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153A_Lec11

Course: CHEM 153A, Winter 2008
School: UCLA
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25, Carbohydrates Jan. 2008 1 Overview Monosaccharides Ring structures Glycosidic Bond Disaccharides Polysaccharides 2 Sugars are conformationally variable Five- and six-membered hexose rings are more stable Stability depends on interactions between substituents on the ring Haworth formulas are just models and not what the rings really look like naturally Atoms in the rings are sp3 hybridized so...

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25, Carbohydrates Jan. 2008 1 Overview Monosaccharides Ring structures Glycosidic Bond Disaccharides Polysaccharides 2 Sugars are conformationally variable Five- and six-membered hexose rings are more stable Stability depends on interactions between substituents on the ring Haworth formulas are just models and not what the rings really look like naturally Atoms in the rings are sp3 hybridized so tetrahedrally shaped 3 Conformations of a cyclohexane ring Substituents are eclipsed e = equatorial a = axial 4 a and e are interconvertable The two alternative chair conformations of -D-glucopyranose. Predominant conformation Bulky substituents are equatorial e = equatorial a = axial 5 Monosaccharides are reducing sugars 6 Some Hexose Derivatives 7 The reversible oxidation of Lascorbic acid to Ldehydroascorbic acid. 8 9 Glycosidic Bond Links monsaccharides disaccharides polysaccharides Glycosidic bond: anomeric hydroxyl group of a sugar reversibly condenses with an alcohol of another sugar Carb analog of the peptide bond Glycosidases hydrolyze this bond Bond is acid catalyzed Stable in basic and neutral conditions in absence of glycosidase 10 Disaccharides Two monosaccharides linked by a glycosidic bond Several common disaccharides Sucrose, lactose, maltose Described by: Component monosaccharides Ring types Anomeric forms linkage May or may not be reducing 11 Maltose. Reducing Sugar 12 Lactose. Milk sugar (0 to 7% of milk) Reducing sugar because free anomeric C on glucose is present Most Lactase 13 Sucrose abundant Found in plant kingdom Table sugar Not a reducing sugar Hydrolysis leads to a change in optical rotation from dextro to levo and this is referred to as invert sugar Invertase (alpha-D-glucosidase) 14 Homo- and Heteropolysaccharides 15 Structural Polysaccharides Cellulose Primary structural component of plant cell walls Accounts for over half of C in biosphere Predominately in plants Also found in tunicates (invertebrates) Chitin Primary structural component of Exoskeletons: Crustaceans, insects, spiders Cell walls: Fungi, algae Almost as abundant as cellulose 16 The primary structure of cellulose Linear polymer of up to 15,000 D-glucose residues Beta(1-->4) glycosidic bonds No defined size Vertebrates cannot hydrolzye beta(1-->4) glycosidic bond, but microbes and termites can using cellulase 17 Electron micrograph of cellulose fibers The fibers are held together by a matrix of polysaccharides In wood the matrix is called lignin (plasticlike phenolic polymer) Cellulose fibers are held in position by intra- and interchain H-bonds Cotton balls are 100% cellulose 18 Structure of chitin Homopolymer of N-acetyl-D-glucosamine, 1-->4 linkage 19 Storage Polysaccharides Starch Glycogen Plant synthesized Animal synthesized Food reserve More branched than starch Deposited in plant cell cytoplasm Present in all cells but most as insoluble granules prevalent in skeletal muscle and liver Main carb source in human diet -amylose, amylopectin 20
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EE 306 F06A MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet State any assumptions and show all work. (1 point) Print Your Name:BraunPart 1 (12 Points) Part 2A (8 Points) Part 2BC (11 Points) Part 3 (9 Points) Total (42 Points)SOLUTIONSBRAUNNo una
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CHAPTER 22.1 Based upon Table 2.1, a resistivity of 2.6 -cm < 1 m-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 -cm > 105 -cm, and silicon dioxide is an insulator. 2.3 I max 2.4 10-8 cm2 7 A = 10 1 = 500 mA (
Cal Poly - EE - 307
CHAPTER 44.1 (a) VG > VTN corresponds to the inversion region (b) VG < VTN corresponds to the accumulation region (c) VG < VTN corresponds to the depletion region 4.2 (a)" ox -14 3.9o 3.9 8.854x10 F / cm F nF C = = = = 6.91x10-8 2 = 69.1 2 -9 Tox T
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CHAPTER 55.1 Base Contact = B n-type Emitter = D 5.2v BC iB + B + E iE C iCCollector Contact = A n-type Collector = FEmitter Contact = C Active Region = EFor VBE > 0 and VBC = 0, IC = F I B or F =IC 275A = = 68.8 4A IBR =0.5 R = =1 1-
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CHAPTER 33.1(1019 cm-3 )(1018 cm-3 ) = 0.979V NA ND j = VT ln 2 = (0.025V )ln ni 10 20 cm -62( 11.7 8.854 x10-14 F cm-1 ) 2s 1 1 1 1 w do = + 19 -3 + 18 -3 (0.979V) j = -19 10 cm q NA ND 1.602x10 C 10 cm w do = 3.73 x 10-6 cm = 0
Cal Poly - EE - 307
Cal Poly - EE - 307
Cal Poly - EE - 307
Cal Poly - EE - 307
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CHAPTER 77.1' n -14 cm 2 (3.9) 8.854x10 F / cm 3.9o K = nC = n = n = 500 Tox Tox V - sec 10x10-9 m( 100cm / m) " oxox()F A A = 173 x 10-6 2 = 173 2 V - sec V V p ' 200 A A " K 'p = pCox = Kn = 173 2 = 69.1 2 n V V 500 ' Kn =
Cal Poly - EE - 307
CHAPTER 88.1(a) 256Mb = 28 210 210 = 268,435,456 bits (b) 1Gb = 210 = 1,073,741,824 bits8 10 10 28( )( ) (c) 256Mb = 2 (2 )(2 )= 2I pA 1mA = 3.73 28 bit 2 bits( )3| 128kb = 2 7 210 = 217 |( )228 = 211 = 2048 blocks 17 28.28.3(a
Cal Poly - EE - 307
CHAPTER 99.1 Since VREF = -1.25V , and v I = -1.6V , Q1 is off and Q2 is conducting.vC1 = 0 V and vC 2 = - F I EE RC -I EE RC = -(2mA)(350) = -0.700 V9.2 V IC 2 0.995 F I EE = exp BE VBE = 0.025ln = 0.132V IC1 0.005 F I EE VT (a) v I = VR
Cal Poly - EE - 307
CHAPTER 1010.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage ampli
Cal Poly - EE - 307
CHAPTER 1111.1v O = vS iS = v 1M 1k (1000)1k + 0.5 | Av = vO = 990 or 59.9 dB 1M + 5k S | Ai = iO 990 6 = 10 = 9.9x105 or 120 dB iS 1000 vO 5V = = 5.05 mV AV 990 vS 990vS and iO = 1M + 5k 1kAP = Av Ai = 990 9.9x105 = 9.8x108 or 89.9 dB | v S =(
Cal Poly - EE - 307
CHAPTER 1212.1(a) A = 10 20 = 2.00x104 | Av-ideal = 1+A Av = = 1+ A FGE =86150k = 13.5 12k2.00x10 4 = 13.49 4 12k 1+ 2.00x10 162k 1 13.5 -13.49 = 6.75x10-4 or 0.0675% | Note : FGE = 6.75x10-4 A 13.5 2.00x10 4 = 125 1.2k 4 1+ 2.00x1
Cal Poly - EE - 307
CHAPTER 1313.1 Assuming linear operation : vBE = 0.700 + 0.005sin 2000t V 5mV vce = (-1.65V ) sin 2000t = -1.03sin 2000t V 8mV vCE = 5.00 -1.03sin 2000t V ; 10 - 3300IC 0.700 IC 2.82 mA 13.2 Assuming linear region operation : vGS = 3.50 +
Cal Poly - EE - 307
CHAPTER 1414.1 (a) Common-collector Amplifier (npn) (emitter-follower)RIQ1viR1R2+RER3vo-(b) Not a useful circuit because the signal is injected into the drain of the transistor.RI viRDM1+R3vo-R1(c) Common-em
Cal Poly - EE - 307
CHAPTER 1515.1(a) IC= F IE =VCE = VC - (-0.7V ) = 5.87V | Q - Point = (20.7A, 5.87V )1 F 12 - VBE 1 100 12 - 0.7 = = 20.7 A | VC = 12 - 3.3x105 IC = 5.17V 5 2 F + 1 REE 2 101 2.7x10 (b) Add= -g m RC = -40(20.7A)(330k)= -273
Cal Poly - EE - 307
CHAPTER 1616.1 Av (s) = 50 s2 s2 | Amid = 50 | FL (s)= | Poles : - 2,-30 | Zeros : 0,0 (s + 2)(s + 30) (s + 2)(s + 30) s rad | L 30 s (s + 30) | fL = Yes, s = -30 | Av (s) 50 fL = L 30 = 4.77 Hz 2 22 2 1 302 + 22 - 2(0) - 2(0) = 4.79 Hz 2 50
Cal Poly - EE - 307
CHAPTER 1717.1(a) T = A = (b) A = 10Av =80 20|Av =1=5|FGE = 0= 10000 | T = 10000(0.2)= 2000A 10000 100% 100% = = 5.00 | FGE = = = 0.05% 1+ A 1+ 2000 1+ A 2001 A 10 100% (c) T = 10(0.2)= 2 | Av = 1+ A = 1+ 2 = 3.33 | FGE = 1+ 2