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University of Texas, KEY BIO 325 Genetics Second exam. A+ GRADED

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BIO 325 Genetics 2nd exam. Name_____________________¬¬¬¬¬____________ UTEID________________________________ Read each question very carefully and make sure you are answering the quest... ion asked. Please make your answers are extremely clear and as short as possible. Place a circle or box around your final answer. You will not get credit if nothing is circled or boxed. Circle EVERY answer, not just the multiple guess question. You may use a calculator if necessary. If you want to have the possibility of a regrade, write your answers in pen. The test is over at 10:45. GOOD LUCK! --------------------------------------------------- 1. Mendel test-crossed pea plants grown from yellow, round F1 seeds to plants grown from green, wrinkled seeds and obtained the following results: 205 yellow, round; 297 green, round; 298 yellow, wrinkled; and 200 green, wrinkled. If you perform a χ2 test—assume 4 classes. Chi Square table is at end of test. A. Pick single letter gene symbols and write out the genotypes of the crossed plants: F1 X test plant (3 pt) Yy Rr x yy rr B. In Genetics we might perform a χ2 test on this data. In this context, what would a χ2 test tell you. (3 pt) a. Probability the cross actually worked i.e. it was not accidental pollination b. Probability that yellow and round are enzymes encoded by genes c. Probability that yellow and round assort independently d. Probability that green and wrinkled are dominant or recessive C. Calculate the χ2 value (Show your work.) below (5 pt) D. How many degrees of freedom? (2 pt) 4 classes -1 = 3 df preferred, especially since you don’t know which are parentals and recombinants. You’ve only been given the F1 phenotype. E. What is the approximate p value? (3 pt) Way less than 0.001 F. The χ2 test involves a hypothesis. What is this hypothesis called? (2 pt) The null hypothesis. G. Can you reject this hypothesis for Mendel’s experiment? (2 pt) Null can be rejected. H. Are the genes linked? (3 pt) Yes, evidence is that they are linked. I. What is the genetic distance between these genes. (3 pt) Parentals 298+ 297 = 595. Recombinants 205 + 200 = 405. Recombinants/Total = 405 / 1000 = 0.405 = ~40 % or ~40 cM 26 points total Refer to the table below to answer the following questions. The numbers in the boxes are the recombination frequencies between the genes in percent. [Show More]

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