2009 Exam 2

2009 Exam 2 - NAME_KEY MIDTERM EXAM II BIOLOGY 231 19...

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NAME: __________ KEY ________________ MIDTERM EXAM II BIOLOGY 231 19 October 2009 ________________________________________________________________________________ -1- GENERAL INSTRUCTIONS: 1. Answer all questions IN INK. 2. NO CALCULATORS. 3. Confine each answer to the BOX or SPACE provided. You are free to use the backs of the pages for scratch work, but ONLY what is written in the box or space will be graded! 4. SHOW YOUR WORK in all calculations and render fractional answers into DECIMAL form. 5. Make sure that your name is PRINTED LEGIBLY on EACH PAGE. 6. This exam consists of 10 problems plus one extra-credit on 14 pages and it is worth a total of 100 points (104 w/ extra credit). 7. Take your time, check your work, good luck! Information that might be useful: ! 1M = 10 mM = 10 ì M; 1V = 10 mV 36 3 ! 10 /10 = 10 ; log(10 ) = X; 10 = X; log(10 /10 ) = log(10 ) x y x-y X log(X) X Y X ° Y ! because of “wobble” pairing between mRNA codons and tRNA anticodons, U in the 5' position of an anticodon can pair with A or G in the mRNA codon, and G in the 5' position of an anticodon can pair with C or U in the mRNA codon. ! for a transmembrane gradient of any solute X, the Gibbs free energy of the gradient is found by: X in out m in out m Ä G = RT ln([X] /[X] ) + zFV = 5.7 kJ/mol ± log([X] /[X] ) + zFV “in” = cytoplasmic; “out” = extracellular or non-cytoplasmic space; z = valence of the ion; F = Faraday constant = 10 kJ/V ± mol = 0.1 kJ/mV ² mol 2 X out in ! the equilibrium potential for an ion, X: V = (59mV/z) ² log([X] /[X] ) Xm X ! for a gradient of a charged molecule, X: Ä G = zF(V ° V ) ! for a proton gradient: Ä G = F( Ä p), where the protonmotive force, Ä p = ø ° (59mV ± Ä pH) m IMS matrix and where ø = V of inner mitochondrial membrane, and Ä pH = (pH ° pH ) ! IMM/OMM = inner/outer mitochondrial membrane; IMS = intermembrane space ! On this date in 1812, French forces under Napoleon Bonaparte began a retreat from Moscow. Having already lost 75% of his troops to combat and sub-zero weather during the invasion, Napoleon ended this retreat with <5% of his original force of 400,000 men alive.
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NAME: __________ KEY ________________ MIDTERM EXAM II BIOLOGY 231 19 October 2009 ________________________________________________________________________________ -2- THE GENETIC CODE (Single letter abbreviations for amino acids) 2nd U C A G 1st 3rd U F S Y C U F S Y C C L S term. A L S W G C L P H R U L P H R C L P Q R A L P Q R G A I T N S U I T N S C I T K R A M (init.) T K R G G V A D G U V A D G C V A E G A V A E G G
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NAME: __________ KEY ________________ MIDTERM EXAM II BIOLOGY 231 19 October 2009 ________________________________________________________________________________ -3- (1) 8 pts. Below are 8 examples of a nucleic acid template (at left) and a possible result of synthesis by polymerase activity on that template (at right). For each, choose whether the enzyme most likely to have generated the product at right from the template at left is: (A) a DNA-dependent DNA polymerase (for replication) (B) a DNA-dependent RNA polymerase (for transcription) (C) RNA primase (makes primer for strand replication) (D) a combination of more than one of these 3 enzymes (E) needs additional enzymes not listed here WRITE the CORRECT LETTER (A-E) for each in the BOX at the far right.
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2009 Exam 2 - NAME_KEY MIDTERM EXAM II BIOLOGY 231 19...

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