1. Wheeler High School The Center for Advanced Studies in Science, Math & Technology Post-AP DNA/Genetics – Ms. Kelavkar Course Introduction Genetics Lecture 3: Mendelian Genetics

2. A Little Background on our Homeboy Gregor Mendel? 1822-1884 Studied Physics & Botany at the U of Vienna Influential publication was Experiments with Plant Hybrids –This publication went unnoticed until well after his death! Was the first to discover the basis for the transmission of heredity traits! Post-AP DNA/Genetics – Ms. Kelavkar

3. Result of Mendel’s monohybrid crosses of Pisum sativum. Post-AP DNA/Genetics – Ms. Kelavkar

4. Mendel’s Suggestions Mendel suggested that heredity resulted in discontinuous variation, as opposed to the existing continuous variation hypothesis of his time—in which offspring were thought to be a blend of the parental phenotypes. Discontinuous variation is controlled by alleles of a single gene or a small number of genes. The environment has little effect on this type of variation. –Basic Mendelian Genetics (one trait, one gene) Continuous variation is a complete range of measurements from one extreme to the other. –Polygenic Inheritance Post-AP DNA/Genetics – Ms. Kelavkar

5. Mendel’s First Three Postulates (collectively considered his 1 st law) 1.Unit Factors In Pairs 1.(Ex: AA or Aa or aa) 2.Dominance/Recessiveness 1.TT = tall, Tt = tall, tt = short 3.Segregation Post-AP DNA/Genetics – Ms. Kelavkar

6. The Monohybrid Cross Reveals How One Trait Is Transmitted from Generation to Generation… Thus confirming Mendel’s first 3 postulates! Post-AP DNA/Genetics – Ms. Kelavkar

8. An Analytical Approach Aaahhhh….the Punnett Square! Reginald D. Punnett was a biologist who came up with this rather simple approach of determining the probability of a cross. Post-AP DNA/Genetics – Ms. Kelavkar

10. The Testcross This is a one character test cross Simple method used today to determine the genotype of plants & animals. You can cross the offspring with a homozygous recessive dwarf plant to determine the genotypes of the parents. Testcross = organism in question x homozygous recessive

11. Independent Assortment –States that allele pairs separate independently during the formation of gametes. Therefore, traits are transmitted to offspring independently of one another.gametes This can be shown using a dihybrid cross. The Testcross: Two Characters Post-AP DNA/Genetics – Ms. Kelavkar

14. Some Key Terms Genotype – genetic makeup of organism Phenotype – physical result of genetic makeup (what you see) Testcross - used to determine if an individual exhibiting a dominant trait is heterozygous or homozygous for –Heterozygous = Hh –Homozygous = HH or hh

15. Any Questions?

16. Wheeler High School The Center for Advanced Studies in Science, Math & Technology Post-AP DNA/Genetics – Ms. Kelavkar Course Introduction Genetics Lecture 4: Probability and Genetics

17. Laws of Probability Help to Explain Genetic Events The probability of two independent events occurring at the same time can be calculated using the product law: –the probability of both events occurring is the product of the probability of each individual event –Look for the word ‘ and ’ in the problem! Post-AP DNA/Genetics – Ms. Kelavkar

18. Laws of Probability Help to Explain Genetic Events The sum law is used to calculate the probability of a generalized outcome that can be accomplished in more than one way. The sum law states that the probability of obtaining any single outcome, where that outcome can be achieved in two or more events, is equal to the sum of the individual probabilities of all such events. –Look for the word ‘ or ’ in the problem! Post-AP DNA/Genetics – Ms. Kelavkar

19. Otherwise Known as the ‘and’ ‘or’ rules! 1)Product rule = ‘and’ =multiply the events 1)Remember, the product rule describes the probability that event A and event B will occur 1)Sum rule = ‘or’ = add the events 1)The sum rule describes the probability that event A or event B will occur Post-AP DNA/Genetics – Ms. Kelavkar

20. The Product Rule (multiply) Suppose we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die. P(4) = 1/6 P(even) = 3/6 P 4 and even = (1/6)(3/6) = 3/36 or 1/12

22. Chi Square & The Null Hypothesis Data from genetic crosses are quantitative. –Geneticists use statistics to understand the significance of their results. Hypothesis developed is called the null hypothesis Null Hypothesis states there is no real difference between the observed data and the predicted data. –Is the genetic event due to chance? If not, then the null hypothesis is rejected and a new hypothesis must be developed to explain the data. Post-AP DNA/Genetics – Ms. Kelavkar

23. Null Hypothesis Example (H 0 ) In a clinical trial of a new drug, the null hypothesis might be that the new drug is no better, on average, than the current drug. We would write: H 0 : There is no difference between the two drugs on average.

24. Chi-Square Analysis Evaluates the Influence of Chance on Genetic Data A simple statistical tool used to test the null hypothesis is called the chi-square (  2 ) test. –“goodness of fit” test Post-AP DNA/Genetics – Ms. Kelavkar

25. Table 3 ‑ 3 shows the steps in  2 calculations for the F 2 generation of a monohybrid cross. Post-AP DNA/Genetics – Ms. Kelavkar

26. Degree of Freedom Chi-square analysis requires that the degree of freedom (df) be taken into account, since more deviation is expected with a higher degree of freedom. –df = n-1 Post-AP DNA/Genetics – Ms. Kelavkar

27. Chi-Square Probabilities Once the number of degrees of freedom is determined, the  2 value can be interpreted in terms of a corresponding probability value (p). Post-AP DNA/Genetics – Ms. Kelavkar

28. Example Let’s analyze the theoretical progeny data from a testcross of smooth, yellow double heterozygote (SsYy) with a wrinkled, green homozygote (ssyy). The progeny data: 154 smooth yellow 124 smooth green 144 wrinkled yellow 146 wrinkled green Total 568 Post-AP DNA/Genetics – Ms. Kelavkar

29. We hypothesize that a testcross should yield a 1:1:1:1 ratio of the 4 phenotypic classes (that is if independent assortment is taking place). (1)(2)(3)(4)(5)(6) Phenotype s Observed # (o) Expected # (e) d (= o-e) d2d2 d 2 /e Smooth, yellow 154142+121441.01 Smooth, green 124142-183242.28 Wrinkled, yellow 144142+240.03 Wrinkled, green 146142+4160.11 Total568 0n/a3.43 (7)  2 = 3.43 (8) df = 3 Post-AP DNA/Genetics – Ms. Kelavkar

30. Chi-Square Probability Chart  2 = 3.43 This means that with the hypothesis being tested, 30 to 50 out of 100 trials (so…30-50% of the time) we could expect chi- square values due to chance. Fail to reject Null Reject Null

31. Any Questions? Post-AP DNA/Genetics – Ms. Kelavkar

32. Figure 3-4 Copyright © 2006 Pearson Prentice Hall, Inc. 3.2The Monohybrid Cross Reveals How One Trait Is Transmitted from Generation to Generation

33. Figure 3-6 Copyright © 2006 Pearson Prentice Hall, Inc. 3.3Mendel’s Dihybrid Cross Revealed His Fourth Postulate: Independent Assortment

34. Figure 3-7 Copyright © 2006 Pearson Prentice Hall, Inc.

35. Figure 3-9 Copyright © 2006 Pearson Prentice Hall, Inc. 3.3Mendel’s Dihybrid Cross Revealed His Fourth Postulate: Independent Assortment- Two Characters

36. 3.4The Trihybrid Cross Demonstrates That Mendel’s Principles Apply to Inheritance of Multiple Traits 3.4.1 The Forked-Line Method, or Branch Diagram

37. Figure 3-11 Copyright © 2006 Pearson Prentice Hall, Inc.

38. Figure 3-11a Copyright © 2006 Pearson Prentice Hall, Inc.

39. Figure 3-11b Copyright © 2006 Pearson Prentice Hall, Inc.

40. Figure 3-11c Copyright © 2006 Pearson Prentice Hall, Inc.

41. Table 3-1 Copyright © 2006 Pearson Prentice Hall, Inc.

42. Table 3-2 Copyright © 2006 Pearson Prentice Hall, Inc.