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HW_1 2007 Solutions - M ii Prob L a m =< CSH‘OOS)302...

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Unformatted text preview: M» ii: \ Prob. L ,. a. m =< CSH‘OOS +)302+vMH4oH-vcn..ggNo..3Oo.4o +c° C02+ EHQO Q 8Q: L4: —— B C: 5°20=1+i> Hi IO“+S‘6=I.QQ+QE O: Sol+ 28+ 3: 0.4.O+2%-+E N= “Hal; gem: LI—fl eqlwfihovss For gtkkhaw—s.‘ Q£5°L+2 {5+ 7; 2040+ 99mg] 6 :2 was 0,9355 __ i‘odw $7 = Leg *‘ 2&1 ('Hyk'sw 53:? Hwfil Vr‘ob ‘ r- Yx : (‘FOWMAK we=5¥¢¥ GQQJBI') é “(Fox-NA A way" 06' “‘5’” Wk) :. \ M,“ (an) + (meg +(l4x.13\ +(\6~<.osld 0.2813 (\th’fl +(\‘\05+(\6~‘S3 st YES. : O.S\“l t yeas'i- lg) ardastg, \O"Y§; 231%: X—. X5" {or M X,S=O 5 - 8 So— 3 r Wake—SP ><—x° X 3' \f% (80— O) +XO :: 57““ 0 8‘7 5% (l05fifabwc\ + 88 0+ 2.... . Mul-aq .32....— IMd‘S Hwfil l) rob. 1 P Q ‘ Plot Time (hr) vs. ln(X): Growth C urve Plot Exponential Phase Points ONLY and calculate slope to determine umax Growth Curve Exponential Phase ON LY Alum: 0- ’04- L‘FI Time (hr) . SINCE M‘MMK ik 9.de Mouook's Mokl’ [AF km S Shw'l'k “RX S>>ks,~me ‘ 3K |A5 ol‘ §¥R419L41ypw k:*$ . S’lk‘5r\a~cr7 place is 91M” £90m M‘l’k SWCK So ‘ Hwtfl Prob Q | r \0. Mom); MO‘MV fl_ _ :fiéfi :5 Jr M’Mk > °,>< 2;") +9 1% xm—wow (Jr—+03 M: \‘A )(m " \VN X0 + ~ +0 Grow” 15 s\owcs\- @ \06 Max 6&3?me @461 $.31?” Q‘A.“ 3’}, ovdy S‘Bf'fi‘*fov~6.¥r r Phksc is _ WC“ we»: SCkM , :. ms: MA Q "\’= 2C *QO \/\“$, 5H 3‘75 M: ‘V\%‘ \V\@ _ O 0123 \/\-‘< — . r 4 40 ‘BQ, Mm” N How, w— ”W” —> MUM-:Mms \i S g 4’ \(SW: ”MM-LS " AS A r kS19~Q~O§L . L. 9/2; TV‘S 15 \ocfch. 51‘0me Q, Cm5¥av~+ was\¥431'z Com.‘ + 0.. X: X0 61"! \vx “(MA M: + FOT RVk\1 pg: \k (333,“)0} 3-0.03 \N‘“ _____,_._—-————- 14' r Y3" 6““? M:M; 0.009\~\~“ g4 Mowl‘s Moaxc\" UV: WMMLS '§ \(g: MNLS‘MS k1, + s “1;“— ks .‘s A—kc, sum: -Eh \oo\—L~ Eula f, \LSQ (Md-A Cram \ooiru ka‘s 3m so‘vc 4‘19“ “mm ”Ck ks. NV) M2 Amfig‘"y\|gl __ “Nakgg-‘M 61 M9,, (l a) -. (0.03 w“) ( 12 ,. W W“ (OJ—acOQOMW K O_ \«I:‘ 03 O .00 94¢" ‘MNVFOJOI \mf‘ \ (/1 Grow the bacteria in shake flasks with varying initial levels of substrate. Measure the biomass concentration at early time points and use the Monod model, d)Udt=uX, to get u vs. S data. (The lag phase can be neglected if the bacteria are subcultured first). By taking the inverse of both sides of the other Monod equation, u=umaxS/(KS+S), you get an equation in y=rnx+b form so you can plot l/u vs. US. The slope = Ks/umax and y- intercept = l/umax ...
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