Exam Questions Flashcards
(2 points) The left null space of a matrix A is spanned by:
a. The columns of A that correspond to the pivots of Ared
b. The row of A that correspond to the pivots of Ared
c. The fundamental solutions to the system of linear equations Ax = 0
d. None of the above.
(2025_1, 2024_1, 2024_2, 2020_test)
d. None of the above.
The left null space (also called the null space of π΄^π or the kernel of π΄^π) consists of all vectors π¦ such that:
π΄ππ¦ = 0
Understanding the options:
(a) The columns of π΄ that correspond to the pivots of π΄red
β Incorrect. The left null space is related to the rows of π΄, not the columns.
(b) The rows of π΄ that correspond to the pivots of π΄ red
β Incorrect. The pivot rows form a basis for the row space, not the left null space.
(c) The fundamental solutions to the system of linear equations π΄π₯ = 0
β Incorrect. The fundamental solutions to π΄π₯ = 0 span the null space of π΄ (not the left null space).
(d) None of the above
β
Correct. None of the provided options accurately describe the left null space.
Final Answer: (d) None of the above
For the reaction in equilibrium (π = πΎeq)β, what is the Gibbs free energy change?
- positive
- negative
- zero
ΞπΊ = 0 at equilibrium
- (2 points) The GPR rule of a reaction catalyzed by a protein complex composed of two units, U1 and U2, where U1 is encoded by two genes g1 and g2, resulting in isoenyzmes, while U2 is encoded by gene g3 is given by
a. gl OR (g2 AND g3).
b. gl AND g2 AND g3.
c. (g1 AND g2) OR g3.
d. (gl OR g2) AND g3.
(2025_1, 2024_2, 2020_test)
d. (gl OR g2) AND g3.
The canonical form of the mathematical program with objective Ξ£(1<=i<=m)|vi|. includes? options with this combination:
2m non-negative variable and 2m real variables.
(2025_1)
TBC
- (2 points) Sink reactions in a stoichiometric representation of a metabolic network contain:
a. Only non-positive stoichiometric coefficients.
b. At least one pair of stoichiometric coefficients with opposite signs.
c. Only non-negative stoichiometric coefficients.
d. All of the above.
(2024_2, 2024_1 2020_test)
a. Only non-positive stoichiometric coefficients.
- (2 points) Stoichiometric coefficients in the biomass reaction of a given model are
a. Unitless integer numbers.
b. Integer numbers with units (mol / gDW * h)
c. Real numbers with units (mol / gDW)
d. None of the above.
(2024_2, 2020_test)
Stoichiometric coefficients in the (except biomass) reaction of a given model with β¦β¦ unit ?
(2025_1, 2024_2, 2024_1)
exam doc said:
c. Real numbers with units (mol / gDW)
(maybe it is this? need to research more)
Perplexity said:
d. None of the above.
Explanation
Stoichiometric coefficients in biomass reactions represent the amount of metabolites consumed or produced per unit of biomass synthesized. Key points:
Real numbers: Coefficients are derived from experimental data (e.g., cell composition) and often include fractional values (e.g., ATP hydrolysis per biomass unit).
Units: While coefficients implicitly relate to units like mmol/gDW or mol/gDW, the stoichiometric matrix itself is typically unitless. Units are instead associated with reaction fluxes (e.g., mmol/gDW/h).
Flux context: The actual metabolite consumption/production rate is calculated by multiplying the coefficient (unitless ratio) by the flux (which carries units like mmol/gDW/h).
Why Other Options Are Incorrect:
a. Coefficients are not restricted to integers (they often include decimals).
b. Units like βmol / gDW * hβ apply to fluxes, not coefficients.
c. While coefficients reflect molar ratios per biomass, they are unitless in the matrix (units are contextual, not explicit).
Thus, none of the listed options fully describe biomass reaction coefficients in standard metabolic models.
- (2 points) A reaction is hard-coupled to objective if its flux value:
a. varies exactly like the objective.
b. can take a value of zero.
c. must be non-zero, but can vary.
d. takes only a value of zero.
a. varies exactly like the objective.
Perplexity:
A reaction is considered hard-coupled to the objective in a metabolic network if its flux value is proportional to the flux of the objective function (e.g., biomass production). This means that the reactionβs flux must scale directly with the objective and cannot be independent or zero as long as the objective has a non-zero value.
Why Other Options Are Incorrect:
b. A hard-coupled reaction cannot take a value of zero if the objective is non-zero.
c. The flux of a hard-coupled reaction cannot vary independently; it must be strictly proportional to the objective.
d. A hard-coupled reaction does not always take a value of zero; it depends on the objectiveβs flux.
- (2 points) A reaction can take place if:
a. The Gibbs free energy is positive.
b. The reaction quotient is at least ten times as large as the equilibrium constant.
c. The Gibbs free energy is negative.
d. None of the above.
(2025_2, 2020_test,
on exam doc answer was:
c. The Gibbs free energy is negative. (only for spontaneous reaction. This question is bs)
Perplexity:
A reactionβs ability to βtake placeβ depends on thermodynamic and biochemical context:
Gibbs free energy (ΞG):
A negative ΞG (option c) indicates spontaneity under standard conditions, but reactions with positive ΞG can still proceed if coupled to energy sources (e.g., ATP hydrolysis) or enzymatic catalysis.
Example: Biosynthetic reactions in cells often have positive ΞG but are driven by energy input.
Reaction quotient (Q) vs. equilibrium constant (K):
If Q>K (option b), the reaction favors the reverse direction under equilibrium assumptions, but this does not preclude it from occurring if energy or enzymes alter the system.
Real-world metabolic networks:
Reactions are often part of interconnected pathways, where thermodynamics are managed via coupling (e.g., ATP/ADP ratios, redox balances).
Why Other Options Are Incorrect:
a and c: ΞG alone does not universally determine feasibility in biological systems.
b: A high Q/K ratio does not inherently block a reaction if energy or enzymes are present.
Thus, none of the options fully capture the conditions under which reactions can occur in biological systems.
- (2 points) xQxβ (w^T Q w)is convex if Q is:
a. indefinite
b. positive semidefinite
c. positive definite
d. negative semidefinite
(2025_1, 2020_test)
Answer
b. positive semidefinite (for convex)
(exam doc also said c for strictly convexβ¦)
Perplexity:
Explanation
A quadratic function
f(x)=xTQx is convex if and only if the matrix Q is positive semidefinite (Qβͺ°0)1367. This follows directly from the definition of convexity:
Convexity condition: For all Ξ»β[1] and vectors x,y:
(Ξ»x+(1βΞ»)y)TQ(Ξ»x+(1βΞ»)y) β€ Ξ»xTQx+(1βΞ»)yTQy
Simplifying this inequality reduces to verifying (xβy)TQ(xβy) β₯ 0, which holds for positive semidefinite Q.
Role of definiteness:
- Positive semidefinite (Qβͺ°0): Ensures xTTQx β₯ 0 for all x, satisfying convexity.
- Positive definite (Qβ»0): Guarantees strict convexity (stronger condition).
- Indefinite or negative semidefinite: These lead to non-convex behavior (e.g., saddle points or concavity).
Why Other Options Are Incorrect:
a. Indefinite: Allows xTQx to take positive and negative values, violating convexity.
c. Positive definite: Implies strict convexity, but convexity only requires Qβͺ°0.
d. Negative semidefinite: Results in a concave function.
Thus, xTQx is convex precisely when Q is positive semidefinite.
- (2 points) Given a linear program
min cTx
s.t.
Ax β€ b
with A β Rnxm, b β Rnx1, c β R^nx1, x β Rmx1 has a dual with:
a. m + n variables.
b. n variables.
c. m variables.
d. none of the above.
(2025_1, 2024_2, 2024_1, 2020_test same question or similar)
b. n variables.
Perplexity Explanation
In linear programming duality:
- The primal problem has m variables (size of x) and n constraints (rows of A).
- The dual problem will have one variable per primal constraint. Since the primal has n constraints (Ax β€ b), the dual has n variables (denoted as y β Rn).
Duality Structure:
Primal (minimization):
min x cTx
s.t. Ax β€ b
Dual (maximization):
max y bTy
s.t. ATy = c (if primal variables are free)
y β₯ 0
Why Other Options Are Incorrect:
a. m+n variables would imply combining primal and dual variables, which is not the case.
c. m variables would correspond to dual constraints (not variables).
d. βNone of the aboveβ is incorrect because b is valid.
Thus, the dual has n variables, one for each primal constraint.
- (2 points) The set of elementary flux modes of a given metabolic network form a convex basis in the network:
a. TRUE.
b. FALSE.
b. FALSE.
Perplexity Explanation
Elementary flux modes (EFMs) represent minimal, non-decomposable pathways in a metabolic network under steady-state conditions. However, they do not inherently form a convex basis.
Key distinctions:
Convex Basis Definition:
- A convex basis consists of extreme rays (or vectors) that generate the flux cone through non-negative combinations. These vectors must be unique and irreducible.
EFMs vs. Extreme Pathways:
- EFMs include all possible minimal pathways, even those involving reversible reactions or redundant combinations.
- Extreme pathways (a subset of EFMs) form the convex basis. They are derived by removing redundancies caused by reversible reactions or exchange fluxes, ensuring uniqueness.
Key Difference:
- EFMs can overlap or include non-unique decompositions, whereas a convex basis requires vectors that cannot be expressed as non-negative combinations of others.
- Example: In networks with reversible reactions, EFMs may include both directions of a pathway, but the convex basis (extreme pathways) omits redundant directions.
Why Other Options Are Incorrect:
a. TRUE: While EFMs span the flux cone, they are not always minimal or unique enough to qualify as a convex basis. Extreme pathways, not EFMs, fulfill this role in constrained networks46.
Thus, EFMs describe all feasible pathways, but only extreme pathways (a subset of EFMs) form the convex basis for the flux cone.
- (2 points) Given a reaction in a metabolic network, its lower and upper feasible bounds determined from flux variability analysis are:
a. Unique.
b. Necessarily positive.
c. Necessarily negative.
d. None of the above.
(2024_2, 2020_test)
d. None of the above.
Perplexity Explanation
Flux variability analysis (FVA) calculates the range of feasible fluxes for each reaction in a metabolic network under steady-state conditions while maintaining near-optimality (e.g., maximizing biomass). Key points:
Non-uniqueness of flux values:
While FVA determines fixed bounds (minimum and maximum fluxes) for each reaction under a given model setup, the actual flux values within these bounds can vary across alternate optimal solutions.
Bounds can include zero or span positive/negative values:
Blocked reactions: May have bounds fixed at zero (e.g., inactive reactions).
Reversible reactions: Can have negative lower bounds and positive upper bounds (e.g., bidirectional transport).
Dependence on network constraints:
Bounds are determined by the stoichiometric matrix, reaction directionality, and objective function. They are not inherently positive, negative, or universally unique across all models or perturbations.
Why Other Options Are Incorrect:
a. Unique: While bounds are fixed for a specific model configuration, they are not universally unique (e.g., altering constraints or optimality criteria changes bounds).
b/c. Positivity/Negativity: Bounds depend on reaction reversibility and network context (e.g., reversible reactions span negative and positive ranges).
Thus, none of the options universally apply to all reactions in metabolic networks.
- (5 points) Given the following Simplex table for a canonical linear programming problem
Basic Variable / x / x1 / x2 / s1 / s2 / s3 / RHS / Upper Bound on Entering Variable
z 1 -13 -11 0 0 0 0 -
s1 0 4 5 1 0 0 1500 375
s2 0 5 3 0 1 0 1575 315
s3 0 1 2 0 0 1 420 420
a. x1 is a entering variable, s2 is a leaving variable.
b. x2 is a entering variable, s1 is a leaving variable.
c. z is a entering variable, s3 is a leaving variable.
d. x is a entering variable, s3 is a leaving variable.
(2024_2, 2020_test)
google doc said:
a. x1 is a entering variable, s2 is a leaving variable.
perplexity said:
b. x2 is an entering variable, s1 is a leaving variable.
Perplexity Explanation of the Correct Answer:
To determine the entering and leaving variables, we typically look for the most negative coefficient in the objective function row (z row) to identify the entering variable and then find the leaving variable based on the minimum ratio of RHS to the coefficient of the entering variable in the constraint rows.
Entering Variable:
The most negative coefficient in the z row is -13 for x1, but since x2 is also a candidate with a coefficient of -11, we need to consider which one is chosen based on additional criteria (e.g., degeneracy resolution). However, in this context, x2 is selected as the entering variable because it is often chosen based on its smaller absolute value or other tie-breaking rules.
Leaving Variable:
Once x2 is chosen as the entering variable, we calculate the ratios for each constraint row (excluding the z row) by dividing the RHS by the coefficient of x2.
For s1:
1500/5 = 300
For s2:
1575/3 = 525
For s3:
420/2 = 210
The smallest ratio is for s1, making s1 the leaving variable.
Thus, the correct answer is b. x2 is an entering variable, s1 is a leaving variable.
- (5 points) Which of the following statements is/are TRUE for the network given in Figure 1 (https://drive.google.com/file/d/1L6dkMciEyT0EuHESJkNBGC0omte_feyD/view?usp=sharing)
a. Reactions 1, 2, and 3 are blocked.
b. Reactions 4 and 5 are blocked.
c. Reactions 3 and 5 are blocked.
d. None of the above.
(2025_1, 2024_2, 2020_test)
google doc:
d. None of the above.
- (5 points) Given the network in Figure 1,
https://drive.google.com/file/d/1L6dkMciEyT0EuHESJkNBGC0omte_feyD/view?usp=sharing
where the flux of every reaction has an upper bound of 1000, which of the non-blocked reactions have a flux of zero in every optimal solution that maximizes the flux through VD:
a. V4, V5, VF.
b. V1, V2, V3, VB.
c. All but VD
d. None of the above.
(2025_1, 2024_2, 2020_test)
a. V4, V5, VF.
- (5 points) Given the network in Figure 1 (https://drive.google.com/file/d/1L6dkMciEyT0EuHESJkNBGC0omte_feyD/view?usp=sharing) , fixing the fluxes through VD and VF fixes the fluxes:
a. V1, V2, V3, VB.
b. V3, V4, V5.
C. V4, V5, VA.
d. All of the above.
(2025_1, 2024_2, 2020_test)
google doc:
b. V3, V4, V5.
- (5 points) Given the network in Figure 1 (https://drive.google.com/file/d/1L6dkMciEyT0EuHESJkNBGC0omte_feyD/view?usp=sharing), which of the following pairs (i,j) satisfy that reaction i is directionally coupled with j:
a. (4, 5)
b. (5,3)
c. (2,3)
d. (3, 1)
(2025_1, 2024_2, 2020_test)
google doc:
c. (2,3)
?
- (5 points) Given the network in Figure 1 (https://drive.google.com/file/d/1L6dkMciEyT0EuHESJkNBGC0omte_feyD/view?usp=sharing), which of the following are/is NOT elementary flux mode(s) ([VA, VB, VD, VF, V1, β¦, V5]): (Not relevant for exams after 2023/topic in 2024 not covered)
a. [0, 1, 1, 0, 0, 1, 1, 0, 0].
b. [1,1,1,1,1,1, 1, 1, 1].
c. [1, 0, 2, 0, 1, 1, 2, 0, 0].
d. [1,0,0, 1, 0, 0, 0, 1, 1].
(2025_1, 2024_2, 2020_test)
google doc:
b. [1,1,1,1,1,1, 1, 1, 1].
- (5 points) The canonical form of the mathematical program with objective Ξ£(1<=i<=m)|vi| and additional n non-negative variables in linear constraints includes altogether:
a. n+m non-negative variables.
b. 2m non-negative variables.
c. 2n non-negative variables.
d. 2m+n non-negative variables.
(2024_2, 2020_test)
- (5 points) The canonical form of the mathematical program with objective Ξ£(1<=i<=m)|vi| and additional n non-negative variables in linear constraints includes altogether:
a. n+m non-negative variables.
b. 2m non-negative variables.
c. 2n non-negative variables.
d. 2m+n non-negative variables.
(2024_2, 2020_test)
d. 2m+n non-negative variables.
(google doc and perplexity and cgpt)
cgpt explanation
Explanation:
The given mathematical program has an objective function Ξ£(1β€iβ€m) |vi|, which involves absolute values. When absolute values appear in an optimization problem, they are typically rewritten using auxiliary variables.
To convert |vi| into a linear program, we introduce two non-negative variables for each vi:
One variable for the positive part of vi (letβs call it viβΊ)
One variable for the negative part of vi (letβs call it viβ»)
We enforce the constraint: vi = viβΊ - viβ», with viβΊ, viβ» β₯ 0
Since there are m absolute value terms, this transformation results in 2m non-negative variables.
Additionally, the problem states that there are n non-negative variables originally present in the constraints. These remain in the problem.
Thus, the total number of non-negative variables in the canonical form is:
2m + n, which corresponds to option d.
β
perplexity explanation
To convert the given mathematical program into canonical form, we need to consider two key aspects:
Handling the absolute value terms in the objective function
Accounting for the additional non-negative variables in linear constraints
Absolute value terms:
For each |vi| in the objective function, we introduce two non-negative variables:
vi+ and vi- such that vi = vi+ - vi-
This transformation requires 2m new non-negative variables (2 for each of the m absolute value terms)
Additional non-negative variables:
The problem statement mentions n additional non-negative variables in linear constraints
These n variables are already non-negative, so they donβt require further transformation
Therefore, the total number of non-negative variables in the canonical form is:
2m (from transforming absolute value terms)
n (original non-negative variables)
The sum of these is 2m + n, which corresponds to option d.
This transformation ensures that:
The objective function becomes linear (sum of vi+ and vi-)
All variables are non-negative
The problem adheres to the standard canonical form for linear programming
Part 3 | Essay question
- (10 points) Provide a brute-force algorithm to determine knock-outs that maximize a given flux (5 points). Provide a description of optKnock and discuss the difference to the previous algorithm (5 points).
(2020_2, )
cgpt
Brute-Force Algorithm for Knock-Out Optimization (5 points)
A brute-force approach to determine knock-outs that maximize a given flux involves systematically testing all possible gene deletions and evaluating their impact on the objective flux.
Algorithm Steps:
Define the metabolic model β Load the stoichiometric matrix, reaction bounds, and biomass or target flux.
Enumerate all possible knock-outs β Generate all combinations of single, double, or higher-order knock-outs.
Evaluate each knock-out β For each set of knock-outs:
Remove the corresponding reactions/genes.
Solve Flux Balance Analysis (FBA) to determine the objective flux.
Select the best set β Identify the knock-out combination that results in the highest flux.
Drawback: This method is computationally expensive, as the number of possible knock-outs grows exponentially with the number of genes.
OptKnock and Comparison (5 points)
OptKnock is a computational method that formulates knock-out selection as a bi-level optimization problem to enhance a desired flux while maintaining metabolic feasibility.
How OptKnock Works:
- Outer optimization: Maximizes the target flux (e.g., product yield) by selecting knock-outs.
- Inner optimization: Uses FBA to determine the metabolic flux distribution that maintains growth while obeying mass balance constraints.
- Solves using MILP (Mixed-Integer Linear Programming) to efficiently find optimal knock-out strategies.
Difference from Brute-Force:
β
More efficient β avoids exhaustive search by using optimization techniques.
β
Finds global optima rather than evaluating all possible solutions.
β
Bi-level structure accounts for cellular adaptation post knock-out.
β Requires a well-formulated metabolic model and is harder to implement.
Part 3 | Essay question
- (10 points) Explain what information is provided by flux variability analysis and to what end can it be used (2 points). Provide the mathematical program for flux variability analysis (5 points). Is there a relationship between flux variability and flux coupling analysis? If so, specify it mathematically (3 points).
(2020_test, 2018_test)
Flux Variability Analysis (FVA) β Information and Uses (2 points)
Flux Variability Analysis (FVA) determines the range of possible flux values for each reaction while maintaining a given optimality criterion (e.g., biomass maximization). It identifies:
β
Alternative pathways that can achieve the same objective.
β
Essential reactions (those with fixed flux values).
β
Redundant reactions (those with large variability).
Uses:
Identifying flexibility in metabolic networks.
Finding reactions with strict flux constraints vs. those with alternative pathways.
Detecting drug targets by identifying essential reactions.
Mathematical Formulation of FVA
FVA solves two linear programs per reaction π to determine its minimum and maximum flux:
π£πmin = minπ£π
π£πmax = maxπ£πβ
subject to:
Sv = 0(Massbalance)
π£min β€ π£ β€ π£max (Fluxbounds)
ππ π£ β₯ πΌπβ (Enforcingoptimality,whereπβistheoptimalobjectivevalue)
where:
- π is the stoichiometric matrix.
- π£ is the flux vector.
- ππ π£ is the objective function (e.g., biomass production).
- πΌ ensures optimality constraint (typically πΌ = 0.9 or 1.0 to explore feasible solutions at optimal growth).
Relationship Between Flux Variability and Flux Coupling Analysis (3 points)
Flux Coupling Analysis (FCA) determines dependencies between fluxes. The relationship between FVA and FCA can be mathematically expressed as:
π£πmin / β‘π£πmaxβ‘ β€
π£π/π£π β€ π£πmax
π£πmin
β‘
where:
- π£π and π£π are fluxes of two reactions.
- If π£πminβ‘ / π£πmax =
π£πmaxβ‘ /π£πmin, the reactions are fully coupled (always operate in a fixed ratio).
- If π£πminβ‘ = π£πmax = 0, the reaction is blocked (cannot carry flux).
Key Insight: FVA provides the flux ranges needed to compute coupling coefficients in FCA.
Part 3 | Essay question
- (10 points) Clearly define and mathematically specify Gibbs free energy of a reaction (2 points). Use the definition to specify the constraints in thermodynamic metabolic flux analysis (TMFA) (8 points).
(2020_test )
cgpt Summary (ask for more details):
- Gibbs Free Energy defines reaction spontaneity based on ΞπΊ = ΞπΊβ + π
πlnπ
- TMFA imposes directionality constraints, chemical potential relationships, and log-concentration bounds to ensure thermodynamic feasibility in FBA.
- MILP formulation is often required to enforce irreversibility constraints.
Part 4 - Synthesis (15 points)
- (15 points) Provide a mathematical program that predicts a parsimonious steady- state flux distribution under the assumption that the modelled system optimizes biomass yield and the total flux through each metabolic pool is provided as input (15 points).
(2020_test)
from google doc, see it for more details
i guess the PFBA of lecture 10 is a hint
But Pfba is more applicable for the extra credit question and i donβt think he will repeat the same concept twice
but in 22 you could technically use a random network
personally i think the minmax in Pfba suits it
it absolutely does
i just think the questions is supposed to see how well your able to build your own program without a specific underlying topic, but this is 100% my personal guess
[ cgpt screenshot see doc]
no guarantee that this is correct thats 21, 22 and 22 in detail