MCB 142. GSI Review Packet. Exam 1

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1 MCB 142 GSI Review Packet Exam 1 This packet is not meant to be a comprehensive review, but rather a supplementary study guide to go over the main topics covered in lecture and discussion. The packet is divided into the following sections: Mendel 1 Pedigrees Hardy Weinberg Chi Squared Mendel 2 Linkage Analysis On the last page you will find a page of equations SIMILAR to the one you will receive on exam day.

2 Mendel 1 Mendel's First Law (The Inheritance of Single Genes) Mendel's Law of Segregation During meiosis: homologous pairs of chromosomes (and the genes that compose them) separate from one another and are packaged into separate gametes. At fertilization, gametes combine at random to form the individuals of a new generation. In this figure, for trait 1 the individual is aa, and the gametes are all a; for trait 2, the individual is BB, and the gametes are all B; and for trait 3, the individual is heterozygous Dd, and the gametes are ½ D and 1/2d. Note also that there was a recombination event, but since this individual is homozygous at the other two loci, only 2 gametic types are produced: abd and abd in a 50:50 ratio. In addition to discovering the basic laws of genetic inheritance, Mendel made two innovations to the science of genetics: 1. Developed pure lines 2. Counted his results and kept statistical notes Pure Line - a population that breeds true for a particular trait [this was an important innovation because any non-pure (segregating) generation would and did confuse the results of genetic experiments]. 2

3 Results from Mendel's Experiments Parental Cross F 1 Phenotype F 2 Phenotypic Ratio F 2 Ratio Round x Wrinkled Seed Round 5474 Round:1850 Wrinkled 2.96:1 Yellow x Green Seeds Yellow 6022 Yellow:2001 Green 3.01:1 Purple x White Flowers Purple 705 Purple:224 White 3.15:1 Tall x Short Plants Tall l787 Tall:227 Short 2.84:1 Terms and Results Found in the Table Phenotype - literally means "the form that is shown"; it is the outward, physical appearance of a particular trait Mendel's pea plants exhibited the following phenotypes (as well as 3 other traits he studied): - round or wrinkled seed phenotype - yellow or green seed phenotype - purple or white flower phenotype - tall or short plant phenotype What is seen in the F 1 generation? We always see only one of the two parental phenotypes in this generation. But the F 1 possesses the information needed to produce both parental phenotypes in the following generation. The F 2 generation always produced a 3:1 ratio where the dominant trait is present three times as often as the recessive trait. Mendel coined two terms to describe the relationship of the two phenotypes based on the F 1 and F 2 phenotypes. Dominant - the allele that expresses itself at the expense of an alternate allele; the phenotype that is expressed in the F 1 generation from the cross of two pure lines Recessive - an allele whose expression is suppressed in the presence of a dominant allele; the phenotype that is not seen in the F 1 generation from the cross of two pure lines but reappears in the F 2 generation Mendelian Genetics Definitions Allele - one alternative form of a given allelic pair; tall and short are the alleles for the height of a pea plant; more than two alleles can exist for any specific gene, but only two of them will be found within any individual Allelic pair - the combination of two alleles which comprise the gene pair Homozygote - an individual which contains only one allele at the allelic pair; for example DD is homozygous dominant and dd is homozygous recessive; pure lines are homozygous for the gene of interest Heterozygote - an individual who contains one of each member of the gene pair; for example the Dd heterozygote Genotype - the specific allelic combination for a certain gene or set of genes Using symbols we can depict the cross of tall and short pea plants in the following manner: 3

4 The F 2 generation was created by selfing the F 1 plants. This can be depicted graphically in a Punnett square. From these results Mendel coined several other terms and formulated his first law. First the Punnett Square is shown for the selfing of F1 individuals. Union of Gametes At Random D d D DD (Tall) Dd (Tall) d Dd (Tall) dd (Short) Punnett Square The Punnett Square allows us to determine specific genetic ratios. Genotypic ratio of F 2 : 1 DD : 2 Dd : 1 dd (or 3 D_ : 1 dd) Phenotypic ratio of F 2 : 3 tall : 1 short Mendel's First Law - the law of segregation; during gamete formation each member of the allelic pair separates from the other member to form the genetic constitution of the gamete Confirmation of Mendel's First Law Hypothesis With these observations, Mendel could form a hypothesis about segregation. To test this hypothesis, Mendel selfed the F 2 plants. If his law was correct he could predict what the results would be. And indeed, the results occurred as he expected. F2 Phenotype Self Tall (D-) Self Short (dd) F3 Phenotype 1/3 All Tall: 2/3 Segregating All Short 3 Tall: 1 Short From these results we can now confirm the genotype of the F 2 individuals. Phenotypes Genotypes Genetic Description F 2 Tall Plants 1/3 DD 2/3 Dd Pure line homozygote dominant Heterozygotes F 2 Short Plants all dd Pure line homozygote recessive Thus the F 2 is genotypically 1/4 Dd : 1/2 Dd : 1/4 dd This data was also available from the Punnett Square using the gametes from the F 1 individual. So although the phenotypic ratio is 3:1 the genotypic ratio is 1:2:1 Mendel performed one other cross to confirm the hypothesis of segregation --- the backcross (in fact it was a testcross). Remember, the first cross is between two pure line parents to produce an F 1 heterozygote. 4

5 At this point instead of selfing the F 1, Mendel crossed it to a pure line, homozygote short plant. Backcross: Dd x dd D Female Gametes d Male Gametes d DD (Tall) dd (Short) Backcross One or (BC 1 ) Phenotypes: 1 Tall : 1 Short BC 1 Genotypes: 1 Dd : 1 dd Backcross - the cross of an F 1 hybrid to one of the homozygous parents; for pea plant height the cross would be Dd x DD or Dd x dd; most often, though a backcross is a cross to a fully recessive parent (which is a testcross). Testcross - the cross of any individual to a pure-breeding homozygous recessive individual; used to determine if the individual is homozygous dominant or heterozygous. So far, all the discussion has concentrated on monohybrid crosses. Monohybrid cross - a cross between parents that differ at a single gene pair (usually AA x aa) Monohybrid - the heterozygote (Aa) offspring of two parents that are homozygous for alternate alleles of a gene pair Remember --- a monohybrid cross is not the cross of two monohybrids. Monohybrids are good for describing the dominance/recessive relationships between alleles (also codominance and incomplete dominance). When an allele is homozygous it will always show its phenotype. It is the phenotype of the heterozygote which permits us to determine the relationship of the alleles. Dominance - the ability of one allele to express its phenotype in the presence of an alternate allele; generally the dominant allele will make a gene product that the recessive does not; therefore the dominant allele will express itself whenever it is present Mendel 1 Practice Questions: 1. A white-flowered plant is crossed with a pink-flowered plant. All of the F1 offspring from the cross are white. a. Which phenotype is dominant? 5

6 b. What are the genotypes of the original parent plants? c. What is the genotype of all the F1 offspring? d. What would be the percentages of genotypes & phenotypes if one of the white F1 plants is crossed with a pink-flowered plant? 2. Explain how it is possible for a cross between two plants, one with red flowers and one with white flowers, to produce all pink flowered offspring, when a cross between two pink flowered plants produces pink, white, and red flowered offspring? Answers: 1. a. white b. WW x ww c. Ww d. ½ Ww white, ½ ww pink 2. incomplete dominance 6

7 Human Genetics and Pedigree Analysis Mendelian Inheritance in Man The same principles of problem solving that apply to peas and fruit flies also apply to humans. Use Punnet Squares to predict the outcome of genetic crosses. Alleles can be dominant, recessive, co-dominant, incompletely penetrant, or lethal Dihybrid Cross: Inheritance of two traits at two separate genes, e.g., ABO blood type and Rh factor Review: Inheritance of ABO human blood type alleles. What are the possible blood types of the offspring of a cross between individuals that are type AB and type O? Blood type O is recessive. Pedigrees: Family Trees One of the central tasks of the human geneticist Pedigree analysis is the construction of family trees A pedigree is used to trace inheritance of a trait over several generations. Three primary patterns of inheritance: 1. autosomal recessive 2. autosomal dominant 3. Sex-linked recessive or dominant (X-chromosomal) Symbols used in pedigree charts: Autosomal Recessive Pedigree 7

8 Recessive: One way to distinguish recessive from dominant inheritance is to see if neither parent has the characteristic phenotype (disease) displayed by the child, this is an indication that the trait is recessive, in the absence of a new mutation. Autosomal: Gene is on one of the autosomes (Chromosomes 1-22). Male and female offspring equally likely to inherit trait. A typical pedigree: Note: Trait can appear in offspring of related individuals Parents of affected children in pedigree shown above are first cousins Autosomal Dominant Pedigree Dominant: Affected individuals appear in every generation unless there is incomplete penetrance. Autosomal: Gene is on one of the autosomes (Chromosomes 1-22). Male and female offspring equally likely to inherit trait. A trait that appears in successive generations is normally due to a dominant allele if the trait is rare. In this pedigree, an affected father passes the trait to half of his 6 children, including 2 daughters and a son. One of the affected daughters passes the same trait to one of her 3 children. Sex-Linked Pedigree 8

9 X-linked recessive: The trait is preferentially seen in males, who are hemizygous. Females are heterozygous "carriers. Most X-linked traits are recessive. Inheritance of red-green color blindness, an X-linked, recessive trait: In the pedigree above:all the daughters of a color blind male will be carriers of the trait or will be affected if the mother is also color blind. When mother is carrier, ½ will be affected and ½ will be carrier when father is color blind. 9

10 Practice Questions on Human Pedigrees: 1. The following pedigree is the result of a dominant condition. What is the risk for individual III 6 of having a child affected with this condition? 2. All questions pertain to the Pedigree below (Assume the trait is fully penetrant and rare). a. What is the mode of inheritance? Explain your answer. b. If an affected male marries an unaffected female, what are the chances the first female child will be affected? 3. Tom and Jane, who have two, healthy children (Liam and Leah) are the parents of a third child, Leander. The front of Leander's diapers show a brown coloration characteristic of alcaptonuria, a trait inherited as an autosomal recessive disorder. Draw a pedigree of this family using the standard symbols. Indicate which members of the pedigree must be carriers of alcaptonuria using a half-filled symbol. If Tom and Jane have another child, what is the probability it will have alcaptonuria? 10

11 Answers: 1. a. 0% 2. a. This trait seems to be X-linked dominant, but would need more affecteds to rule out autosomal dominant. b. ½ (if trait is x-linked dominant) 3. a. both parents must be heterozygous for the alcaptonuria allele, b. 1/4 Hardy-Weinberg Review Let s start with thinking about evolution in allele frequency terms. Evolution is the change in the allele frequencies in the gene pool of a population. Even though evolutionary forces operate on single individuals, effects of evolution are seen in the form of allele frequency changes at the population level. If a certain pair of alleles A and a have frequencies f (A) = 90% (p=0.9) and f (a) =10% (q=0.1) in one generation, and the next generation has p=0.92 and q=0.08, then evolution has occurred. The population s gene pool has evolved in the direction of a higher frequency of the A allele. Assuming that this change is a statistically significant change (i.e., it s not some random fluctuation; in the limit of an infinite population, we can be sure that it is significant), this leads us to think that there might be some selective force causing the change. If the change is not statistically significant, then it could be due to genetic drift effects. Now consider the (hypothetical) converse case, where no external forces are operating upon the population. It can be shown that allele frequencies are inherently stable in the absence of external forces, and so must be the same from generation to generation. (Note that it is allele frequencies and genotype frequencies that are stable, and not the numbers of individuals of each genotype. These will change as the population size changes) An example serves to illustrate the law: Suppose you are setting up an aquarium with blue, green, and yellow fish. Also imagine that the fish color is determined by two alleles at a single locus, such that BB is blue, Bb is green, and bb is yellow. One aquarium shop sells only blue fish and another shop only sells yellow ones. You buy 12 blue (75%) and 4 yellow (25%) fish. Thus, BB=0.75, Bb=0, and bb=0.25% (in the parent generation); additionally, p=0.75 and q=0.25. Under the conditions of the Hardy-Weinberg Law, after one generation, the proportion of blue (BB) fish will be 9/16 (0.5625), green fish (Bb) will be 6/16 (0.375), and yellow (bb) fish will be 1/16 (0.0625). Furthermore, these ratios or proportions will not change in future generations; they will remain the same. Of course, given that the populations are of finite size, there will be random changes in allele and genotype frequencies from expected under Hardy Weinberg proportions. Note that the heterozygous proportion increased at the expense of the homozygous proportions (D and R). Also note that p 1 =p 0 ; in other words the initial allele frequency does not change. The Law In the absence of disrupting factors (unbalanced mutation, unbalanced gene flow, natural selection, and genetic drift), the allelic and genotypic frequencies at a locus in a randomly interbreeding population (of diploids) will be repeated faithfully from generation to generation. Should the frequencies be perturbed for any reason, they will 11

12 come to expected equilibrium frequencies after one generation of random mating. (That is, even if the parents were not in equilibrium, the Hardy-Weinberg equilibrium will be restored in one generation under random mating and the other assumptions of H-W law.) The law is entirely theoretical; it represents a static situation wherein the genetic structure (composition) of the population does not change; it describes the situation in which there is no evolution. By revealing the conditions under which evolutionary change cannot occur, it also shows the forces that could operate to cause change in the genetic composition of a population. Hardy Weinberg Assumptions: 1. mutation is not occurring 2. natural selection is not occurring 3. the population is infinitely large 4. all members of the population breed 5. all mating is totally random 6. everyone produces the same number of offspring 7. there is no migration in or out of the population Deviations from Hardy Weinberg: 1. Mutation rates are usually low, but mutation is important in creating new variation 2. Strong selective forces 3. Small population sizes may show genetic drift and therefore deviations from HW, but these will not be significant with a chi-square test fit to HWP. 4. Non-random mating: Positive Assortative mating (applies to traits influencing selection of mates in humans: height, IQ); Self-fertilization in plants produces more homozygotes than expected from HW; Inbreeding 5. Relative fitness of different genotypes different number of offspring from individuals from different genotype classes 6. Migration effects Hardy and Weinberg went on to develop a simple equation or formula that can be used to discover the genotype frequencies in a population and to track their changes from one generation to another. This has become known as the "Hardy-Weinberg equilibrium equation." In this equation (p² + 2pq + q² = 1), p is defined as the frequency of the dominant allele and q as the frequency of the recessive allele for a trait controlled by a pair of alleles (A and B). In other words, p equals all of the alleles in individuals who are homozygous (AA) and half of the alleles in people who are heterozygous (AB) for this trait. In mathematical terms, this is: p = AA + ½AB Likewise, q equals all of the alleles in individuals who are homozygous (BB) and the other half of the alleles in people who are heterozygous (AB). q = BB + ½AB 12

13 Because there are only two alleles in this case, the frequency of one plus the frequency of the other must equal 100%, which is to say p + q = 1 Since this is logically true, then the following must also be valid: p = 1 - q q = 1 - p If random mating (panmixis) were to occur: The frequency of homozygous (AA) people = p 2 The frequency of homozygous (BB) people = q 2 The frequency of heterozygous (AB) people = pq + qp = 2pq. From observations of phenotypes, it is sometimes only possible to know the frequency of homozygous recessives, or q² in the equation, since they will not have the dominant trait. Those who express the dominant trait in their phenotype could be either homozygous dominant (p²) or heterozygous (2pq). Assuming Hardy-Weinberg Proportions, allows us to estimate the respective frequencies of these two classes. We cannot then do gene (allele) counting. We must estimate the using the square root formula: p=f (aa) 1/2. Knowing p and q, it is a simple matter to plug these values into the Hardy-Weinberg equation (p² + 2pq + q² = 1). This then provides the frequencies of all three genotypes for the selected trait within the population. However it makes no sense to try and test these values for being in HW equilibrium, since that was an assumption we made in order to calculate q and p. By comparing genotype frequencies from the next generation with those of the current generation in a population, one can also learn whether or not evolution has occurred and in what direction and rate for a selected trait. However, the Hardy-Weinberg equation cannot determine which of the various possible causes of evolution were responsible for the changes in gene pool frequencies. Testing for HW equilibrium One way to test whether a given population is in Hardy Weinberg equilibrium for a given trait is to take a sample from the population and count up the different phenotypes in that population. If the trait is such that one can distinguish heterozygotes from both homozygotes, (and thus the genotype frequencies are the same as the phenotype frequencies) then we can get the values of p and q from (homozygous dominants + ½ heterozygotes) and so on. We then calculate p 2, q 2, and 2pq; these multiplied by the total sample size give us the theoretical estimates of the numbers of individuals expected for each genotype class (if the population were in HWE). We already have observed a set of numbers of individuals for each genotype class; we need to test whether the deviation of the observed numbers from the expected numbers is statistically significant. A simple test to use is the chi-square test. 13

14 Degrees of freedom For the case of a biallelic locus, there are 3 genotype classes; of which one is not a free parameter. Also we estimate one of the allele frequencies from the population; hence in this case df = = 1. Refer to Handout #3 for discussion of: Relative fitness, Heterozygote advantage, and the various selective forces operating on alleles (frequency-dependant selection, balanced polymorphism, directional selection), and their counterparts for quantitative traits (stabilizing selection, directional and disruptive or diversifying selection); genetic drift, common mitochondrial and nuclear ancestor, founder effects, and evolutionary bottlenecks Notes on the above: Relative fitness: This is obtained by converting the average number of offspring from each genotype class (AA, AB, and BB for a two allele locus) to ratios. If the homozygote AA > het AB > homozygote BB i.e., then the number of AA individuals will increase and increase (unless the fitness values change at some later time point) until eventually the population consists only of AA individuals. This means that the A allele has been fixed (we have ignored chance effects). AA AB BB A fitness model, where AA is fitter than the other two: AA AB BB 1 1 hs 1 s The value of h depends on whether the gene is dominant or recessive or additive: if h = 0, then AA is as fit as AB and the A allele is dominant; if h=1, then AB is the same as BB and the B allele is dominant; for intermediate values of h, AB has intermediate fitness values. s decides how less fit the BB allele is with respect to AA, and therefore the strength of selection against the B allele; and the strength of selection for the A allele. It is possible to model this system and predict the number of generations that will be required for the A allele to change from f (A) = 0.01 to In our example above, s = 0.2; and h = 0.5; and the number of generations (refer table in handout #3) = 83. The heterozygote has a higher fitness than the homozygotes; both alleles will be maintained in the population: these alleles will be balanced. AA AB BB 1 - s s2 14

15 In these cases, the allele frequencies will reach certain equilibrium values; the equilibrium values can be shown to be p (A) = s2/ (s1 + s2); (Notice p (A) higher if more selection against B) p (B) = 1- p (A); Evolutionary trees: It is a basic tenet of biology that all life diversified from some common ancestor, and this pictorially looks like a tree with the root representing a common ancestor and the branches going away from the tree are the various species undergoing evolution. The farther away a species is from the root, the more different it is (having undergone evolution for a longer time; the branch length is proportional to the passage of time). This tree could be built by grouping all the life forms with similar body anatomies, for example, and some of the ancestors can be placed on the tree using the fossil record, and other ancestors could be inferred. Note that in general, we know of no direct way to build this tree to our satisfaction the ancestral life forms are no longer extant. Fossil records help us in some cases, but they are woefully inadequate, for several reasons. Also fossil records can only give us information about the morphological characters of the ancestral life forms. Evolution, as we understand it now, is a process of mutation and change in the genetic material of the organism (and we would like to understand it as such). Thus the descendant species of an ancestor species will be similar in a large portion of their genomes, but different in a subset (which is the subset that makes them different species, in some sense). As a corollary, all life forms share several basic biochemical mechanisms, and all life forms have counterparts of basic cellular machinery. When these counterpart proteins can be shown to have come (evolved, diversified) from an ancestral protein, they are known as HOMOLOGS. Homologs often have very similar amino acid sequence and function. Homologs are useful to understand the interrelationship of the extant (existing) species at a molecular level. If a protein in one organism is different from its homolog in another animal, it is possible, using various ideas that we shall not discuss here, to infer the sequence of an ancestor of the two organisms. These trees correlate well with the fossil record, and allow us to extend our understanding of evolution much further than would have been possible with the fossil record alone. From building many such trees, and from studying genetic material in several life forms, and from various statistical and biochemical principles, we can make educated guesses about how often different kinds of changes occur in the genes and proteins of various organisms. This rate is fairly regular, and this is termed a molecular clock. Note that there is no ONE molecular clock, the rate of change, and therefore the rate at which the clock runs, is different for different organisms, proteins, genes, position on chromosomes, function of proteins, and so on. Thus if we see a change, we can say things like On average, this change occurs once every 50 million years and therefore organism A is 50 15

16 million years away from organism B. This would correspond to the branch lengths from A to B on our tree. Practice problems: Solved problems pg 583, problems 2, 3, 10, 11(a, b) from the text. Also problems from practice problem set #2 and Handout #3. Chi squared review The chi squared test is a statistical test which can measure the probability of observing a certain set of data, given your expectations about what the data should look like. For instance, we can ask if we flip a coin 100 times, what is the probability that it will come up with 60 heads. Our expectation in this case is 50 heads if we start out with the expectation that it s a fair coin. But maybe it isn t fair. So is that a significant difference? What if we had 70 heads in 100 flips? We can set up our calculation as follows: Heads tails Observed Expected O-E (O-E) (O-E) 2 /E 2 2 chi squared is then 2+2=4. but what does this tell us? Nothing until we figure out how to use our chi squared table. The table is shown here: Significance Degrees 95% 99% 99.9% Of Freedom We will get to how to calculate degrees of freedom (df) in a minute. For now, let s just figure out how to use the table once we know the degrees of freedom. Let s say we have our chi squared=4 and that we know it has one df. Then we look at the df=1 row, and find if 4 is greater than our first number at probability p=.05, which is It is, so we know that our observed data are off from our expectation by an amount that should happen by chance <5% of the time. Another way to say this is that we have >95% confidence that our hypothesis (being that the coin is fair) is incorrect. We then check to see if our chi squared = 4 is greater than the next number in the row, It isn t, so we stop there and say we have >95% confidence in our rejection of the hypothesis, but not >99% confidence. So in general, the bigger our chi squared value is, the worse our original hypothesis is, and the greater confidence we have in its rejection. If we had seen 70 heads in 100 flips, our chi squared value would be 16 (try the calculation 16

17 yourself for practice if you want), and this is greater than even the.001 level for df=1. this means that we are even more sure that our hypothesis that the coin is fair is wrongwe now have >99.9% confidence. Now, what about degrees of freedom? This is a tricky concept, so I m going to make it as simple as I can. It may seem like an arbitrary thing, since for simplicity I m not going to go into any of its derivation. You ll just have to believe that it works, and if you want to know more feel free to look in up in a stats book. To calculate df, follow these 3 easy steps: 1. Check if your expected numbers depend at all on the data. For instance, if we re looking at a cross and we re testing if we have close to a 3:1 ratio for some trait in the offspring, then our expected numbers don t depend on the data. If this is the case, df= (number of classes)-1. The number of classes is just the number of phenotypes in a cross, so it ll be 2 classes if there are two variants of a trait (like tall stems vs. short stems), 3 if there are 3 variants (like blue, brown, and green eyes), and so on. 2. If your expected numbers do depend on the data, then it s a bit more complicated. For instance, let s say we re given the frequencies of AA, AB, and BB genotypes in a population, and we want to check if it s close to Hardy-Weinberg (HW) equilibrium. We don t know our expected numbers right off the bat- we need to use the data to calculate them. So we count our number of A alleles and B alleles to get our p and q values, and then our expecteds are p 2, 2pq, and q 2. When we have to do this, we take our df found in #1, and we subtract the number of parameters we had to estimate from the data. In this case, there are two parameters (p and q), but since q = 1-p, we really only had to estimate one of them from the data. So we subtract 1 from AA, AB and BB df. Since there are 3 classes in this case, df=3-1-1=1. 3. Now there s just one last step-clumping. Since chi squared is just an approximationnot an exact calculation- we need to take into account that it doesn t do so well with small numbers in the expected frequencies. We use a somewhat arbitrary cutoff of E<5 to say that if any expected class is <5, then we need to combine it with another class. We combine it with the class that has the next lowest expected value. When we do this, we lose one class, and thus one df. We don t lose any parameters though, since we ll still need the same number of parameters to calculate our expected numbers. Let s do an example of this: AA AB BB total Observed Thus frequency of A =p=0.1, and q=1-0.1=0.9. so our expected numbers are AA AB BB total Expected As you can see, expected number of AA is only 1, so we have to combine it with the next lowest expected class, that being AB. We then have a new class of AA+AB of 19 expected and 10+5=15 observed. We now have 2 classes, so our df is then 2-1-1=0, since 17

18 we still have one parameter we are estimating. This means we can t do the calculation, since we don t have any df. For an example where you combine classes and still have df left, see pg 8 of the 2 nd lecture handout. And that s it! See, that wasn t so bad. Now you just need to practice it. Here are some problems you should try if you want: Problems 1+2 from pg 8-9 of the 2 nd lecture handout Problem 7 from the lectures 5-12 practice questions Confidence limits review A topic related to chi squared is confidence limits. What if we don t want to compare a specific set of observed data to expected data, but instead just want to figure out what is the range of values the true answer is going to be within most of the time, based on our observed data? For instance, if we observe in a set of 100 organisms that frequency of some alleles p=.2 and q=.8, what can we say about the values of p and q for the entire population, if the population is much bigger than those 100 organisms? For this we need confidence limits, a much simpler concept than chi squared. If we want to know, based on our observed p and q for the 100 organisms, what two numbers will the p and q of the entire population be within 95% of the time, these two numbers are our 95% confidence limits. To calculate them, simply use the equation: Standard deviation = VTUW[p (1 - p) / 2n] Where p is either allele frequency [since p(1-p)=q(1-q)], and n is our sample size. For the example above, sd = sqrt (.2*(1 -.2) / (2*100) = sqrt (.16/200) = sqrt (.0008) =.028. one sd gives our 68% confidence limits, and 1.96*sd (or 2*sd when we round off) gives our 95% confidence limits. So 2*.028=.056. We add this to our p=.2 to get the upper bound of p=.256, and subtract it to get the lower bound of p=.144. This tells us that 95% of the time, our true population p frequency will be between.144 and.256. Likewise, the q will be between.744 and % of the time. Note that the more observations we have the more confidence we have- if we observe p=.2 for 200 organisms, our confidence limits shrink to between.18 and.22 for p. That s all there is to it. Try problem 1 from pg 16 of lecture handout 1 and problem 11 from lectures 2-4 practice problems if you want to practice it. 18

19 Dihybrid Cross: The independent assortment of alleles Question: How are two traits inherited? Experimental Approach: A cross involving two true-breeding traits. System: Pea Plants. Seed color (Y or y) and seed shape (S or s). F1 Generation All F1 plants of an SSYY x ssyy were yellow and smooth. F1 Genotype is SsYy What gametes are produced by the F1 Generation? 19

20 Each of the male gametes types of (SY, Sy, sy, sy) can fuse with each of the types of female gametes (SY, Sy, sy, sy). 16 possible combinations of gametes are possible. We will see that there are 9 possible genotypes and 4 possible phenotypes. The two parental phenotypes, and two new phenotypes were obtained. Pleiotropy and Epistasis Some genes affect more than one feature in the phenotype. For example a plant that lacks a functional gene involved in synthesis of abscisic acid has leaves that wilt easily and seeds that do not become dormant in the normal way. This is an example of a pleiotropic effect. Phenotypic characters may require appropriate alleles of several genes in order to be expressed. Flower color is often determined by enzymes that hydroxylate anthocyanin molecules at various positions. However, if genes for one or more enzymes early in flavonoid metabolism are not functional, the pigment may not be formed at all. The flavonoid biosynthesis genes are epistatic to the hydroxylating enzyme genes. Dihybrid Cross Practice Problems: Note: Please realize that even though I am giving questions that are multiple choice, there will be no multiple choice on the exam!! 1. A phenotype ration of 9:3:3:1 in the offspring of a mating of two organisms heterozygous for two traits is expected when: A. the genes reside on the same chromosome B. each gene contains two mutations C. the gene pairs assort independently during meiosis D. only recessive traits are scored E. none of the above 20

21 2. Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1? A. SSYY x ssyy B. SsYY x SSYy C. SsYy x SsYy D. SSyy x ssyy E. ssyy x ssyy 3. The expected phenotypic ratio of the progeny of a SsYy x ssyy test cross is: A. 9:3:3:1 B. 1:1:1:1 C. 3:1 D. 1:2:1 E. 3:1:1:3 4. In a dihybrid cross, AaBb x AaBb, what fraction of the offspring will be homozygous for both recessive traits? A. 1/16 B. 1/8 C. 3/16 D. ¼ E. ¾ 5. Two unlinked loci affect mouse hair color. CC or Cc mice are agouti. Mice with genotype cc are albino because all pigment production and deposition of pigment in hair is blocked. At the second locus, the B allele (black agouti coat) is dominant to the b allele (brown agouti coat). A mouse with a black agouti coat is mated with an albino mouse of genotype bbcc. Half of the offspring are albino, one quarter are black agouti, and one quarter are brown agouti. What is the genotype of the black agouti parent? A. BBCC B. BbCc C. bbcc D. BbCC E. BBcc 6. Two unlinked loci affect mouse hair color. AA or Aa mice are agouti. Mice with genotype aa are albino because all pigment production is blocked, regardless of the phenotype at the second locus. At the second locus, the B allele (agouti coat) is dominant to the b allele (black coat). What would be the result of a cross between two agouti mice of genotype AaBb? A. 4 agouti: 4 black: 8 albino B. 9 agouti: 3 black: 3 albino: 1 grey C. 9 agouti: 3 black: 4 albino D. 8 agouti: 4 black: 4 albino 21

22 Three point crosses: 1. If the total is not given, do that first. 2. Label the parentals (LARGEST CLASSES) and double recombinants (smallest classes). 3. Order the genes. Compare the double recombinants to the parentals. The allele that is different is the middle allele. 4. Find the distance between the two end alleles by adding up all the recombinants and two times the number of double recombinants. Divide by the total number of offspring: # of recombinants RF= 100 x = % of recombinants # of offspring 5. Find the distance between the middle and one of the ends. 6. Find the distance between the middle and the other end. 7. Check that the three pairwise recombination distances are consistent. 3 pt cross practice problems: b+ wx+ cn 382 b wx cn+ 379 b+ wx cn 69 b wx+ cn+ 67 b+ wx cn+ 48 b wx+ cn 44 b wx cn 6 b+ wx+ cn a. What is the order of the 3 genes? b. What is the correct distance in map units between genes b and wx? c. What is the distance between b and cn? d. Draw a map to illustrate the distances between the three loci. 22

23 8. Using the test cross data below, make a map of these 3 genes (pr = purple eyes, dp = dumpy body, and bl = black body color). Phenotype Count pr dp bl 21 pr dp + 23 pr + bl dp bl 60 pr dp bl 750 Total 2500 Linkage Disequilibrium (LD) Two loci are in linkage equilibrium if genotype frequencies at one locus are independent of genotype frequencies at the second locus, otherwise the two loci are in linkage disequilibrium. Linkage disequilibrium can arise from history (a new mutation occurs on a specific haplotype, and hence LD is immediately created), genetic drift, and selection on multilocus genotypes. Significant LD is usually only seen for very closely linked loci. One can test whether or not two loci are in linkage equilibrium by comparing known twolocus genotype frequencies with two-locus genotype frequencies calculated from onelocus genotype frequencies. A chi-square test can be performed to determine if the known and expected differ significantly. D, the coefficient of linkage disequilibrium measures the amount of disequilibrium between two loci. For example, for a population of 500 individuals, with two alleles at each gene, you might find the following combinations of alleles at loci A and B. AB 251 Ab 249 ab 255 ab 245 The four haplotypes involving these 4 alleles occur at approximately equal frequencies. Thus neither of the alleles at locus A (A or a) occur preferentially with the alleles at locus B (B or b). These are said to be in linkage equilibrium. However, consider the following population: AB 490 Ab 1 ab 2 ab

24 Here, allele A is almost always with allele B and allele a is almost always with allele b. The combinations Ab and ab are extremely rare. One says that loci A and B are in linkage disequilibrium. In this example, P A = 0.491, P B = 0.492, and D = (0.491)(0.492) = (see equations below). How do we measure linkage disequilibrium? Expected gametic frequencies if two alleles are independently associated can be obtained from allelic frequencies in the population: Allele A a allele p A 1-p A freq. B p B p A p B (1-p A )p B B 1-p B p A (1-p B ) (1-p A )(1-p B ) Sum of frequencies = 1 We can measure the non-randomness of the gametic frequencies by means of a deviation from two locus equilibrium: D is the linkage disequilibrium coefficient, or measure of deviation from 2 locus equilibrium, as follows: Gametic = random deviation frequencies p AB = p A p B + D p Ab = p A (1-p B ) - D p ab = (1-p A )p B - D p ab = (1-p A )(1-p B ) + D Obviously, the sum p AB + p Ab + p ab + p ab = 1 D = 0 alleles at the 2 loci are randomly associated (linkage equilibrium) D > 0 A more frequent with B, than expected by chance (and a with b) D < 0 A & b are preferentially associated (and also a and B) It is important to realize that linkage disequilibrium is that not the same as genetic linkage. Genetic linkage refers to the fact that two loci are physically close together on a chromosome, whereas linkage disequilibrium refers to a statistical association among alleles at those loci. LD Practice Problem: 9. Below is the data from two different populations. Suppose that these two populations are mixed, in equal numbers. What would be the new frequency of the A1, B1 and A1B1 haplotypes? Is there linkage disequilibrium (hint: calculate D)? Population 1: Population 2: f(a1) = 0.1 f(a1) = 0.7 f(b1) = 0.2 f(b1) = 0.6 f(a1b1) = 0.02 f(a1b1) =

25 Answers: 1. C. the gene pairs assort independently during meiosis: The parental organisms have the same phenotype, but their offspring have 4 different phenotypes. The 9: 3: 3: 1 ratio can only occur if the two traits assort independently during meiosis. 2. C. SsYy x SsYy: The dihybrid cross was invented by Mendel to discover the independent assortment of alleles during gamete formation. 3. B. 1:1:1:1: SsYy, ssyy, Ssyy, ssyy are predicted to occur in a ratio of 1:1:1:1. 4. A. 1/16: ¼ of the gametes of each parent will be ab. The fraction of the offspring homozygous for both recessive traits will be ¼ times ¼, or 1/ B. BbCc 6. C. 9 agouti: 3 black: 4 albino: This is a variant of the normal 9:3:3:1 ratio for a normal dihybrid cross without epistasis. 7. a) b cn wx b) 25 mu c) 10 mu d) b cn wx Map:.25 pr bl dp Both populations are originally in linkage equilibrium, LD = 0. After mixing: f(a1) = ( )/2 =.4, f(b1) = ( )/2 = 0.4 and the haplotype f(a1b1) = ( )/2 =.22. If in linkage equilibrium D = 0, this would mean that f(a1b1) = f(a1) x f(b1) = (0.4)(0.4) = In this case, D = = 0.06, thus the newly mixed population is in Linkage Disequilibrium, created by the mixing of two populations with different allele frequencies. 25

26 Mendelian and Population Genetics Equations, Tables, Formulas, etc. confidence limits of an allele frequency estimate: denoting by p the true frequency of allele A, the variance of the allele frequency estimated by the method of gene (allele) counting is p(1-p)/(2n) where n is the sample size. An estimate of this variance is obtained using this formula and the estimated allele frequency. The standard deviation (sd) is the square root of the variance. The 95% confidence interval for the range of the true allele frequency p is twice the sd on either side of the estimate. Theoretical Chi-square values with different probabilities: Degrees Probability Of Freedom 0.05* 0.01** 0.001*** degrees of freedom (df): when the expected values are independent of any features of the observed data, then the df is given by the number of genotype classes considered (with expected value >=5) minus 1. When the observed data is used in calculations to determine the expected values, e.g., for testing HWP the allele frequencies must be estimated from the observed data to determine the expected HWP for that data set, then the df is as above minus the number of independent parameters estimated. linkage disequilibrium (LD): Consider two genes with two alleles each (A, a and B, b), and hence four gametic types (referred to as haplotypes): AB, Ab, ab, and ab. The frequencies of the four gametes (denoted f(ab) etc.) can be described in terms of the allele frequencies p A (p a = 1 - p A ) and p B (p b = 1 - p B ) at the two loci, and the LD parameter D, as follows: f (AB) = p A p B + D, f (Ab) = p A p b - D, f (ab) = p a p B - D, f (ab) = p a p b + D, where D = f(ab) - p A p B. heterozygote advantage: the heterozygote has a higher relative fitness than both homozygotes, this leads to a balanced polymorphism. If we denote the relative fitness values of the 3 genotypes in a 2-allele system as follows: A1A1 A1A2 A2A2 relative fitness 1 - s s2 then the equilibrium frequency for allele A1, denoted p1* = s2/(s1 + s2), and that for allele A2 is p2* = s1/(s1 + s2), with p1* + p2* = 1. 26

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