READING ASSIGNMENT GENETIC ANALYSIS OF DROSOPHILA POPULATIONS I. HOW DO MITOSIS AND MEIOSIS COMPARE?

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1 READING ASSIGNMENT GENETIC ANALYSIS OF DROSOPHILA POPULATIONS I. HOW DO MITOSIS AND MEIOSIS COMPARE? II. HOW CAN WE DETERMINE EXPECTED RATIOS OF OFFSPRING? What rules can we learn from Mendel s work with pea genetics? III. HOW DO WE KNOW IF OUR OBSERVED RESULTS SUPPORT THE GENETICS PROPOSED? How does the Chi Square Test help? IV. WHAT DO WE KNOW ABOUT THE DROSOPHILA LIFE CYCLE? Reading Assignment - Genetics

2 I. HOW DO MITOSIS AND MEIOSIS COMPARE? Read the sections in your text book that describe the cell cycle and the stages of mitosis and meiosis. After reading these you should know: 1) which cells in your body undergo mitosis and for what reasons do they undergo mitosis. 2) the various stages of mitosis and how they can be recognized. 3) which cells undergo meiosis and for what reasons they undergo meiosis rather than mitosis. 4) the stages of meiosis and how they can be recognized. 5) the major differences between mitosis and meiosis. 6) what relationship(s) Mendel's laws of independent assortment and segregation have to specific meiotic events. A. Based on your reading, complete the following table. C Fill in the major events which occur in each phase of mitosis, meiosis I and meiosis II. C In the last row of the table indicate which events of a given stage are common (that is which occur in all three of the division cycles: mitosis, meiosis I and meiosis II) and which are different or unique (occur in only one of the division cycles). Then answer the following questions. Mitosis Interphase Prophase Metaphase Anaphase Telophase Cytokinesis Meiosis I Meiosis II Common events vs. differences B. Answer the following summary questions. 1. How does mitosis differ from meiosis? 2. What are the major differences between meiosis I and meiosis II? 3. Why is it essential that the resulting cell products of meiosis (which will become gametes) have half the number of chromosomes of the parental cells? Reading Assignment - Genetics

3 II. HOW CAN WE DETERMINE EXPECTED RATIOS OF OFFSPRING? What rules can we learn from Mendel s work with pea genetics? A. Who was Mendel? Genetics, as a field of scientific inquiry, began with the studies of Gregor Mendel, an Austrian monk. In 1865, he reported the results of eight years of breeding experiments with plants to the Naturalist Society of Brun. Mendel's experiments, and those of many scientists since, demonstrated that many characteristics of individuals are inherited as individual units, now referred to as genes. Mendel's analysis is especially impressive since the importance of chromosomes and the mechanics of mitosis and meiosis had yet to be discovered. B. What kinds of research did Mendel do? Mendel discovered the genetics of specific traits by doing controlled crosses of known parents. He began by establishing pure breeding lines of plants which displayed contrasting genetic characteristics such as tall vs. short plants, yellow vs. green seeds, smooth vs. wrinkled seeds. A pure breeding line of tall plants mated with each other will always produce tall offspring. Similarly, a pure breeding line of plants with red flowers when crossed with each other always produce offspring with red flowers. Today we would say that these lines are homozygous for the gene in question. In one set of experiments, Mendel crossed a pure tall line to a pure short line and made the following observations. 1. Parental generation (P) -- Tall plants crossed with short plants 2. First filial generation (F 1 ) -- All offspring are tall 3. Tall F 1 offspring crossed with each other 4. Second filial generation (F 2 ) tall plants and 277 short plants Similar results were obtained for all seven pairs of characteristics Mendel studied. He concluded that: a) Inherited characteristics were controlled by pairs of factors (alleles of genes). b) Each individual contains a pair of such factors (alleles) for a particular gene. c) During the formation of gametes, this pair is separated or segregated, so that only one member of the pair appears in any one gamete. d) At fertilization, the single allele in the sperm and the single allele in the egg are combined so that the new individual again has a pair of alleles for that trait. In the example above, each of the tall plants contained a pair of alleles for tall stature and the short plants contained a pair of alleles for short stature. When the mating occurred, each parent (the P generation) contributed a single allele of the gene for height to the offspring. The tall plant, TT, contributed one T allele. The short plant (tt) contributed one t allele. The F 1 plants therefore had the genotype, Tt. These plants appeared tall because, in this case, the tall allele (T) was dominant over the short allele (which is then said to be recessive). When two F 1 plants are crossed, Tt crossed to Tt, each parent will contribute a T allele half of the Reading Assignment - Genetics

4 time and a t allele the other half of the time. This will result in three different genotypes, but two different phenotypes as shown in the offspring below: (½) T Sperm (½) t Eggs (½) T TT (1/4) Tt (1/4) (½) t Tt (1/4) tt (1/4) This results in 1/4 TT, (1/4 + 1/4) Tt, and 1/4 tt. Since TT and Tt individuals are phenotypically indistinguishable (all look alike; all are tall), the tall offspring would make up 3/4 of the total and the short, 1/4. This was just what Mendel observed. This is an illustration of what is now known as Mendel s law of segregation. Here, it is the individual alleles of a gene which are segregated. We know now that this occurs during anaphase I of meiosis when homologous chromosomes migrate to opposite poles of the spindle. As a result, the two alleles of each gene are physically separated or compartmentalized into gametes. C. Dihybrid Crosses Go one step further, as Mendel did, and deal with two specific traits, size and color, e.g. tall vs. short and yellow vs. green. Here, tall is dominant to short and yellow is dominant to green. In this dihybrid cross the parental genotypes would be TTYY and ttyy. The F1 genotypes would all be TtYy. We can predict the number of different genotypes that should appear in the F2 by using a Punnet square. To do this we must first determine how many different kinds of sex cells (gametes) an individual with the genotype, TtYy, can produce. The answer is four, because each gamete must have one allele from each pair but it makes no difference which one. The number of possible gametes can be calculated from the formula: 2 n = number of possible gametes. Here n = the number of heterozygous allele pairs (genes on separate chromosome pairs) under consideration. The four possible gamete combinations for this example would be: T t Y TY ty y Ty ty Next we use a Punnet square to determine expected types and probabilities of offspring for the cross. A Punnet square lists male gametes on one side and female gametes on the other. This allows you to visualize all possible combinations of male and female gametes. Therefore, the hypothetical or expected proportions of various genotypes and phenotypes can be determined. SAMPLE PUNNET SQUARE : Reading Assignment - Genetics

5 Sperm TY Ty ty ty TY Eggs Ty ty ty 1. Use this Punnet square to combine the gametes in all possible ways. 2. What expected phenotypic and genotypic ratios result in the offspring? 3. How were you able to determine this? This illustrates Mendel's Law of Independent Assortment. This law states that the alleles of different pairs of genes will assort (segregate) independently of each other during gamete formation. (Note: This is true, only if the genes in question are on different pairs of homologous chromosomes, i.e. are not linked on the same chromosome. In addition, genes that are very far apart on the same chromosome have a high probability of cross over. As a result, they may appear to be unlinked and to segregate independently.) If the genes in question assort independently, we can also solve problems like this by solving for each gene separately. For example, we can treat the cross of TtYy X TtYy as two separate crosses: Tt X Tt and Yy X Yy. We know that these crosses will produce: Tt X Tt 3/4 tall 1/4 short Yy X Yy 3/4 yellow 1/4 green In combination, these would yield: 3/4 tall 1/4 short 3/4 yellow 1/4 green 3/4 x 3/4 = 9/16 tall, yellow 3/4 x 1/4 = 3/16 tall, green 1/4 x 3/4 = 3/16 short, yellow 1/4 x 1/4 = 1/16 short, green Keep in mind that every offspring has to have both size and color and that the most frequent size and most frequent color will be observed together most often. Reading Assignment - Genetics

6 Terms to Know: Phenotype - the expression of the genes in an individual organism; what the individual looks like, e.g. tall, green, blonde. Genotype - the combination of genes found in an individual organism, e.g. TT, Tt, tt. If T (tall) is dominant, both TT and Tt individuals will be tall (have the same phenotype), but their genotypes will differ. Allele - different or contrasting forms of the same gene. For example, T (tall) and t (short) are alleles of the gene for overall height of the organism. Homozygous - an individual is said to be homozygous for a given gene when the two alleles in the genotype are the same. For example, TT is homozygous tall and tt is homozygous short. Heterozygous - an individual is said to be heterozygous for a given gene when the two alleles in the genotype are different, e.g. Tt. If T (tall) is dominant, this individual would display the tall phenotype. Hemizygous - the special condition seen when an otherwise diploid organism has only one representation (allele) of certain genes. For example, in mammals, females are XX (have two X chromosomes) and males are XY (have an X and a Y chromosome). For most genes on the X chromosome, there is no allele on the Y. For these genes, males would be hemizygous, i.e. have only one allele of these genes. III. HOW DO WE KNOW IF OUR OBSERVED RESULTS SUPPORT THE GENETICS PROPOSED? How does the Chi Square Test help? Expected ratios from a given genetic cross can be calculated. For example, if two genes are assorting independently, the offspring of individuals heterozygous for both genes are expected to occur in a 9:3:3:1 phenotypic ratio. However, observed ratios of offspring often differ from expected ratios. This occurs because each cross, like each flip of a coin, is subject to errors of sampling, or chance deviation from ideal or expected values. If you think about it, the significance of a given deviation is clearly related to sample size. If chance alone is causing the observed deviation from expected, larger samples should result in less deviation than smaller samples. For example, you are more likely to come closer to 500 heads and 500 tails out of 1000 flips of a coin than you are to 5 heads and 5 tails out of 10 flips of the same coin. On the other hand, if the deviation is consistently large, it is likely that something other than chance alone is operating. For example, if you flipped a coin 100 times and turned up 90 heads (a deviation of 40 from the expected 50), you would suspect a biased coin. Reading Assignment - Genetics

7 Chance variation can also affect the outcome of genetics experiments. In a cross of Tt (heterozygous tall) by tt (homozygous recessive short) you would expect a 1:1 ratio of tall to short in the offspring. In Experiment 1, you evaluate only 40 offspring, and find 30 tall and 10 short, the deviation of 10 may seem significant. Experiment 1: Tall Short Observed: Expected: Deviation: In Experiment 2, when you evaluate 200 offspring from the same cross, a deviation of 10 may seem less significant. Experiment 2: Tall Short Observed: Expected: Deviation: Using the Chi Square statistical test allows you to determine if the deviation you observe is significantly different from that expected on the basis of chance deviation alone. The Chi Square test is calculated as follows. 1. Observed values (number of actual offspring from a cross) are counted and expected values are calculated based on the assumed genetics. If you assume you are crossing a heterozygote with a homozygous recessive you would expect ½ the offspring to be dominant and ½ to be recessive for phenotype. To calculate the expected values multiply the total number of offspring actually counted by ½ to get each expected value. 2. Subtract the observed number of offspring from the expected number then square the resulting difference or deviation. Experiment 1 (30-20) 2 + (10-20) 2 = (10) 2 + (-10) 2 = 200 Experiment 2 (90-100) ) 2 = (-10) 2 + (10) 2 = Divide each squared deviation by the number you expected for that class. (In the example above, there are two classes, tall and short.) Experiment 1 100/ /20 = 10 Experiment 2 100/ /200 = 1 4. The resulting quotients (values from 3) are added together to yield the Chi Square value (O 2 ) for the distribution. O 2 = 3(d 2 /e) Here 3 means the sum of ; d = deviation = (observed - expected) and e = expected. In the examples above, two traits (classes) were possible. Where four offspring classes are possible, it becomes reasonable to allow the value of O 2 to be larger before questioning the hypothesis that chance alone explains the deviation. Conventionally, the effect of the number of independent classes is recognized as degrees of freedom. The number of degrees of freedom in tests of genetic ratios is almost always one less than the number of classes possible. Thus, in tests of 1:1 or 3:1 ratios, there would be one degree of freedom (two possible classes of offspring minus one = 1 degree of freedom). A test of a 1:2:1 ratio would have two Reading Assignment - Genetics

8 degrees of freedom and so on. To help you understand why degrees of freedom are calculated as they are, consider the situation encountered when you put your shoes on. You have two shoes, but only one degree of freedom. This occurs because you have one decision to make, that is, which shoe to put on the first foot. Once that decision is made, right or wrong, the other shoe has to go on the remaining foot; there is no other option. The first decision (or value) can be thought of as arbitrary; the last, as fixed by assigning a value (foot) to the first. After calculating O 2 and determining degrees of freedom, you can determine the probability that your deviation from expected is due to chance alone by consulting a O 2 Table. To use the table, look down the column of degrees of freedom to locate the calculated degrees of freedom for your problem. The values in the row next to your value for degrees of freedom are the O 2 values which have been calculated for that number of degrees of freedom. At the top of the table you will see column headings which indicate the probability that a given O 2 value could result from chance deviation alone. To show you how to use the table, assume you have calculated that O 2 equals 2.56 for a problem where there are two degrees of freedom. 1. Go to the degrees of freedom column and read down it until you find 2". 2. Read across the row for 2 degrees of freedom to determine where your calculated O 2 value of 2.56 would fall. In this example, 2.56 falls between and Read up the columns which contain the numbers and These correspond to p values of 0.5 and 0.2 respectively. 4. This indicates that between 20 and 50% of the time, the deviation you observed in your data could be accounted for by chance alone. Such a deviation is not significant enough to indicate anything other than chance is operating to cause the deviation. In most scientific investigations, p values must be less than 0.05 to be considered significant and less than 0.01 to be considered highly significant. What are the p values for Experiment 1 and Experiment 2 above? How many degrees of freedom are there in each of these experiments? Reading Assignment - Genetics

9 O 2 TABLE Degrees of Freedom p = X 2 ANALYSIS - SAMPLE FORM CLASS OR PHENOTYPE OBS EXP O-E (O-E) 2 (O-E) 2 /E Total: X 2 = Degrees of Freedom = p= Reading Assignment - Genetics

10 IV. WHAT DO WE KNOW ABOUT THE DROSOPHILA LIFE CYCLE? Much of our knowledge of genetics has been gathered from extensive studies of the fruit fly, Drosophila melanogaster. The life cycle of the fruit fly is completed in 10 days at 25 C, the optimal temperature. There are four stages in the life cycle: egg, larva, pupa and adult (see figure below). The duration of each of these stages depends on many environmental factors, especially temperature. The life cycle is completed more quickly at higher temperatures and more slowly at lower temperatures. During the larval stage, there are two molts and three larval periods or instars. Pupation usually occurs on a dry surface. Upon emerging, the adults appear lightly colored with folded wings. Within a short time, they darken, their exoskeletons harden and their wings expand. Males can be distinguished from females because males are usually shorter with the posterior region of the abdomen more blunt. Females have a more pointed abdomen. On the female s dorsal abdominal surface, there are alternating dark and light color bands which extend to the tip of the abdomen. In the male the posterior third of the abdomen is a solid dark color. Female fruit flies can store sperm for relatively long periods of time. Therefore, it is important that females used in genetics experiments be virgins to control for the source of the sperm and/or paternal genotype. Newly emerged males will not mate for 12 hours. This allows us to collect virgin females for mating experiments by removing all flies from breeding vials at time t = 0. Eight hours later, the flies which have hatched are collected and separated by sex. Using this technique, these hatchlings will all be virgins. Reading Assignment - Genetics

11 Drosophila life cycle Reading Assignment - Genetics