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# Exam2Key - BCB 444/544 Fall 2005 Exam 2 Key Friday October...

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BCB 444/544 Fall 2005 Exam 2 Key Friday, October 28 This closed-book, closed-notes 50-minute test consists of 7 questions. The number of points for each problem is indicated below. Read all questions carefully before starting. Write all your answers on the space provided in the exam paper. Remember that concise answers are preferable to wordy ones. Clearly state any simplifying assumptions you make in solving a problem. The last question is only for students enrolled in BCB 544. Name: 1 2 3 4 5 6 7 total 14 18 18 18 16 16 18 100 or 118

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1. (14 points) HMMs Consider the occasionally dishonest casino example seen in class. The system has three states: B denotes the start state, F denotes the state where a fair die is used, and L denotes the state where a loaded die is used. The transition probabilities between these states are shown below. 0.5 F 0.01 0.2 L 0.8 0.99 B 0.5 The emission probabilities for state F are e F (1) = e F (2) = ··· = e F (6) = 1 / 6 , while for state L they are e L (1) = e L (2) = = e L (5) = 0 . 1 and e L (6) = 0 . 5 . (a) (7 points) What is the most probable sequence of states, starting from state B , to pro- duce the sequence of die tosses h 1 , 6 i ? Show your work. 1 6 B 1 0 0 F 0 1 2 × 1 6 = 1 12 max 1 12 × 0 . 99 × 1 6 = 0 . 0138 1 20 × 0 . 2 × 1 6 = 0 . 001667 ± = 0 . 0138 L 0 1 2 × 1 10 = 1 20 max 1 12 × 0 . 01 × 1 2 = 0 . 00042 1 20 × 0 . 8 × 1 2 = 0 . 02 ± = 0 . 02 Answer. The most probable sequence of states is B L L (b) (7 points) What is the total probability of the sequence h 1 , 6 i ? Show your work. 1 6 B 1 0 0 F 0 1 2 × 1 6 = 1 12 1 12 × 0 . 99 × 1 6 = 0 . 0138 1 20 × 0 . 2 × 1 6 = 0 . 001667 ± = 0 . 015467 L 0 1 2 × 1 10 = 1 20 1 12 × 0 . 01 × 1 2 = 0 . 00042 1 20 × 0 . 8 × 1 2 = 0 . 02 ± = 0 . 02042 Answer. Total probability of the sequence h 1 , 6 i = 0 . 015467 + 0 . 02042 = 0 . 035887 2
2. (18 points) Proﬁle HMMs For the proﬁle Markov model shown below, indicate three different ways in which the se- quence ATG could have been emitted, along with their respective probabilities. For ease of reference, assume that the match states are labeled M 1 ,M 2 3 , that the deletion states are labeled D 1 ,D 2 3 , and that the insertion states are labeled I 1 ,I 2 3 4 . Answer. Here are 5 possible paths; you only needed thre e : P ( B M 1 M 2 M 3 E ) = 0 . 7 × 0 . 1 × 0 . 7 × 0 . 2 × 0 . 7 × 0 . 4 × 0 . 9 = 0 . 0024696 P ( B M 1 I 2 M 2 D 3 E ) = 0 . 7 × 0 . 1 × 0 . 1 × 0 . 25 × 1 × 0 . 1 × 0 . 2 × 1 × 1 = 0 . 000035 P ( B M 1 D 2 M 3 I 4 E ) = 0 . 7 × 0 . 1 × 0 . 2 × 1 × × 1 × 0 . 1 × 0 . 1 × 0 . 25 × 1 = 0 . 000035 P ( B I 1 M 1 M 2 D 3 E ) = 0 . 1 × 0 . 25 × 1 × 0 . 5 × 0 . 7 × 0 . 2 × 0 . 2 × 1 × 1 = 0 . 00035 P ( B I 1 M 1 →→ D 2 M 3 E ) = 0 . 1 × 0 . 25 × 1 × 0 . 5 × 0 . 2 × 1 × 1 × 0 . 4 × 0 . 9 = 0 . 0009 3

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3. (18 points) BLAST and FASTA (a) (6 points) Explain why neither FASTA nor BLAST are guaranteed to return a sequence that has maximum similarity score with respect to the query sequence.
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Exam2Key - BCB 444/544 Fall 2005 Exam 2 Key Friday October...

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