Corn Genetics Lab

Corn Genetics Lab Notes From the teacher Day 1: Before class: • Review the lab. • Complete the pre-lab. o Title and date of the lab (remember to add this lab to your table of contents). o Purpose → 1-2 sentences describing the overall goal of the lab; use complete sentences o Pre-Lab Questions→ you need to number the question, rewrite the question, and then answer it for full credit o Lab Procedure and Data Tables → ▪ Do a SUBTITLE for PART A, write the procedure (numbered list) and cut out and tape in Data Tables 2 & 3. Leave room for all the analysis questions for Part A (you will be doing calculations here too) ▪ Do a SUBTITLE for PART B, write the procedure (numbered list) and cut out and tape in Data Table 4. Leave room for all the analysis questions for Part B (you will also be doing calculations here) ▪ Do a SUBTITLE for PART C, write the procedure, and cut out and tape in Data Tables 5, 6, &7. Make sure you have room for the analysis questions. In class: • Complete the Lab CORN GENETICS LAB LEARNING OBJECTIVES In this laboratory, you will • Use corn to study genetic crosses • Recognize contrasting phenotypes • Collect data from F1 ears of corn, and analyze the results from monohybrid and dihybrid crosses • Study dominance, segregation, and independent assortment of alleles in corn • Compare predicted results with results obtained from actual data At the completion of this laboratory, you should be able to • Collect and organize data from genetic crosses • Analyze the data using Chi-square analysis techniques • Predict patterns of inheritance given relevant data BACKGROUND Mendelian Genetics and Corn Mendelian traits refer to phenotypical features whose pattern of inheritance follows Mendel’s theories about the inheritance of traits. Corn — a diploid organism — has been widely used to study and illustrate Mendelian traits. In this lab, we are going to look at two traits: kernel color and kernel texture. In corn, the dominant gene R determines the presence of colored aleurone. Aleurone is a protein found in maturing seeds that can give a kernel its color. Individuals possessing one copy of the dominant gene (R) will exhibit purple kernels. Recessive phenotypes (rr) result in yellow kernels. Also, in corn, the dominant gene T produces the endosperm phenotype smooth. Smooth kernels appear hard and starchy. The recessive phenotype produces wrinkled kernels, which are shrunken in appearance. You will see both monohybrid (single trait) and dihybrid (two traits) crosses of the color and texture traits. We will examine the mode of inheritance (HOW it is inherited) of these two genes by looking at the offspring of a fertilization event between a male corn flower and a female corn flower. Each kernel is the progeny (offspring) of one of such fertilization events. An ear of corn shows the results of hundreds of such events. By counting the different kernel types, we should be able to determine their pattern of inheritance. For instance if you observed an ear of corn where 100% of the kernels were yellow what could you deduce about the parents? Think about it. Yellow is a recessive trait so therefore the parents must be yellow. How about an ear that exhibited 300 smooth kernels and 100 wrinkled kernels? This 3:1 ratio is indicative of a Tt x Tt cross. Probability and the Chi Square Analysis Genetics, like gambling, deals with probabilities. When you flip a coin, you have the same chance of getting a head as a tail: a one-to-one ratio. That does not mean that if you flip a coin 100 times you will always get 50 heads and 50 tails. You might get 53 heads and 47 tails. That is probably close enough to a one-to-one ratio that we would accept it without a second thought. But what if you got 61 heads and 39 tails? At what point do you begin to suspect that something other than chance is at work in determining the fall of your coin? The probabilities calculated in a Punnett Square assume that chance (and chance only) has been operating in the assortment and recombination of alleles that gave rise to that offspring. Thus, any variation of the observed results from the expected results are due to chance. This is known as the null hypothesis. Therefore, if you ACCEPT your null hypothesis, you are saying that the genes are showing typical Mendelian inheritance and any difference between the observed and expected values is simply due to chance. If you REJECT your null hypothesis, you are saying that the genes do NOT show typical Mendelian inheritance and therefore the genes are linked. Linked genes are genes that are located on the same chromosome and therefore are inherited together. So, to determine if your data is due to chance, and chance alone, perform a chi square analysis test to see if the null hypothesis is accepted or rejected. The Chi-square test, x2, is a statistical test used to determine how well observed ratios fit expected ratios. The x2 value consists of the summation of these values for all phenotypes. X2 = (Observed – Expected)2 Expected Table 1: Chi-Square Values and Probabilities Probability Degrees of Freedom (df) 1 2 3 4 5 0.05 3.84 5.99 7.82 9.49 11.1 0.01 6.64 9.21 11.3 13.2 15.1 0.001 10.8 13.8 16.3 18.5 20.5 Remember that degrees of freedom (df) is n-1. Monohybrid cross df = 1 Dihybrid cross df = 3 --------------------------------------------------------------------------------------------------------------------------------------PRELAB QUESTIONS: 1. What organism will we be using to study genetics in this activity? 2. What traits will we be examining? 3. What is the difference between a monohybrid and dihybrid cross? 4. Two F1 parents are crossed: Rr x rr. Predict the offspring. (Do a Punnett Square) 5. The following offspring are recorded: 52 smooth, 48 wrinkled. Determine the parental genotypes. 6. How can you test if your data is due to the chance assortment of alleles during meiosis? 7. You collected the following data: 433 kernels, including 334 purple and 99 yellow. This ear of corn was produced by monohybrid cross (Rr x Rr). Do your expected results match your observed results? Calculate a chi square value and determine the probability. (df = 1) 8. How can you tell if genes are linked or not? 9. Why are Drosophila often used in labs for genetics experiments? 10. What is a “wild type” fly? --------------------------------------------------------------------------------------------------------------------------------------PART A: Analysis of Ear A & B MATERIALS → PART A: Corn Ear A Corn Ear B PROCEDURE → PART A: 1. Obtain an ear of corn labeled A. 2. Determine whether it is a monohybrid (one trait) or dihybrid (two traits) cross. 3. In Table 2, record one phenotype in each column. 4. Count 10 rows of kernels and record how many kernels express each phenotype. 5. Put your data into the Class Data Chart on Mrs. Willis’ computer. 6. Once all groups have recorded their data, fill in the class data portion of Table 2. Table 2. Corn Ear A Data Number of kernels with phenotype: Number of kernels with phenotype: Total number of kernels: Your Data (OBSRVED) Class Data (OBSERVED) Class Data (EXPECTED) 7. Obtain an ear of corn labeled B. 8. Repeat steps #2-6 above and record your data in Table 3. Table 3. Corn Ear B Data Number of kernels with phenotype: Number of kernels with phenotype: Total number of kernels: Your Data (OBSRVED) Class Data (OBSERVED) Class Data (EXPECTED) ANALYSIS QUESTIONS → PART A: Use THE CLASS DATA for your analysis! 1. Calculate the percent of each phenotype present on Ear A. 2. Determine the parental phenotypes and genotypes based on the data for Ear A. 3. Make and solve a Punnett Square for the parental cross to determine the expected offspring for Ear A. 4. Based on your total number of kernels for Ear A, and the expected ratios you found in your Punnett square, what are you expected numbers for each phenotype on Ear A? Fill these numbers in on Table 2. 5. Using your expected numbers you calculated in #4, and the observed from your data table, do a chi square test to determine if the observed results match the expected results for Ear A. What is your chi square value? Show your work in your lab notebook. 6. What is the critical value of p for this chi square test? 7. What is your null hypothesis for Part A, Ear A (this is also your null for Ear B)? 8. Based on your chi square value, would you accept or reject the null hypothesis? 9. Calculate the percent of each phenotype present on Ear B. 10. Determine the parental phenotypes and genotypes based on the data for Ear B. 11. Make and solve a Punnett Square for the parental cross to determine the expected offspring for Ear B. 12. Based on your total number of kernels for Ear B, and the expected ratios you found in your Punnett square, what are you expected numbers for each phenotype on Ear B? Fill these numbers in on Table 3. 13. Using your expected numbers you calculated in #12, and the observed from your data table, do a chi square test to determine if the observed results match the expected results for Ear B. What is your chi square value? Show your work in your lab notebook. 14. What is the critical value of p for this chi square test? 15. Based on your chi square value, would you accept or reject the null hypothesis? --------------------------------------------------------------------------------------------------------------------------------------PART B: Analysis of Ear C MATERIALS → PART B: Corn Ear C PROCEDURE → PART B: 1. Obtain an ear of corn labeled C. This corn was created by crossing two individuals heterozygous for both traits. 2. Determine whether it is a monohybrid (one trait) or dihybrid (two traits) cross. 3. In Table 4, record one phenotype in each column. (You will have 4 phenotypes here) 4. Count ALL of the kernels and record how many kernels express each phenotype (you can ignore the ones under the sticker that you cannot see). Fill in the OBSERVED DATA in Table 4. Table 4. Corn Ear C Data Number of kernels with phenotype: Number of kernels with phenotype: Number of kernels with phenotype: Number of kernels with phenotype: Total number of kernels: Your data OBSERVED DATA EXPECTED DATA ANALYSIS QUESTIONS → PART B: Use YOUR DATA for your analysis! 1. Make and solve a Punnett Square for the parental cross to determine the expected offspring. HINT: look back at Step 1 for the parents genotypes. 2. Using your total number of kernels and the expected ratios you calculated in #1, what are the expected values for each of the 4 phenotypes? Record these values in Table 4 under EXPECTED DATA. 3. Using your expected values from #2 and your observed values from Table 4, perform a chi square analysis test. Make sure to show all work in your lab notebook. What is your chi square value? 4. What is your null hypothesis for Part B, Ear C, of this lab? 5. What is your p value for a dihybrid cross (df = 3)? 6. Would you accept or reject your null hypothesis and why? 7. Are the color and texture genes linked? --------------------------------------------------------------------------------------------------------------------------------------PART C: Analysis of Drosophila Drosophila melanogaster, the common fruit fly, is a model organism for genetic studies. You will often be asked questions dealing with this organism. The reason it is so widely used is because it is easily cultured in the lab, has a short generation time, can produce many offspring, and has traits that can easily be seen. In fruit fly genetics, the normal fly is called a "wild type" and any fly exhibiting a phenotypic mutation is called a "mutant". Mutant flies are given names that generally denote the type of mutation the fly exhibits. For example, the mutant "ebony" has a much darker body than the wild type fly. • • Mutant (w) – ebony body color Wild Type (W) – normal body color PROCEDURE → PART C: 1. Answer all the ANALYSIS QUESTIONS for this section of the lab. Make sure to show all calculations. BACKGROUND INFORMATION → PART C: Assume you identify 100 flies and record the following data for the offspring of an unknown cross involving a single trait. Body color in Drosophila is found on chromosome 3 (it is not x-linked ➔ so you do not have to take into account males vs. females and can just use the TOTAL numbers to do your chi square. If it WAS xlinked, we would expect very different numbers between males and female flies). Table 5. Fruit Fly Data – Unknown Cross MALE FEMALE TOTAL EXPECTED OBSERVED Wild Type 40 33 73 Mutant 15 12 27 ANALYSIS QUESTIONS → PART C: 1. Assuming the parents are both heterozygous for body color, make and solve a Punnett Square for the parental cross to determine the expected results. 2. Using the totals from Table 5 and the expected ratios from #1, what are the expected numbers for each phenotype? Record those figures in Table 5. 3. Complete a chi square analysis to test the null hypothesis. Make sure to show all your work. What is your chi square value? Do you accept or reject the null? You identify 532 flies and record the following data for the offspring of a cross involving two traits. Parent one was heterozygous for both traits. Parent two was mutant for both traits. • • • • Mutant (w) – ebony body color Wild Type (W) – normal body color Mutant (a) – short wings Wild Type (A) – long wings Table 6. Fruit Fly Data – Dihybrid Heterozygote crossed with Dihybrid Mutant OBSERVED DATA EXPECTED DATA Wild type color and mutant wings Mutant color and wild type wings Wild type for both traits 87 101 298 Mutant for TOTAL both traits 46 4. Determine the genotypes for both parents. 5. Make and solve a Punnett Square for the parental cross to determine the expected results. Find the ratio and then determine the expected values for each genotype. Fill in Table 6. 6. Use the chi square test to determine if color and wings are linked genes. Show all your work and make sure to include your chi square value, whether you accepted or rejected the null, and if the genes are linked or not. 7. The percent recombination between two genes indicates the map distance between them. 1% recombination = 1 map unit (m.u.). To determine the map distance between a pair of loci, complete Table 7: Table 7. Calculating Map Units Separating Body Color and Wing Length Genes Answer the Question or SHOW YOUR WORK Here Identify the PARENTAL PHENOTYPES: (What is the phenotype of each parent? There will be 2) Identify the RECOMBINANT PHENOTYPES: (What are the other 2 phenotypes that are not expressed by either parent? There will be 2) Count the # of recombination events (# of offspring that show a recombinant phenotype) Calculate the recombination frequency: (# recombinants/ total # of offspring) x 100 So, how many map units separate the body color gene and the wing length gene on the chromosomes of Fruit Flies?