Heredity and Meiosis AP BIOLOGY. Heredity. Slide 1 / 142 Slide 2 / 142. Slide 4 / 142. Slide 3 / 142. Slide 6 / 142. Slide 5 / 142.

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1 Slide 1 / 142 Slide 2 / 142 AP BIOLOGY Heredity Slide 3 / 142 Slide 4 / 142 Heredity Unit Topics Click on the topic to go to that section Heredity & Meiosis Law of Independent Assortment What Mendel Didn't Know Heredity and Meiosis Probability, Statistics, and Genetics Return to Table of Contents Slide 5 / 142 Heredity Heredity is the passing of traits to offspring from parents. This is the process by which an offspring organism acquires, or becomes likely to have, the characteristics of its parent organisms. Through heredity, variations exhibited by individuals can accumulate and cause some species to evolve. The study of heredity in biology is called genetics. Two individuals have a genetic relationship if one is the ancestor of the other, or if they share a common ancestor. In evolutionary theory, species which share an evolutionary ancestor are said to be of common descent. Slide 6 / 142 Common Descent Chimpanzee Bonobo Human The gene pool of the chimpanzee and bonobo share 99.6% of the same genes. 98.7% of human genes are common with the bonobo. Common Ancestor

2 Slide 7 / 142 Heredity Review Fill in the blanks below as a reminder of what you know... DNA is the molecule that is used to store information. Genes are the basic unit of heredity. A chromosome is defined as a discrete package of genes that is used to transfer information. The nucleus is where genes are kept in eukaryotic organisms. Slide 8 / 142 Heredity and Reproduction Reproduction is a requirement for heredity. The 2 methods of reproduction that are most frequently used: 1) Asexual binary fission in prokaryotes. This method is simple but produces minimal variation. 2) Sexual reproduction in multicellular eukaryotes This method is highly complex requiring massive amounts of energy. The trade off is more variation. Single celled eukaryotes can take advantage of asexual mitosis to produce offspring daughter cells. Slide 9 / 142 Sexually Reproducing Beetles Let's look at an example to explore the laws of heredity: A researcher was curious about a new species of beetle. These beetles have two versions of the trait body color, black or red. In other words, their body color phenotype can be black or red. Slide 10 / 142 Sexually Reproducing Beetles The researcher did some breeding experiments to try to determine how the alleles were passed from generation to generation. He wanted to know the heredity pattern of the phenotypes. First he did crosses of red individuals and got 3 reoccurring outcomes (repeatable on numerous occasions). Record the observations you see on the following slides in your notes. Slide 11 / 142 Slide 12 / 142 Outcome 1 of Red Crossed with Red Outcome 2 of Red Crossed with Red a b c d

3 Slide 13 / 142 Slide 14 / 142 Outcome 3 of Red Crossed with Red There Was Only 1 Outcome When He Crossed Black with Black e f g h No matter which individuals he chose to breed or how many times this was the result Slide 15 / 142 Sexually Reproducing Beetles The question: How is it genetically determined whether a beetle will be black or red? Assuming that beetles always mix their genes the same way, describe a mechanism that can account for all the outcomes that the researcher observed. Use the numbers you recorded to hypothesize. Suggest the alleles for body color that each of the parent beetles possess using your hypothesis. This is known as their genotype. Slide 16 / 142 Sexually Reproducing Beetles Individual beetle's genotype from the sample crosses: a- b- c- d- e- f- g- h- Slide 17 / 142 Sexually Reproducing Beetles The researcher recognized that the allele for black body color must be "hiding" in the genes of some of the red individuals but not all. Slide 18 / 142 Sexually Reproducing Beetles The researcher was curious about the offspring of the red and black beetle cross. He took 2 of the offspring from that cross and then crossed them. To further understand the inheritance of this gene he crossed a beetle that he knew didn't have a hiding black body gene (e or f) with a black beetle. P generation try to predict the result F1 generation Predict the result Do you want to revise any of your previous hypotheses? Click to reveal result

4 Slide 19 / 142 Mechanism of Sexual Reproduction The preceeding experiment was similar to how Gregor Mendel figured out his postulates (hypotheses) of sexual genetic heredity. Compare your theories with his... Expanding on Mendel's Postulates... Hypothesis # 1: Alleles Slide 20 / 142 Meiosis and Mendel There are alternative forms of genes that account for variations in inherited characteristics. Slide 21 / 142 Meiosis and Mendel Hypothesis # 2: All Organisms have 2 alleles For each characteristic, an organism inherits two alleles (one from each parent). The two alleles may be the same or they may be different. Technically the offspring is inheriting a group of alleles for many different traits contained on a chromosome. Each gamete contributes a full set of haploid chromatids containing one allele for every possible trait. Slide 22 / 142 Meiosis and Mendel Hypothesis # 3: Dominant and Recessive If the two alleles of a pair are different (heterozygous), one determines the appearance and is called the dominant allele. We now understand that the "dominant" allele does not overcome the "recessive" allele. It simply masks the phenotype of the recessive. Both alleles are active in the diploid offspring. This often leads intermediate to traits that are a combination of the allele variants. Slide 23 / 142 Meiosis and Mendel Hypothesis # 4: Law of Segregation A sperm or egg carries only one allele for each trait because allele pairs separate from each other during sex cell formation. In meiosis, a complex shuffling of chromosomes and the production of 4 genetically unique sex cells from one diploid cell leads to massive amounts of possible variations. This leads to a practically infinite number of trait combination and a healthier population. Slide 24 / 142 Mendelian Problems When Mendel chose his pea plant traits he did so very carefully. In fact, sometimes he would not get the desired results and omit them from his studies. Explain in terms of modern understanding of genes why the following may happen: AaBb AaBb Mendel would expect this cross to yield a phenotype ratio of... 9 both dominant: 3 A dominant B recessive: 3 A recessive B dominant: 1 both recessive Actual Results... 3 both dominant :1 both recessive

5 Slide 25 / 142 Mendel's Experiments Click here for a review of Mendel's Experiments Slide 26 / In a classic Mendelian cross, what would be the ratio of the offspring in the F1 generation? A 1:1 B 2:1 C 3:1 D no ratio, only 1 kind If further review is needed, please see NJCTL's first year biology course. Genetics First Year Course Slide 27 / In a classic Mendelian cross, what would be the ratio of the offspring in the F2 generation? A 1:1 B 2:1 C 3:1 D no ratio, only 1 kind 3 Aa x AA Slide 28 / 142 If this cross produces 500 offspring what is the expected number of individuals who would have the dominant phenotype? 4 Aa x aa Slide 29 / 142 If this cross produces 500 offspring what is the expected number of individuals who would have the dominant phenotype? 5 Aa x Aa Slide 30 / 142 If this cross produces 500 offspring what is the expected number of individuals who would have the dominant phenotype?

6 Slide 31 / 142 Law of Independent Assortment Slide 32 / 142 Mendel's Laws As we have seen, the Law of Segregation is based on Mendel's monohybrid (single trait) crosses. To understand his second law, the Law of Independent Assortment he had to look at 2 traits simultaneously, a dihybrid cross. The following example will be used to demonstrate and practice the second law. Return to Table of Contents Slide 33 / 142 While observing samples of pond water from a remote area of New Jersey, a researcher discovers a microbe. She determines that they are capable of sexual reproduction and devises a way to selectively breed individuals of the species. The researcher concentrates her study of their heredity pattern on two easily observable traits. Their color and their shape. Slide 34 / 142 Fill in the data. Possible colors: Possible shapes: Possible combinations of color and shape: 40 Magnification Light Microscope First the researcher wants to determine which traits are dominant. Suggest some tests that will help determine dominance for both of these traits. Test 1: Test 2: Slide 35 / 142 The researcher does the following experiments: She takes a pool of only orange individuals and isolates them. After enough time for offspring to be produced she samples and observes the population. This is what she sees: Slide 36 / 142 Test 3; Test 4: 40 Magnification Light Microscope 40 Magnification Light Microscope Test 5:

7 Slide 37 / 142 Slide 38 / 142 She then takes a pool of only green individuals and isolates them. After enough time for offspring to be produced she samples and observes the population. This is what she sees: 40 Magnification Light Microscope Using what we learned in the beetle crosses, try to determine which color is dominant, green or orange. Click below to see the researcher's conclusion and explanation Since the first test cross produced orange and green, the orange organisms must have had "hidden" recessive green alleles that they passed to their offspring. She concludes that the orange color is dominant to green. The second test confirms this because none of the green individuals carry a hidden orange allele. In other words, to be orange your genotype can be heterozygous (Oo) or homozygous dominant (OO). But to be green there is only one possible genotype, homozygous recessive (oo). The researcher does a similar experiment to determine shape. Using the same logic she determines that rectangle is dominant to oval. Slide 39 / 142 Slide 40 / What is the genotype of the organism? A B oorr C oorr D OORR What was in the original pool that produced the results you see in this microscope field? 40 Magnification Light Microscope Slide 41 / What is the genotype of the organism? A B oorr C oorr D OORR Slide 42 / What is a possible genotype of the organism? A B oorr C oorr D oorr

8 Slide 43 / In order for the pond organism's phenotype to be orange their genotype must be: A heterozygous only B heterozygous or homozygous dominant C heterozygous or homozygous recessive D homozygous dominant only Slide 44 / An organism with the genotype OOrr would appear to be: A Orange and Oval B Orange and Rectangle C Green and Oval D Green and Rectangle Now that the researcher knows the dominance of these traits she can learn more by doing individual test crosses. She wants to know: Slide 45 / 142 She isolates 2 individuals of known phenotype and genotype and crosses them. These will be the parent (P) generation. Slide 46 / 142 Male Female P oorr OORR Does the inheritance of color effect the inheritance of shape? For example: If one is green, is it more likely to be oval? If one is orange is it more likely to be rectangular? 40 Magnification Light Microscope Slide 47 / 142 Slide 48 / 142 They produce thousands of offspring. All of them are the same: Orange and Rectangular. We will call all of their offspring the F 1 Generation Male oorr OORR Female P F 1 To determine why all the F 1 generation are the same we need to consider the Mendelian Law of Segregation. Alleles segregate from one another when sperm and egg are produced. So each sex cell only has one allele for each trait. Male oorr OORR Female P F 1

9 Slide 49 / 142 Slide 50 / 142 If we look at the male in this cross we will see that he can only give the recessive allele for color (o) and the recessive allele for shape (r) to the offspring. All of his sperm have the genotype: or Male oorr OORR Female P F 1 For the female she only has the dominant for both traits so she can only give the dominant alleles to her offspring. All of her eggs have the genotype: OR Male oorr OORR Female P F 1 Slide 51 / 142 Slide 52 / 142 Since the sperm and egg combine into one offspring there is only one possible result: Male oorr OORR Female P The next step for our researcher is to determine the expected result of a cross between 2 individuals from the hybrid (F 1) generation. Male Female F 1 Egg OR Sperm or F 1 She hypothesizes that there is no connection between the 2 traits. She figures out the ratio of possible offspring that she should observe if this is true. Fertilization produces the geneotype of offspring Slide 53 / 142 Slide 54 / 142 The next step for our researcher is to determine the expected result of a cross between 2 individuals from the hybrid (F 1) generation. Male Female F 1 Since these individuals are heterozygous (hybrids) they are able to produce many different combinations of alleles in their gametes. Male Female F 1 She assumes that there is no connection between the 2 traits. She figures out the ratio of possible offspring that she should observe if this is true. sex cells: OR Or or or

10 Slide 55 / 142 Slide 56 / 142 To determine the ratio of offspring we need to compare the possibilities for combinations of sperm and egg in fertilization. We do this just like in the P generation, but it is a more complex setup. Male Female F 1 First put in all the possibilities for sperm above the top row Sperm OR Or or or sex cells: OR Or or or Slide 57 / 142 Then put the possibilities for eggs along the side Sperm OR Or or or Slide 58 / 142 Now fill in the table by combining each egg and sperm (fertilization) in the appropriate square. Click each to reveal the result. Sperm OR Or or or OR OR OORR OORr OoRR Egg Or Egg Or OORr OOrr Oorr or or OoRR oorr oorr or or Oorr oorr oorr Slide 59 / 142 Since we are comparing observable traits we need to convert each genotype into its phenotype. OR OR Or Sperm or or Slide 60 / 142 Finally we can count the possibilities and determine the expected ratio OR OR Or Sperm or or 9 Orange Rectangle Egg Or Egg Or 3 Orange Oval or or 3 Green Rectangle or or 1 Green Oval

11 Slide 61 / 142 Slide 62 / 142 Now the researcher is ready to do the actual cross. She breeds 2 individuals from the F 1 and records the number of offspring. This is the F 2 generation. Male Possible Phenotypes Female F 1 F 2 Conclusion: The original question was: Does the inheritance of color effect the inheritance of shape? If you discover observed results fit closely with the expected result then you should conclude that color does not effect shape and therefore these are unlinked genes. If they do not conform to your expected then they are linked (on the same chromosome) and their inheritance is not independent. number of offspring Does this fit with the expected ratio? Slide 63 / 142 Law of Independent Assortment The preceding slides demonstrated Mendel's second law: The Law of Independent Assortment states that separate genes for separate traits are passed independently of one another from parents to offspring. Slide 64 / 142 Independent Assortment To demonstrate independent assortment lets look at 5 nondescript traits inside of one organism. Each has 2 alleles. Trait 1: Allele A Allele B The biological selection of one gene's allele pair for a particular trait has nothing to do with the selection of any other genes for any other trait. Trait 2: Allele A Trait 3: Allele A Trait 4: Allele A Trait 5: Allele A Allele B Allele B Allele B Allele B Slide 65 / 142 Independent Assortment Just because allele A of trait 1 is selected in a particular gamete (sperm or egg) that does not mean it will be the same for the other traits. Each is selected independent of the others. 11 AABB aabb Slide 66 / 142 What would be the expected number of offspring that have both dominant traits if the above cross produced 1000 offspring? Trait 1: Allele A Trait 2: Allele A Trait 3: Allele A Trait 4: Allele A Trait 5: Allele A Allele B Allele B Allele B Allele B Allele B

12 Slide 67 / 142 Slide 68 / AaBb aabb What would be the expected number of offspring that have both dominant traits if the above cross produced 1000 offspring? 13 AaBb AaBb What would be the expected number of offspring that have both dominant traits if the above cross produced 1000 offspring? Slide 69 / 142 What Mendel Didn't Know Slide 70 / 142 Non-Mendelian Traits Over the years since Mendel's discovery of genetics, modern science has discovered many traits that show exception to Mendel's Law. We can term these non-mendelian traits. For example, many traits are linked which means they do not follow the law of independent assortment. Examples of linked traits are red hair and freckles. Return to Table of Contents Molly Ringwald Slide 71 / 142 Linked Genes The main new discovery that changed how we interpret Mendel's results is chromosomal inheritance. Slide 72 / 142 Linked Genes Red hair and Freckles are on the same chromosome Sexual reproducers pass their genes on small bundles of DNA called chromosomes. Each chromosome contains many genes. The chromosomes DO independently assort but the gene for a few, several, or many traits may be on the same chromosome. These linked genes are inherited as a package. Genes Chromosome Molly Ringwald

13 Slide 73 / 142 Non-Mendelian Genetics Click here for a review video of Non-Mendelian Genetics If further review is needed, please see NJCTL's first year biology course. Slide 74 / ABO blood type is not a Mendelian "either /or" trait. Which of the following allelic relationships is responsible for this phenotype (trait)? A Codominance B Multiple genes C Incomplete dominance D Multiple alleles E Both A and D Genetics First Year Course Slide 75 / Sickle-cell disease is caused by a single trait but causes multiple effects on its victim. This is an example of A Epistasis/Pleiotropy B Multiple genes C Incomplete dominance D Codominance E Multiple Alleles 16 Skin color is an example of A Epistasis B Multiple genes C Incomplete dominance D Codominance E Multiple Alleles Slide 76 / 142 Slide 77 / A white mouse and black mouse produce an offspring mouse that is grey. This is most likely due to A Epistasis B Multiple genes C Incomplete dominance D Codominance E Multiple alleles Slide 78 / A white mouse and a black mouse produce an offspring mouse with black and white spots. This is most likely due to A Epistasis B Multiple genes C Incomplete dominance D Codominance E Multiple alleles

14 Slide 79 / Fur color of a particular organism could be orange, purple, green or blue. This is most likely due to A Epistasis B Multiple genes C Incomplete dominance D Codominance E Multiple alleles Slide 80 / A researcher does a cross to determine the heredity of a sexually reproducing organism. He uses 2 F 1 individuals (hybrids) that are heterozygous for 2 traits. He expects a 9:3:3:1 ratio. Out of 2,000 offspring he gets 800 individuals that are recessive for both traits. Does this fit his expected results of this dihybrid cross? Yes No Slide 81 / In the previous question what would the expected number of homozygous recessive individuals be? (2,000 offspring 9:3:3:1 ratio) Slide 82 / What is the reason for the inconsistency in the last example? A Multiple alleles B Incomplete dominance C Codominance D Linked genes E Any of the above Slide 83 / 142 Linked Genes Mendel didn't know was that alleles are grouped together on chromosomes. In this case the recessive are on one of the chromosome pairs of the parent and the dominants on the other. Slide 84 / 142 Linked Genes The previous slide illustrated the concept of linked genes. They turned out to be an important focus of study that lead to a more advanced understanding of chromosomes. a b A B Take notes as you go through the next 4 slides so that you can answer a question about the practice experiment. ab After meiosis there is only 2 possibilities for gametes AB

15 Slide 85 / 142 Beetle Genes A geneticist is working with blister beetles and happens to find a new mutation in one individual. This individual is striped while the rest of the population is one color. Slide 86 / 142 Beetle Genes Curious to see if the new allele was dominant or recessive he conducted a Mendelian style cross. The mutant was male so he crossed it with a female wild type. Beetle A represents the wild type, or the common variant found in nature Beetle B represents the mutant, or the unusual variant. As expected all the offspring were all wild type (F1 generation) Slide 87 / 142 Slide 88 / 142 Beetle Genes Next he took a male and female from the F1 generation and crossed them. As expected he got a 3 to 1 ratio (F2 generation). However, out of more than 10,000 offspring, all the striped individuals were male. Beetle Genes Finally he took a striped male and an F 2 generation female and crossed them. Take a look at his results. With a small group theorize why he got these results. Start with the original pairing and work to this cross. F2 25% 25% 25% 25% only and 15,000 total offspring produced Slide 89 / 142 Slide 90 / 142 Thomas Hunt Morgan The previous experiment mimics the one done by Thomas Hunt Morgan in Look through the below slide show and compare Morgan's result to your own. Click here for a synopsis of Morgan's experiment Sex-linked Genes The sex chromosomes have genes for many characteristics unrelated to sex. A gene located on either sex chromosome is called a sex-linked gene. Sex-linked genes follow specific patterns of inheritance. Many sex-linked disorders are located on the chromosome. Typically these disorders are inherited through the mother, as she gives an chromosome to any child she has. Sex-linked disorders show up in males more often than females because they only have 1 copy of an chromosome.

16 Some disorders caused by recessive alleles on the chromosome in humans: Color blindness Duchenne muscular dystrophy Slide 91 / 142 -linked Disorders Slide 92 / linked disorders show up more often in males because in terms of sex chromosomes: A B C D they are likely to inherited the recessive allele from their father they only inherit one -chromosome the Y-chromosome from their mother carries the recessive allele the father can pass the damaged allele on Hemophilia Male pattern baldness Slide 93 / 142 Slide 94 / In reference to the parental genotypes below, % of children are likely to be male. A 100 B 25 C 50 D cannot be determined 25 In reference to the parental genotypes below, % of children are likely to be female. A 100 B 50 C 25 D cannot be determined father mother father mother Y h Y h Slide 95 / 142 Slide 96 / In reference to the parental genotypes below, there is likelihood that a child will have a hemophilia phenotype. 27 In reference to the parental genotypes below, there is likelihood that a female child will have a hemophilia phenotype. A no A no B some B some father Y mother h father Y mother h

17 Slide 97 / 142 Slide 98 / In reference to the parental genotypes below, there is likelihood that a male child will have a hemophilia phenotype. 29 The likelihood that a male child with hemophilia will be born is %. A B no some father mother father mother Y h Y h Slide 99 / 142 Genetic Linkages Take notes: Another researcher was studying a type of caterpillar. She noticed that a very few individuals in the population had blue spots. After careful screening, she also observed that if a caterpillar had blue spots it was also very likely to have long antennae. Slide 100 / 142 Genetic Linkages After extensive work she determined that the 2 mutant traits are recessive and linked. She notated the wild type genes as and the recessives as a and s Wild Type Mutant Wild Type Mutant Short Antennae = No spots = Long Antennae = a Spots = s Slide 101 / 142 Slide 102 / 142 Genetic Linkages Genetic Linkages She preformed the following cross: Determine expected results: as as, as as, as as, as as, as, as Heterozygous wild type female as, as Mutant Male 1 wildtype : 1 mutant Her actual results differed from her expected. Actual Results: 22,000 offspring produced 10,855 both traits wild type 10,930 both traits mutant 110 blue spots short antenae 105 no spots long antenae What has taken place that caused these linked genes to recombine? Parental type Recombinations

18 Slide 103 / 142 Recombination Frequency Crossing over can separate genes that are typically on the same chromosome. In this case the maternal chromosome containing the wild type crossed over with homologous pair containing the mutant alleles a a a a a a Slide 104 / 142 Recombination Frequency Recombination frequency is the average percentage of cross over events that occur between linked genes. number of recombinations number of offspring = Recombination Frequency Use the actual results from the previous experiment to calculate the recombination frequency of the 2 linked genes. s s s s s s The individuals that received the parental type genomes had no crossover event. The ones the received recombined traits received cross over chromosomes 22,000 offspring produced 10,855 both traits wild type 10,930 both traits mutant 110 blue spots short antenae 105 no spots long antenae 215 / 22,000 = % Rf Slide 105 / 142 Genetic Linkage Maps After further observation and experiments a third difference was recognized between the wild type and mutant caterpillars. Slide 106 / 142 Genetic Linkage Maps When placed near a bright light source the wild type moved toward it. The mutant moved away. Wild Type Mutant Wild Type Mutant Short Antennae = Long Antennae = a Short Antennae = Long Antennae = a No spots = Spots = s No spots = Spots = s Slide 107 / 142 Slide 108 / 142 Genetic Linkage Maps This behavioral trait was recognized as a third linked gene and new notation produced. Genetic Linkage Maps Using the previous experiment as a model, design experiments that will show the recombination frequency of the new trait and each of the previously tested mutant traits. Wild Type Mutant Wild Type Mutant Short Antennae = Long Antennae = a Short Antennae = Long Antennae = a No spots = Spots = s No spots = Spots = s Move to light = Move from light = m Move to light = Move from light = m

19 Slide 109 / 142 Genetic Linkage Maps After you run your experiments you get the following data: Rf a s m a x 1% 3% s x x 2% m x x x What does this tell us about the position of these genes on the common chromosome? Rf a s m a x 1% 3% s x x 2% m x x x Slide 110 / 142 Genetic Linkage Maps Remember that the amount of times that a crossing over event happens between 2 genes correlates to the recombination frequency. a a a a a a s s s s s s Slide 111 / 142 Genetic Linkage Maps Recombination frequency can show the relative positions of genes on a chromosome. The higher the Rf the further apart the genes. In this case... Slide 112 / 142 Genetic Linkage Maps Since recombination frequency directly relates to distance, Rf can be converted to a unit of distance known as a centimorgan (cm), also called a map unit (mu) Rf a s m a x 1% 3% s x x 2% m x x x s a 1% 2% 3% m Rf a s m a x 1% 3% s x x 2% m x x x s a 1cM 2cM m 3cM Slide 113 / 142 Genetic Linkage Maps Construct a linkage map for the following table of map units. Start by drawing a line to represent a chromosome. Then use a ruler to create a scale (ex- 1mu = 1/2centimeter). Slide 114 / 142 Probability, Statistics and Genetics Return to Table of Contents

20 Slide 115 / 142 Genetics is Based on Prediction The study of genetics relies on the comparison of expected values and observed values based on actual experimental observation. Slide 116 / 142 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? 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. 10,000 offspring produced 6,860 purple flowers 3,140 white flowers Slide 117 / 142 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. This is the formula: Slide 118 / 142 Chi Squared Test In other words, it tells us if the difference is due to random chance or if it is due to some other factor that effected the results of our experiment. Chi Squared A simple example: Slide 119 / 142 Chi Squared Test 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). Slide 120 / 142 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. However, when we actually flip a coin 100 times we get (o) =

21 Slide 121 / 142 Chi Squared Test Slide 122 / 142 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. The first step is to get a value for 2. 2 = (56-50) 2 (44-50) = = No that we have our value for Chi Squared, we can compare it to a statistical analysis chart known as critical values. Slide 123 / 142 Chi Squared Test 1.44 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 Slide 124 / 142 Chi Squared Test 1.44 Next we decide our level of significance. Usually it is.05, which means that we are 95% certain of our outcome. If a higher standard is needed another level could be used. Slide 125 / 142 Chi Squared Test 1.44 < 3.84 is our critical value. If 2 is less than the critical value we accept the null hypothesis. If it is more than we reject the null hypothesis Slide 126 / 142 Chi Squared Test null hypothesis- There is no significant difference between the observed and expected frequencies.

22 Slide 127 / 142 Genetics is Based on Prediction Now that you know chi squared, does this observed data meet with the expectations set by Mendel? Slide 128 / 142 Genetics is Based on Prediction What if we look at genotype? Does this observed data coincide with Mendel's postulates? 10,000 offspring produced 6,860 purple flowers 3,140 white flowers 10,000 offspring produced 2,930 Homozygous dominant 3,870 Heterozygous 3,200 Homozygous recessive Slide 129 / 142 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 Slide 130 / 142 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% yellow Pod color - Gg x Gg = 25% yellow 0.75 x 0.25 = = 3/16 Pod color G: green g: yellow Pea color Y: yellow y: green Slide 131 / 142 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 of the different ways. Slide 132 / If we toss two coins at the same time, what is the chance that both coins will land heads up? 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. P PP F1 Generation Pp 25% PP 25% pp = 50% homozygous p Pp pp F2 Generation

23 Slide 133 / In an F 1 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)? Slide 134 / 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 C D Law of Independent Assortment. Law of Probability. Law of Pea Plant Genetics. Slide 135 / An organism heterozygous for two characteristics is a. A B C D dihybrid monohybrid homozygote double dominant Slide 136 / 142 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 Aa: Probability of aa offspring = Bb Bb: Probability of bb offspring = Cc Cc: Probability of cc offspring = Slide 137 / 142 Slide 138 / 142 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. 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? What is the probability?

24 Slide 139 / A plant with genotype AABbCC is crossed with a plant having genotype AaBbCc. What is the probability of an offspring having the genotype AABBCC? Slide 140 / 142 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. Slide 141 / 142 Slide 142 / 142 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. Sexually reproducing populations are able to increase their variation and enable the population to survive more environmental possibilities. This is know as population genetics.