BIOE 349 Metabolic Engineering - Metabolic Engineering: Homework 8. Q&A 100%.

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Metabolic Engineering: Homework 8 (8 points) Assigned: Tuesday 11/19/19 Due: Tuesday 11/26/19 BioE 349/449: all 1. (2 points) Short answer a) (1.5) Mass spectra: Using the NIST website (http://webbook.nist.gov/chemistry/name-ser.html), show the mass spectrum for adenosine. On the spectrum, label the molecular ion, the base peak, and three fragment ions.

The molecular ion is the peak at m/z = 267, which is the molecular weight of adenosine. The base peak is the peak with the highest intensity. The fragment ions are all the other peaks.

b) (0.5) Name one advantage of using MFA with isotopes. What are some general precautions that are needed when handling these compounds? An advantage is that allows one to determine unknown fluxes and flux distributions in non-linear pathways, such as those that are branched or contain cycles. When handling radioactive compounds, one needs appropriate personal protective equipment, training, equipment segregation, and waste management.

2. (3 pts) Metabolic flux analysis in an overdetermined system Growth, product formation, and catabolism in yeast are described in the five reactions below: Reaction

Description

Rate

1 CH1.8O 0.58N0.16 + 0.105CO2 + 0.355NADH - 1.105CH2O - 0.16NH3 - 2 ATP - 0.231 NADPH = 0

cell growth



2 CO2 + 2NADPH – CH2O = 0

pentose phosphate

PP

3 0.5CO2 + CH3O0.5 + 0.5ATP – 1.5CH2O = 0

ethanol production

ret

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4 CH8/3O - 0.333NADH - 0.333ATP - CH 2O = 0

glycerol production

rgly

5 -ATP = 0

ATP consumption for maintenance

mATP

Notes: glucose = CH2O; ethanol = CH3O0.5; glycerol= CH8/3O. The intracellular products are ATP, NADH, and NADPH.

a) (1pt)The rates of biomass formation, ethanol production, and glycerol production are measured.

The results for the measured rates are: =0.07 h-1, reth=0.03 C-mol eth/(C-mol X - h), and rgly=0.07 C-mol gly/(C-mol X - h). Determine: 1. PP and mATP

−2 0.355 −0.231 1−m 0 0 2 2−c G= 0.5 0 0 3−m −0.333−0.333 0 4−m 1 0 0 5−c

[

G m=

]

[

−2 0.355 −0.231 1 3 0.5 0 0 4 −0.333 −0.333 0

]

Gc = 0 0 2 2 −1 0 0 5

[

]

μ 0.07 v m = r eth = 0.03 0.07 r gly

( )( )

Calculate vc using: vc=-pinv(GcT)*GmT*vm

v c = ν 2 = ν ppp = 0.0081 −0.1483 ν5 m ATP

()( )(

)

2. The total rate of glucose consumption =-1.105* - 2 - 1.5reth-rgly=(-1.105*0.07)-0.0081-(1.5*0.03)-0.07 =-0.200 The total rate of glucose consumption is 0.2 C-mol glucose/(C-mol X - h) b) (1pt)Express the system using Stephanopoulos Equation (8.13). Again, the rates of biomass, ethanol, and glycerol are measured and are the same as they are in Part (a). Partition T into T 11, T12, T21, T22. Partition v into v1 and v2.

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Rearrange G so that all the measured components are in the first rows, and are followed by all the calculated components.

−2 0.355 −0.231 1−m 0.5 0 0 3−m G new = −0.333−0.333 0 4−m 0 0 2 2−c 1 0 0 5−c

[

]

Construct T: The bottom rows contain GnewT. The dimension of I is the number of measured rates.

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 T= −2 0.5 −0.333 0 −1 0.355 0 −0.333 0 0 −0.231 0 0 2 0

[

]

Draw the horizontal line such that T21 and T22 contains GnewT and T11 and T12 contain I and 0.

Therefore, T22 needs to be 3x3.

μ r eth v1 v = m = r gly v2 vc ν PPP m ATP

()( )

()

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Partition

( vv ) 1

using: v2 = -T22-1T21v1. Since T22 is 3x3 and T21 is 3x2, then v1 is 2x1 and v2 is 3x1.

2

c) (1pt)Using your result from Part (b), what are the best estimates of v 1 and v2? What are the best estimates of the rate of ethanol formation and the ATP maintenance rate?

Calculate Tr = T11 - T12T22-1T21 =

The best estimate for v1 is

1 0 0 1 1.066 0

[ ]

−1 T T v^1=( T r T r ) T r v m= μ = 0.0677 r eth 0.0300

( )(

)

Therefore, the best estimate of the ethanol rate is 0.03 C-mol eth/(C-mol X - h).

r gly 0.0722 The best estimate for v2 is v^2= ( T 22 ) T 21 v^1= ν ppp = 0.0078 m ATP −0.1444 −1

( )( )

Therefore, the best estimate of the ATP maintenance rate is -0.14 C-mol eth/(C-mol X - h).

3. (3 pts) Sensitivity analysis Using the same metabolic pathway shown in Problem 2: a) (1pt)Calculate the norm of GT using the eigenvalue method (text, Eqn 2, Box 8.1) and by using the “norm” function in MATLAB. Compare your results.

Using

‖GT‖=√ λ max (G GT )

>> eig(G*G') ans = -0.0000 -0.0000 0.1676

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3.9093 5.5742 >> sqrt(5.5742) ans = 2.3610 Next, use the norm function in MATLAB. >> norm(G') ans = 2.3610 The results are the same using both methods.

b) (1pt)Calculate the condition number of G T using Eqn 8.28 and the “cond” function in MATLAB. Compare your results. Is GT well conditioned or poorly conditioned?

G (¿ ¿ T )¿ Using ¿ T ( ) c G =‖GT‖ ¿

>> norm(G')*norm((pinv(G'))) ans = 5.7673 Next use the cond function. >> cond(G') ans = 5.7673 The results are the same. Since the condition number is less than 100, the system is well conditioned.

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c) (1pt)Now the rates of biomass and ethanol are measured. Calculate vc/vm . Is the rate of glycerol formation more sensitive to changes in the biomass rate or to changes in the ethanol rate?

Gm= −2 0.355−0.231 1 0.5 0 0 3

[

]

0 0 2 2 Gc = −0.333−0.333 0 4 −1 0 0 5

[

]

0.1155 0 ∂ vc −1 =−( GTc ) G Tm= 1.0661 0 ∂ vm −2.355 0.5

[

]

When the ethanol rate changes, there is no effect on the glycerol rate. Therefore, the glycerol rate is more sensitive to changes in biomass than to changes in ethanol.

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