x 0.25 = 0.1875 = 3/16 19 Content from NJCTL

20. The Rule of Addition The rule of addition states the probability that an event can occur in two or more alternative ways is the sum of the separate probabilities on the different ways. For example, you would use the addition rule if you want to know the probability of a plant (produced from a monohybrid cross) being homozygous dominant for flower color or homozygous recessive for flower color. 25% PP + 25% pp = 50% homozygous 20 Content from NJCTL

21. 30. If we toss two coins at the same time, what is the chance that both coins will land heads up? 21 Content from NJCTL

22. 30. If we toss two coins at the same time, what is the chance that both coins will land heads up? CLASS? 22 Content from NJCTL

23. 31. If an F1 cross between two pea plants that are heterozygous (Pp) for purple flowers and heterozygous for pea color (Gg), what is the probability that a particular offspring of this cross will have the recessive genotype for both traits (white flowers, yellow seeds)? 23 Content from NJCTL

24. 31. If an F1 cross between two pea plants that are heterozygous (Pp) for purple flowers and heterozygous for pea color (Gg), what is the probability that a particular offspring of this cross will have the recessive genotype for both traits (white flowers, yellow seeds)? CLASS? 24 Content from NJCTL

25. 32. Mendel stated that each pair of alleles segregates independently of other pairs of alleles during gamete formation. What is this law known as? A. Law of Segregation B. Law of Independent Assortment C. Law of Probability D. Law of Pea Plant Genetics 25 Content from NJCTL

26. 32. Mendel stated that each pair of alleles segregates independently of other pairs of alleles during gamete formation. What is this law known as? A. Law of Segregation B. Law of Independent Assortment C. Law of Probability D. Law of Pea Plant Genetics 26 Content from NJCTL

27. 33. An organism heterozygous for two characteristics is a _____________. A. dihybrid B. monohybrid C. homozygote D. double dominant 27 Content from NJCTL

28. 33. An organism heterozygous for two characteristics is a _____________. A. dihybrid B. monohybrid C. homozygote D. double dominant 28 Content from NJCTL

29. 33. An organism heterozygous for two characteristics is a _____________. A. dihybrid B. monohybrid C. homozygote D. double dominant 29

30. Trihybrid Crosses We can use the rules of probability to solve complex genetics problems. If we crossed two organisms both having the genotype AaBbCc, what is the probability that an offspring of this cross will have the genotype aabbcc? Since each allele pair assorts independently, we can treat this trihybrid cross as three separate monohybrid crosses: Aa x Aa: Probability of aa offspring: _____ Bb x Bb: Probability of bb offspring: _____ Cc x Cc: Probability of cc offspring: _____ 30 Content from NJCTL

31. Trihybrid Crosses We can use the rules of probability to solve complex genetics problems. If we crossed two organisms both having the genotype AaBbCc, what is the probability that an offspring of this cross will have the genotype aabbcc? Since each allele pair assorts independently, we can treat this trihybrid cross as three separate monohybrid crosses: Aa x Aa: Probability of aa offspring: 25% Bb x Bb: Probability of bb offspring: 25% Cc x Cc: Probability of cc offspring: 25% 31 Content from NJCTL

32. aabbcc Because the segregation of each allele pair is an independent event, we can use the multiplication rule to calculate the probability of an offspring being aabbcc. What is the probability? 32 Content from NJCTL

33. aabbcc Because the segregation of each allele pair is an independent event, we can use the multiplication rule to calculate the probability of an offspring being aabbcc. What is the probability? CLASS? 33 Content from NJCTL

34. 34. If you cross two F 1 individuals both having the Genotype AaBbCc, what is the probability that an F 2 offspring will have the genotype AabbCc? 34 Content from NJCTL

35. 34. If you cross two F 1 individuals both having the Genotype AaBbCc, what is the probability that an F 2 offspring will have the genotype AabbCc? 50% x 25% x 50% = 6.25% 35 Content from NJCTL

36. 35. A plant with genotype AABbCC is crossed with a plant having genotype AaBbCc. What is the probability of an offspring having the genotype AABBCC? 36 Content from NJCTL

37. 35. A plant with genotype AABbCC is crossed with a plant having genotype AaBbCc. What is the probability of an offspring having the genotype AABBCC? You guys…yes … again! 37 Content from NJCTL

38. Why Have Sex? Now that you have learned about Darwin’s descent with modification and Mendel’s patterns of inheritance, we can now combine the two and explore why sex, from a survival perspective, is worth the trouble. 38 Content from NJCTL

39. What We Know Now… Because of Darwin and Mendel we know that individuals are a conduit for population evolution. Individuals get their genes for ancestry, and pass their genes to the next generation in varying frequency based on their ability to survive. This makes the population stronger as a whole. This is known as population genetics. 39 Content from NJCTL

40. What We Know Now… Because of Darwin and Mendel we know that individuals are a conduit for population evolution. Individuals get their genes for ancestry, and pass their genes to the next generation in varying frequency based on their ability to survive. This makes the population stronger as a whole. This is known as population genetics. 40 Content from NJCTL Sexually reproducing populations are able to increase their variation and enable the population to survive more environmental possibilities.