Midterm2AKey

Midterm2AKey - BIS101-001 Simon Chan Winter 2008 Midterm 2...

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BIS101-001 Simon Chan Winter 2008 Midterm 2 Version A Key 1. (6 pts) Are the following components involved in the processes indicated? Answer Yes or No. DNA Packaging Transcription Ribosome ____No ______ _____ No ________ Reverse Transcriptase ____No________ _____ No _______ TATA binding protein _____ No ________ _____ Yes ________ Ligase ____No ______ _____ No ________ Deoxyribonucleic Acid ____Yes_________ ____Yes_________ Histone ____Yes ______ ____Yes_________ 2. (12 pts) The following DNA fragment is a piece of the coding strand of an E. coli gene. 3’ A T C G T T A G C T T T T G C A C A A A T T G A G G G A C T G A T G 5’ a. Draw the mRNA corresponding to this segment, assuming this whole section is transcribed. Be sure to label the 5’ and 3’ ends. (4 pts) 3’ A U C G U U A G C U U U U G C A C A A A U U G A G G G A C U G A U G 5’ b. Draw the corresponding section of protein coded for by this sequence. Label the NH 3 and COOH sides. (See the genetic code on the front page of the exam) (4 pts) NH 3 - val-val-arg-glu-leu-asn-thr-phe-ser-ile-ala…COOH c. If there was a substitution mutation changing the 3 rd C of the coding strand to an A, draw the corresponding section of this protein. (2 pts) NH 3 - val-val-arg-glu-leu-asn-lys-phe-ser-ile-ala…COOH (-1 for each incorrect item) d. The codon 5’-GUC-3’ is recognized by the anticodon 3’-CAG-5’. What is another codon that can be recognized by this anticodon? (2 pts)
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The anticodon 3’-CAG-5’ can also recognize the codon 5’-GUU’, because G in the third position can pair with U. 3. (16pts) In Neurospora, the biosynthetic pathway for methionine is the following with the enzymes carrying out the reactions indicated by numbers above the arrows: 5/6 serine cysteine ------------------------- 1 4 2 3 cystathionine homocysteine methionine homoserine succinylhomoserine ---- a) Predict the growth of mutants of these enzymes with the various compounds by filling out completely the table below using a + if the mutant will grow and a – if it will not. (12pts) methionine serine cystathionine homoserine cysteine succinylhomoserine homocysteine 1 + - + - - - + 2 + - - - - - - 3 + - + - - + + 4 + - - - - - + 5 + - + - + - + 6 + - + - + - + b) Mutant 5 and 6 look the same; you mate the two mutants and the resultant F1 progeny have a wild- type phenotype (prototrophy for methionine). What does this mean and is this consistent with the one gene-one enzyme hypothesis? Explain your answer. (4 pts) Mutant 5 and 6 are mutations in two different genes both of which are required for converting serine to cysteine. Presumably two different polypeptides are required for this enzymatic activity
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Midterm2AKey - BIS101-001 Simon Chan Winter 2008 Midterm 2...

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