# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

4 Pages

### am2

Course: MATH 241, Winter 2004
School: Berkeley
Rating:

Word Count: 1185

#### Document Preview

NAME: PRINT SOLUTIONS Calculus IV [2443002] Midterm II Q1]...[10 points] Consider the double integral 1 0 22y 1y f (x, y ) dx dy Sketch the region of integration. Soln. The limits x = 2 2y and x = 1 y tell us that the region is bounded on the right by the line x + 2y = 2 and on the left by the parabola (left half) y = 1 x2 . The limits y = 0 and y = 1 tell us the upper and lower bounds for this region....

Register Now

#### Unformatted Document Excerpt

Coursehero >> California >> Berkeley >> MATH 241

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
NAME: PRINT SOLUTIONS Calculus IV [2443002] Midterm II Q1]...[10 points] Consider the double integral 1 0 22y 1y f (x, y ) dx dy Sketch the region of integration. Soln. The limits x = 2 2y and x = 1 y tell us that the region is bounded on the right by the line x + 2y = 2 and on the left by the parabola (left half) y = 1 x2 . The limits y = 0 and y = 1 tell us the upper and lower bounds for this region. We see that the parabola and line already intersect at y = 1, so the region is drawn as shown. . .. .. . y y = 1 x2 ................. . . .. .. .. .. . . . . . . . . . . . . . .. .... ........... ....... .......... ....... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... . x + 2y = 2 .. . .. ... x Reverse the order of integration. Soln. Note that reversing the order of integration means building up the region using vertical strips. There are two dierent tops on this region; the parabola top on the left side of the y -axis, and the straight line top on the right side. Thus, we have to divide the region into two pieces along the y -axis. So our answer will be a sum of two iterated integrals as shown. 0 1 0 1x2 2 f (x, y ) dy dx + 0 0 (2x)/2 f (x, y ) dy dx Q2]...[10 points] Consider the following polar coordinates double integral /4 /4 0 sec r3 dr d Sketch the region of integration. Soln. Note that the lines = /4 and = /4 correspond to the cartesian lines y = x and y = x 1 respectively. Also the curve r = sec is just r = cos which rewrites as r cos = 1 or x = 1. [Gotta hate those trig functions!] Thus we get the following region. . .. .. y y = x............... . . . . .. .. .. ... ... ... ... ... ... . . ... ... ... ... .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . x=1 .. . .. ... x y = x Rewrite the integral as an iterated Cartesian coordinates integral (with appropriate limits). You do NOT have to compute this integral. Soln. Remember that dA = rdrd and that the remaining r2 can be written as r2 = x2 + y 2 . Thus we get the following iterated integral. 1 0 x x x2 + y 2 dy dx Q3]...[20 points] Use double integrals in polar coordinates to compute the surface area of the portion of the sphere x2 + y 2 + z 2 = a2 which is above the xy -plane and which lies inside the cone z 2 = x2 + y 2 . Your answer will involve a. Soln. We saw in class that surface area of a portion (over the region R) of the graph of z = f (x, y ) is given by 2 2 fx + fy + 1 dA R In this case we have (by implicit dierentiation of the sphere equation) fx = x/z and (likewise) fy = y/z . We did a sphere example in class! Check out the implicit dierentiation details there! Thus, 2 2 fx + fy + 1 = x2 + y 2 + z 2 a2 =2 z2 z which becomes a2 a2 r2 in polar coordinates. Now the cone equation is z 2 = r2 and this intersects the sphere when r2 + r2 = a2 or r = a/ 2. Thus, the region that we are integrating over in the plane is given by 0 2 and 0 r a/ 2. Filling all this into the area surface integral (and remembering that dA = rdr d) gives 2 0 0 a/ 2 ar dr d 2 r2 a We use the substitution u = a2 r2 (so that rdr = du/2) to evaluate the r integral. Heres what we end up with. 2 2 r 2 ]a/ 2 = (2 )(a2 / 2 + a2 ) = 2a2 (1 1/ 2) []0 [a a 0 Q4]...[20 points] Use the method of Lagrange multipliers to nd the maximum and minimum values of the function f (x, y, z ) = xy + z 2 on the sphere x2 + y 2 + z 2 = 4. Soln. f = y , x, 2z and g = 2x, 2y, 2z so the Lagrange multiplier equations are y x 2z 4 = = = = 2x 2y 2z x2 + y 2 + z 2 So thats it for the calculus. Now we just have to keep our head with all this algebra. First of all, note that the rst two equations tell us that x = 0 precisely when y = 0 (since x and y are multiples of each other). So lets break this analysis into two cases. Case I: [x = 0 and y = 0] In this case the fourth equation becomes z 2 = 4 and so we get z = 2. Thus, we get two points: (0, 0, 2) and (0, 0, 2). Case II: [x = 0 and y = 0] In this case the rst two equations give x/y = 2 = y/x. But this means that 2 must be equal to its own reciprocal (since x/y and y/x are reciprocals) and so must be 1. Thus the 2 third equation becomes z = z which that = 0. Now equation 4 2 implies z becomes 2x = 4 or x = 2. We get four points: ( 2, 2, 0), ( 2, 2, 0), ( 2, 2, 0), and ( 2, 2, 0). Finally, we evaluate f on these 6 points and see that the maximum f -value is 4 (occurs at (0, 0, 2) and (0, 0, 2)), and that the minimum f -value is -2 (occurs at ( 2, 2, 0) and ( 2, 2, 0)). Bonus Question. Let u and v be dierentiable functions of one variable with derivatives denoted by u and v respectively. Let R be the triangular region with vertices at the points (a, a), (b, b) and (b, a). By evaluating the double integral u (x) v (y ) dA R in two dierent ways (as iterated integrals), give a new derivation of the integration by parts formula b a u dv = uv |b a b v du . a Soln. . . .. .. y .. ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... ... ... ... .. ... ... .. .. ... ... ... ... ... ... ... ... .. ... b (b, b) Here is a diagram of the triangular region with its sides labelled. a y=x x=b (a, a) a y=a (b, a) ... . .. .. b x On the one hand, we can integrate with respect to y rst to get b x u (x) v (y ) dA = R a b a u (x)v (y ) dy dx [u (x)v (y )]x dx a u (x)v (x) u (x)v (a) dx v (x)u (x) dx [u(x)v (a)]b a v (x)u (x) dx u(b)v (a) + u(a)v (a) = a b = a b = a b = a On the other hand, we can integrate with respect to x rst to get b b u (x) v (y ) dA = R a b y u (x)v (y ) dx dy [u(x)v (y )]b dy y u(b)v (y ) u(y )v (y ) dy b = a b = a = [u(b)v (y )]b a u(y )v (y ) dy a b = u(b)v (b) u(b)v (a) Finally, setting these two expressions equal to each other gives b a u(y )v (y ) dy a b v (x)u (x) dx u(b)v (a) + u(a)v (a) = u(b)v (b) u(b)v (a) u(y )v (y ) dy a which simplies down to (removing explicit reference to the dummy variables of integration x and y ) b a v du = uv |b a b u dv a and were nished.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Berkeley - MATH - 241
PRINT NAME: SOLUTIONSCalculus IV [2443002] Midterm III 11.1Q1 [15 points]Part 1Write down the change of variables formula for triple integrals.1.2Answer to part 1Suppose the change of variables (x(u, v, w), y (u, v, w), z (u, v, w) takes a region
Berkeley - MATH - 241
PRINT NAME: SOLUTIONSCalculus IV [2443002] Quiz IQ1]. Which one of the four functions listed below has the following level curves?1. 2. 3. 4.f (x, y ) = (x + 1)(y 2). g (x, y ) = (x 1)(y + 2). h(x, y ) = (x + 1)2 (y 2)2 . k (x, y ) = (x 1)2 (y + 2)2 .
Berkeley - MATH - 241
PRINT NAME: SOLUTIONSCalculus IV [2443002] Quiz IIQ1]. State the second derivative test for functions of two variables. Ans: Let (a, b) satisfy fx (a, b) = 0 and fy (a, b) = 0. Dene D(x, y ) = (fxx )(fyy ) (fxy )2 If D(a, b) &gt; 0 and fxx (a, b) &gt; 0, then
Berkeley - MATH - 241
PRINT NAME: SolutionsCalculus IV [2443002] Quiz IIITuesday, April 4, 2000Q1]. Write the following triple integral out as a spherical coordinates triple integral. 3 9x2 9x2 y 2z (x2 + y 2 + z 2 )dzdydx3 00Soln: The region is precisely one quarter o
Berkeley - MATH - 241
Calculus IV [2443004] Midterm IFor full credit, give reasons for all your answers. Q1].[15 points] Draw the level curves f = 0, f = 1, f = 4, and f = 1 for the function f (x, y ) below. Also, sketch the graph of f in a neighborhood of the origin. f (x, y
Berkeley - MATH - 241
Calculus IV [2443004] Midterm IIFor full credit, give reasons for all your answers. Q1].[15 points] For the double integral below, rst sketch the region of integration, and then convert it to a polar coordinares integral. 2 0 2y y 22y y 2f (x, y ) dx
Berkeley - MATH - 241
Calculus IV [2443004] Midterm IIIFor full credit, give reasons for all your answers. Q1].[15 points] Evaluate the following triple integral by rst sketching the region of integration, and then converting it to a spherical coordinates integral. 1 0 1 1y
Victoria AU - CHEMISTRY - 101
NAME _ Student No. _SECTION (circle one): A01 (Codding)A02 (Lee)A03 (Briggs) A04 (Boyer)UNIVERSITY OF VICTORIAVersion AThis test has two parts:CHEMISTRY 101 Mid-Term Test 1, October 17 2008Version APART I is a multiple choice section and is worth
Victoria AU - CHEMISTRY - 101
NAME _KEY_Student No. _SECTION (circle one): A01 (Codding)A02 [morning](Lee) A03 [afternoon](Lee)UNIVERSITY OF VICTORIAVersion AThis test has two parts:CHEMISTRY 101 Mid-Term Test 1, October 16 2009Version APART I is a multiple choice section an
Victoria AU - CHEMISTRY - 101
NAME _KEY_Student No. _SECTION (circle one): A01 (Codding)A02 [morning](Lee) A03 [afternoon](Lee)UNIVERSITY OF VICTORIAVersion BThis test has two parts:CHEMISTRY 101 Mid-Term Test 1, October 16 2009Version BPART I is a multiple choice section an
Victoria AU - CHEMISTRY - 101
NAME _KEY_ Student No. _ SECTION (circle one): A01 (Codding) A02 [morning](Lee) A03 [afternoon](Lee)UNIVERSITY OF VICTORIAVersion AThis test has two parts:CHEMISTRY 101 Mid-Term Test 2, November 20 2009Version APART I is a multiple choice section an
Victoria AU - CHEMISTRY - 101
NAME _ Student No. _SECTION (circle one): A01 (Codding)A02 (Lee)A03 (Briggs) A04 (Boyer)UNIVERSITY OF VICTORIAVersion AThis test has two parts:CHEMISTRY 101 Mid-Term Test 2, November 21 2008Version APART I is a multiple choice section and is wort
Victoria AU - CHEMISTRY - 101
NAME _ Student No. _SECTION (circle one): A01 (Codding)A02 (Lee)A03 (Briggs) A04 (Boyer)UNIVERSITY OF VICTORIAVersion AThis test has two parts:CHEMISTRY 101 Mid-Term Test 2, November 21 2008Version APART I is a multiple choice section and is wort
Victoria AU - CHEMISTRY - 101
NAME _KEY_ Student No. _ SECTION (circle one): A01 (Codding) A02 [morning](Lee) A03 [afternoon](Lee)UNIVERSITY OF VICTORIAVersion BThis test has two parts:CHEMISTRY 101 Mid-Term Test 2, November 20 2009Version BPART I is a multiple choice section an
Maryland - MATH - 111
Sample Midterm Questions Stat111 Summer 2008NOTE: This sample midterm is shorter than the actual midterm will be. Question 1 Do IQ scores of Self-concept affect GPA? Based on 102 eighth-grade students in a rural Midwestern school with their IQ score, Sel
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem 130/MCB 100A, Fall 2006, Lecture 2 We will now take a brief diversion to review the basic kinds of molecular interactio
Berkeley - MCB - MCB100A
Restricted: For students enrolled in CHEM 130/MCB 100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley CHEM 130/MCB 100A. Lecture 3. Fall 2006. The standard (Watson-Crick) structure of DNA is called Bform:The grooves are def
Berkeley - MCB - MCB100A
Restricted: For students enrolled in CHEM 130/MCB 100A, UC Berkeley, Fall 2006 ONLYPage 1John Kuriyan: University of California, Berkeley Lecture 4. Fall 2006.1Restricted: For students enrolled in CHEM 130/MCB 100A, UC Berkeley, Fall 2006 ONLYPage 2
Berkeley - MCB - MCB100A
Purication of Proteins Chem130/MCB100A Fall 2006 University of California, BerkeleyOutline: Proteins have different physical properties These properties can be used to isolate populations of specific proteins from one anotherSource: Matthew YoungProtei
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem 130/MCB 100, Lecture 6, Fall 2006. Stretches of hydrophobic residues in soluble proteins are short (7-10 residues). Thi
Berkeley - MCB - MCB100A
Restricted: Lecture 7. For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYPROTEIN SEQUENCE DETERMINES STRUCTURE The sequence of a protein allows it to adopt its unique fold s pontaneously: n o t emplates o r g uides a re needed.How do
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem 130/MCB 100A, Fall 2006, Lecture 8 SEQUENCE IDENTITY &amp; STRUCTURAL VARIATION The level of sequence identity between two
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2005 ONLYJohn Kuriyan: University of California, Berkeley Chem 130/MCB 100A, Fall 2006, Lecture 9 Amino acid substitution score: Sij = 2 log2 Lij Here i and j are two amino acids, e.
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2005 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2005, Lecture 17 FREE ENERGY To review the concept of free energy:We are interested in determining
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 17Equilibrium ConstantsTo review: the chemical potential, , is an intensive propert
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 18The net result is that ATP is produced due to the high concentration of B outside.
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 19 Dissociation Constants/Association Constants In the last lecture we considered the
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 20 Review of Dissociation Constants/Association Constants In the last lecture we cons
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY.Lecture 21John Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 21 (November 16) Allostery In the last lecture we saw that if the binding
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 22 Stability of DNA basepairing.DNA polymerase make 1 error in 107108 in corporation
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLYJohn Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 23 Chemical Kinetics Definition of rates The rate of a chemical reaction is given by
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY Lecture 24John Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 24 Michaelis-Menten Kinetics When we study an enzyme reaction we monitor t
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY Lecture 25John Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 25 Electrochemistry Redox reactions involving metals or organic compounds
Berkeley - MCB - MCB100A
Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY Lecture 26John Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 26 Review redox/potentialsFor such an electrochemical cell, under standar
Berkeley - MCB - 100A
Berkeley - MCB - 100A
Berkeley - MCB - 100A
October 24, 2006 Corrections to Midterm # 2 solution: Q1B.2: There was a numerical error in the solution p( A) e 1, A 2, A 3, A e (0+1+3) e 4 = = = = 7.42 p( B) e (u1,B +u2,B +u3,B ) e (1+ 4+1) e 6) (u +u +uQ3B.3: There was a numerical error in the solu
Berkeley - MCB - 100A
Berkeley - MCB - 100A
UC Berkeley, CHEM C130/MCBC100A. FALL 2005. MID-TERM EXAM 3. YOUR NAME:_UNIVERSITY OF CALIFORNIA, BERKELEY CHEM C130/MCB C100A MIDTERM EXAMINATION #3. November 17, 2005 INSTRUCTOR: John Kuriyan THE TIME LIMIT FOR THIS EXAMINATION IS 1 HOUR AND 20 MINUTE
Berkeley - MCB - 100A
1. Q1B, unit for delSo should be KJ.mol^-1.K-1 3. Q4A, there is a sign error, the answer should read 1.52 kJ/ mol. It makes sense that way as you have positive entropy of mixing when you have both components rather than pure components. The mistake is
Berkeley - MCB - 100A
Berkeley - MCB - 100A
Berkeley - MCB - 100A
Berkeley - MCB - 110
MCB 110:Biochemistry of the Central Dogma of MBPar t 1. DNA r eplication, r epair and genomics (Pr of. Alber )Par t 2. RN A &amp; p r otein synthesis. Pr of. ZhouPar t 3. M embr anes, pr otein secr etion, tr affick ing and signaling Pr of. N ogalesMCB 110
Berkeley - MCB - 110
Patterns and principles of RNA structureR NAstructurecanbespecific,stableandcomplex. (Asaresult,RNAmediatesspecificrecognition andcatalyticreactions.) Principles/ideasRNAscontaincharacteristic2 and3 motifs Secondarystructurestems,bulges&amp;loops Coaxialstac
Berkeley - MCB - 110
DNA polymerase Summary1. DNA replication is semi-conservative. 2. DNA polymerase enzymes are specialized for different functions. 3. DNA pol I has 3 activities: polymerase, 3-&gt;5 exonuclease &amp; 5-&gt;3 exonuclease. 4. DNA polymerase structures are conserved.
Berkeley - MCB - 110
DNA polymerase summary1. D N A r eplication is semi-conser vative. 2. D N A polymer ase enzymes ar e specialized for differ ent f unctions. 3. D N A pol I has 3 activities: polymer ase, 3-&gt;5 exonuclease &amp; 5-&gt;3 exonuclease. 4. D N A polymer ase str uctur
Berkeley - MCB - 110
Accessory factors summary1. DNApolymerasecantreplicateagenome. Solution Nosinglestrandedtemplate Helicase Thesstemplateisunstable SSB(RPA(euks) Noprimer Primase No3&gt;5polymerase Replicationfork Tooslowanddistributive SSBandslidingclamp Slidingclampcantget
Berkeley - MCB - 110
DNA methods summary1. RestrictionenzymescutatspecificDNAsites.(N) 2. Vectorsallowgenestobeclonedandproteins expressed.(N) 3. GelelectrophoresisseparatesDNAonthebasisof size. 1. DNAscanbesynthesized(upto~100bases commercially).(N) 1. PCRamplifiesanytarget
Berkeley - MCB - 110
DNA packaging summary1. Problemispackaging 2. Levelsofchromatinstructure(nucleosomes, 30nmfiber,loops,bands) 3. Histonecodemarksactiveandinactive sequences 4. DNAelementsforchromosomestructure include(ARS),TELandCEN. 5. CENpromotestheassemblyofthe kineto
Berkeley - MCB - 110
Genomes summary1. 2. 3. 4. 5. 6. 7. &gt;930bacterialgenomessequenced. Circular.Genesdenselypacked. 210Mbases,4707,000genes Genomesof&gt;200eukaryotes(45higher)sequenced. Linearchromosomes Onaverage, ~50%ofgenefunctionsknown. Humangenome:&lt;40,000genescodefor&gt;120
Berkeley - MCB - 110
Genomics and bioinformatics summary1. Gene finding: computer searches, cDNAs, ESTs, 1. Microarrays 2. Use BLAST to find homologous sequences 3. Multiple sequence alignments (MSAs) 4. Trees quantify sequence and evolutionary relationships 5. Protein seque
Berkeley - MCB - 110
Human disease genes summary1. Goals: discover the basis for disease, understand key processes, and develop diagnostics and cures. 2. Finding human disease genes - OMIM 3. Sickle Cell Anemia 4. Inheritance and linkage 5. RFLPs and chromosome &quot;walking&quot; 6.
Berkeley - MCB - 110
DNA damage and repair summary1. Defects in repair cause disease 2. Common types of DNA damage 3. DNA repair pathways Direct enzymatic repair Base excision repair Nucleotide excision repair Mismatch repair Double-strand break repair Non-homologous end joi
Berkeley - MCB - 100B
MCB 100; Fall 2005 Problem Set I Energetics and ATP 1. The structure of ATP explains why its hydrolysis has such a large negative free energy. 2. It has been argued that the use of ATP rather than Pi in the phosphorylation of glucose to G6P is required fo
Berkeley - MCB - 100B
Problem Set- 2-Glycolysis 1. A colleague proposes to get more energy from glycolysis by engineering hexokinase to first allow the attack of glucose on the second P of ATP and then split off one of the Pi of the glucose pyrophosphate. Show in detail the re
Berkeley - MCB - 100B
Problem Set III, Oxidation/Reduction 1. A Washington researcher, distracted by dreams of the Rose Bowl, proposes to increase the yield of energy from glycolysis by engineering cells to reduce pyruvate to lactate with Washox, a new compound that has an E'
Berkeley - MCB - 100B
Problem Set-4Krebs Cycle1. A biochemist wishes to measure the amount of Pi in a solution. He selects an enzyme from the Krebs cycle to add to that solution because, he says, it will use the Pi, and the level of the product, which he can measure, could be
Berkeley - MCB - 100B
Problem Set 5 Lipids/Membranes 1. Botulinum toxin is a multisubunit protein secreted by pathogenic strains of the bacterium, Clostridium botulinum, and is known to bind to a specific ganglioside, GT1B, on the surface of mammalian cells. (a) Describe the g
Berkeley - MCB - 100B
Problem Set 6 Electron Transport/Oxidative Phosphorylation1. A colleague studying oxidative phosphorylation uses isolated, functioning mitochondria and continuously washes the intermembrane space with neutral buffer. He demonstrates that this effectively
Berkeley - MCB - 100B
Figure 13-15 G ly c e r a te /g ly c e r a ld e h y d eEo = -0.55 V N A D -&gt; N A D HEo = -0.32 V G ly c e r a te /g ly c e r a ld e h y d e s tr o n g e r r e d u c in g - r e a c tio n p r o d u c e s N A D H P y r u v a te / la c ta teEo = -0.19