Exam | MCQ Flashcards
Part 1 | concepts (2 pts)
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.
Correct would be:
The left null space of a matrix A…
* … the fundamental solutions to the homogeneous system ATx = 0
* … the vectors orthogonal to all rows of A
* … the basis vectors of the null space of AT
i.e. the left null space consists of all vectors x such that: ATx = 0
Understanding the other options:
(a) The columns of 𝐴 that correspond to the pivots of 𝐴red
❌ Incorrect. The left null space is related to the rows of 𝐴, not the columns.
(b) The rows of 𝐴 that correspond to the pivots of 𝐴 red
❌ 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 𝐴𝑥 = 0
❌ Incorrect. The fundamental solutions to 𝐴𝑥 = 0 span the null space of 𝐴 (not the left null space).
Part 1 | concepts (2 pts)
For the reaction in equilibrium (𝑄 = 𝐾eq), what is the Gibbs free energy change?
- positive
- negative
- zero
Δ𝐺 = 0 at equilibrium
Part 1 | concepts (2 pts)
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.
Part 1 | concepts (2 pts)
The canonical form of the mathematical program with objective Σ1≤i≤m|vi| includes:
a. m real variables
b. m non-negative variables
c. 2m non-negative variables
d. 2m real variables
(2025_1)
c. 2m non-negative variables
Each |vi| is replaced with vi+ + vi−, vi+, vi− ≥ 0
This gives 2m non-negative variables in total
Part 1 | concepts (2 pts)
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.
Part 1 | concepts (2 pts)
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)
c. Real numbers with units (mol / gDW)
Part 1 | concepts (2 pts)
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)
c. The Gibbs free energy is negative.
ΔG < 0 ⟶ Reactionisthermodynamicallyfavorable
… can [spontaneously] take place! if…
[a bit ambiguous, because it doesn’t specify how the reaction can take place… spontaneously? with an enzyme?]
Part 1 | concepts (2 pts)
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.
(2020_test
a. varies exactly like the objective.
[?? b. can take a value of zero → only if the objective is zero! ]
Part 1 | concepts (2 pts)
xTQx is convex if Q is:
a. indefinite
b. positive semidefinite
c. positive definite
d. negative semidefinite
(2025_1, 2020_test)
b. positive semidefinite
positive semidefite: some eigenvalues 0, others +ve
Why the other options are incorrect:
a. Indefinite → ❌ Not convex; the function can curve both up and down
c. Positive definite → ❌ This implies strict convexity, but the question asks about convexity in general — and PSD is the minimal requirement
d. Negative semidefinite → ❌ Makes the function concave, not convex
Part 1 | concepts (2 pts)
Given a linear program
min cTx
s.t.
Ax ≤ b
with A ∈ Rnxm, b ∈ Rnx1, c ∈ Rnx1, x ∈ 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.
The dual variables correspond to the n constraints in the primal.
c should be ∈ Rm, not c ∈ Rn! probably a typo in the exam.
Part 1 | concepts (2 pts)
The set of elementary flux modes of a given metabolic network form a convex basis in the network:
a. TRUE.
b. FALSE.
b. FALSE.
(Not relevant for exams after 2023/topic in 2024 not covered)
Part 1 | concepts (2 pts)
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.
a. Unique → ❌
The bounds themselves (min/max) are not necessarily unique, depending on alternate optima, model formulation, etc.
Also, the flux value for the reaction is not unique — FVA gives a range.
b. Necessarily positive → ❌
Fluxes can be positive, negative, or zero, depending on direction and reversibility of the reaction.
Reversible reactions often have negative lower bounds.
c. Necessarily negative → ❌
Same reason as above — bounds can be positive, negative, or zero, depending on the network.
Part 2 | Hands on calculations (5 pts)
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)
a. x₁ is a entering variable, s₂ is a leaving variable.
🔹 Step 1: Identify the entering variable
From the objective row (z row), the coefficients of decision variables are:
x₁: −13
x₂: −11
Since we’re maximizing z, we choose the most negative coefficient — this gives the most improvement in the objective.
✅ So, x₁ is the entering variable (−13 < −11)
🔹 Step 2: Identify the leaving variable
To do this, we calculate the minimum ratio of: RHS ÷ coefficientofenteringvariable
Only rows where the coefficient of the entering variable (x₁) is +ve are considered.
Let’s compute the ratios:
Row | Coefficient of x₁ | RHS | RHS÷Coefficient
s₁ 4 1500 375
s₂ 5 1575 315 ⬅️ minimum
s₃ 1 420 420
✅ So, s₂ is the leaving variable
Part 2 | Hands on calculations (5 pts)
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)
d. None of the above.
Part 2 | Hands on calculations (5 pts)
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.
Part 2 | Hands on calculations (5 pts)
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)
b. V3, V4, V5.
Part 2 | Hands on calculations (5 pts)
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)
c. (2,3)
Rxn𝑖 is directionally coupled to Rxn𝑗 if for every 𝑣 with 𝑁𝑣 = 0:
- 𝑣𝑖 ≠ 0 implies 𝑣𝑗 ≠ 0
- but not the reverse.
Part 2 | Hands on calculations (5 pts)
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, …, V5]):
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)
PROBABLY NOT RELEVANT ANYMORE !!!
google doc:
b. [1,1,1,1,1,1, 1, 1, 1].
Part 2 | Hands on calculations (5 pts)
The canonical form of the mathematical program with objective Σ1≤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.
Part 1 | concepts (2 pts)
The null space of a matrix A is spanned by:
a. The pivot columns of A
b. The rows of AT that correspond to pivot positions
c. The fundamental solutions to the system Ax = 0
d. The columns of A that correspond to the zero rows in Ared
(based on question in 2025_1, 2024_1, 2024_2, 2020_test )
c. The fundamental solutions to Ax = 0 form a basis for the null space
These are the special solutions found by assigning values to free variables
They span all vectors x such that Ax = 0
a. The pivot columns of A ❌
These form a basis for the column space, not the null space
Pivot columns are linearly independent and span the range (image) of A
The null space is orthogonal to the row space, not related to pivot columns
b. The rows of AT that correspond to pivot positions ❌
The rows of AT are just the columns of A
Pivot positions relate to the row space (and thus the left null space, indirectly)
The null space is made of input vectors x that A maps to 0, not built from rows of AT
d. The columns of A that correspond to the zero rows in Ared ❌
Zero rows in the row-reduced form (Ared) relate to dependencies, not basis vectors
These indicate the presence of free variables, but the corresponding columns do not directly span the null space
The null space is spanned by special solution vectors, not original columns of A
Part 1 | concepts (2 pts)
Stoichiometric coefficients in non-biomass reactions are
a. Unitless real numbers.
b. Integer numbers with units (mol / gDW * h)
c. Real numbers with units (mol / gDW)
d. None of the above.
(2025_1, 2024_2, 2024_1)
a. Unitless real numbers.
Part 1 | concepts (2 pts)
xTQx is strictly convex if Q is:
a. indefinite
b. positive semidefinite
c. positive definite
d. negative semidefinite
(2025_1, 2020_test)
c. positive definite
Part 1 | concepts (2 pts)
Shadow prices of canonical LP correspond to change in the objective resulting from the changes in upper bound of the constraints.
TRUE/FALSE?
(2024_2)
FALSE.
✅ Corrected Statement: Shadow prices in a canonical LP correspond to change in the objective resulting from a 1-unit change in the right-hand side (RHS) of a constraint (not the upper bound of the constraint).
🔍 Explanation:
In a canonical LP:
maxx cTx
s.t.
Ax ≤ b
x ≥ 0
The shadow price (also called dual variable or Lagrange multiplier) for a constraint tells you how much the objective value would change if you relax or tighten the RHS (b) of that constraint by 1 unit.
It does not refer to the upper or lower bound on variables — that’s a different sensitivity concept.
Part 1 | concepts (2 pts)
The function xTQx is concave if Q is:
a. indefinite
b. positive semidefinite (for convex)
c. positive definite (for strictly convex)
d. negative semidefinite
(2024_2)
d. Negative semidefinite
Why the other options are wrong:
a. Indefinite → ❌ could be both convex and concave in parts, not globally
b. Positive semidefinite → ❌ makes it convex, not concave
c. Positive definite → ❌ makes it strictly convex, not concave
