Energies in kT
Warm-upIn E. coli growing on glucose, splitting ATP releases about 47 kJ/mol (Milo and Phillips). How many kT is that at 37 °C? (At 37 °C, kT per mole is RT = 2.58 kJ/mol.)
Test yourself on the math from the lesson. There are 41 kinds of problem, from warm-ups to challenges, and each one gives you new numbers whenever you ask.
RememberA reaction’s rate goes as , with kJ/mol at 37 °C (2.48 at 25 °C). To turn kJ/mol into kT, divide by RT. Every 5.94 kJ/mol off the barrier makes a reaction 10 times faster at 37 °C (5.71 at 25 °C), so a speed-up of needs kJ/mol. A rate constant and a half-time are linked by , with . An enzyme lowers the barrier both ways, so it never moves the equilibrium.
Review this in the lessonIn E. coli growing on glucose, splitting ATP releases about 47 kJ/mol (Milo and Phillips). How many kT is that at 37 °C? (At 37 °C, kT per mole is RT = 2.58 kJ/mol.)
An enzyme makes a reaction 10 times faster at 37 °C. By how much must it lower the energy barrier, in kJ/mol? At 37 °C, every 5.94 kJ/mol lower makes a reaction 10 times faster.
An enzyme lowers the energy barrier of a reaction by 20 kJ/mol. Does that change where the reaction’s equilibrium lies? Answer yes or no.
Two enzymes do the same job at 37 °C, but one holds its substrate in a better shape and lowers the barrier 18 kJ/mol more than the other. How many times faster is it?
One substrate molecule sits in the active site of syn3A’s pyruvate kinase (pyk, JCVISYN3A_0221), which reacts at kcat = 3,200 per second. What is the half-time, the time by which half of such molecules have reacted, in ms?
A reaction (made-up numbers) has a half-time of 3 hours on its own and 2 ms with an enzyme. How many times faster is the reaction with the enzyme?
RememberAn enzyme’s rate is . At it runs at of its top speed: half at Km, 75% at 3 Km, 90% at 9 Km. To reach a fraction f you need . Vmax = kcat × number of enzymes. A typical enzyme has kcat ≈ 10 per second and Km ≈ 0.1 mM. In syn3A, 1 mM is about 20,170 molecules.
Review this in the lessonAn enzyme's substrate is at a concentration 9 times Km. What percent of its top speed Vmax does the enzyme run at?
You measure the rate of an enzyme in a test tube (made-up numbers) at several substrate concentrations. The rate levels off at Vmax = 50 per second. Your measurements (rate in per second): at 0.2 mM, 8.33; at 1 mM, 25; at 2 mM, 33.3; at 9 mM, 45. What is Km, in mM?
syn3A’s glucose-6-phosphate isomerase (pgi, JCVISYN3A_0445) has kcat = 929 per second, and a cell has 266 copies of it. These numbers were measured in related enzymes and are used in syn3A’s model. What is Vmax, the most all the copies can do together, per second?
syn3A’s whole-cell model starts NAD⁺ at 2.18 mM (a model value, estimated from other bacteria). In syn3A, 1 mM is about 20,170 molecules. How many NAD⁺ molecules is that in one cell?
A typical enzyme has kcat = 10 per second and Km = 0.1 mM (Bar-Even et al. 2011). How many reactions per second does one molecule of it carry out with its substrate at 0.1 mM?
Michaelis and Menten’s invertase has Km = 16.7 mM for sucrose. At what concentration of sucrose does the enzyme run at 50% of Vmax, in mM?
syn3A takes up about 15,000 glucose per second, and each one passes through glucose-6-phosphate isomerase once. syn3A’s glucose-6-phosphate isomerase (pgi, JCVISYN3A_0445) has kcat = 929 per second and 266 copies per cell. These numbers were measured in related enzymes and are used in syn3A’s model. What percent of the enzymes’ top speed does the cell use?
A typical enzyme has kcat = 10 per second and Km = 0.1 mM (Bar-Even et al. 2011). kcat/Km measures how well an enzyme catches substrate when it’s scarce. The fastest possible value, set by how quickly molecules meet by diffusion, is about per molar per second. How many times below that limit is this enzyme?
RememberA competitive inhibitor sits in the active site, so it raises the apparent Km: , and Vmax stays the same. A noncompetitive inhibitor locks the enzyme somewhere else, so it lowers Vmax: , and Km stays the same. At the factor is 2. Invertase: Km = 16.7 mM for sucrose, Vmax = 0.76 mM per minute; fructose Ki = 58.8 mM, glucose Ki = 91 mM.
Review this in the lessonA drug (made-up numbers) looks like an enzyme’s substrate and sits in its active site. The enzyme has Km = 0.1 mM, and the drug has Ki = 0.05 mM. What is the apparent Km with 0.025 mM of the drug present, in mM?
Glucose competes with sucrose for invertase’s active site, with Ki = 91 mM (Michaelis and Menten 1913). How much glucose does it take to double the apparent Km, in mM?
Invertase has Vmax = 0.76 mM per minute and Km = 16.7 mM for sucrose, and glucose competes for its active site with Ki = 91 mM (Michaelis and Menten 1913; values from the 2011 reanalysis). How fast does it split 50 mM sucrose when 20 mM glucose is present, in mM per minute?
A metal ion (made-up numbers) binds an enzyme away from its active site and stops it working. Without it, the enzyme has Vmax = 120 µM per second. It has Ki = 20 µM. What is Vmax with 60 µM of it present, in µM per second?
Glucose, one of invertase’s products, takes up room in its active site and slows it, unless there’s plenty of sucrose. Is this a competitive or a noncompetitive inhibitor?
RememberGlucose (6 carbons) splits into two 3-carbon halves. Per glucose: 2 ATP net, 2 NADH, 2 pyruvate. syn3A has no hexokinase: the PTS brings glucose in and pays with a phosphate from PEP, and the net is still 2 ATP. Sending pyruvate to acetate earns 1 more ATP each (2 + 2f per glucose if a share f goes that way) but loses a carbon as CO₂. syn3A has about 44,000 NAD⁺ and uses about 30,000 a second.
Review this in the lessonsyn3A ferments 1 glucose molecule to lactate. How many ATP does it gain, after paying back what it spent?
How many NADH does glycolysis make from 100 glucose?
syn3A’s PTS turns outside glucose into G6P inside the cell, and makes a pyruvate as it does. Which high-energy molecule does it use up? Give its short name.
Suppose 50% of syn3A’s pyruvate goes down the acetate route and the rest to lactate. The acetate route (pdhC, lpdA, pta, ackA) earns 1 extra ATP per pyruvate, at acetate kinase. How many ATP does each glucose give, net?
syn3A’s whole-cell model starts NAD⁺ at 2.18 mM (1 mM is about 20,170 molecules in syn3A). Glycolysis uses up one NAD⁺ (turning it into NADH) for each three-carbon half, and the cell takes up about 15,000 glucose per second (Thornburg et al. 2022). If nothing turned NADH back into NAD⁺, how long would the NAD⁺ last, in s?
RememberTurnover time = pool ÷ flow. If use outruns supply, the pool lasts pool ÷ (use − supply). In syn3A: about 74,000 ATP, made and used at about 35,000 per second. A model flux of 1 mmol gDW⁻¹ h⁻¹ is about 1,706 molecules per second in one cell. Making protein costs about 4 ATP per amino acid. Power in watts = ATP per second × 50,000 J/mol ÷ .
Review this in the lessonA syn3A cell has about 830 PEP molecules, made and used at about 30,000 per second. How long does a PEP molecule last, on average, in ms?
FBA of syn3A’s model gives a flux of 6.9 mmol per gram dry weight per hour (mmol gDW⁻¹ h⁻¹) for acetate secretion. In one syn3A cell, 1 mmol gDW⁻¹ h⁻¹ is about 1,706 molecules per second. How many acetate per second is that?
A syn3A cell has 500 ribosomes. Each adds 4 amino acids per second (a made-up speed), and each amino acid costs 4.32 ATP, the value in syn3A’s FBA model (2 to charge its tRNA, 2.32 for the peptide bond). How many ATP per second does making protein cost?
A syn3A cell has about 44,000 NAD⁺, and glycolysis turns about 30,000 per second into NADH. A drug (made up) slows lactate dehydrogenase, so only 20,000 per second are turned back. If use stays the same, how long until the NAD⁺ runs out, in s?
FBA says syn3A makes about 46,700 ATP per second. Each ATP carries about 50 kJ/mol of free energy in a cell. What power is that, in watts (joules per second)?
RememberA vector is a list of numbers; a matrix is a table. A dot product pairs up two lists, multiplies and adds: . In the stoichiometric matrix S, each row is a molecule and each column a reaction; an entry is how many that reaction makes (+) or uses (−). S·v is worked out one row at a time: each row times the fluxes v gives how fast that molecule changes. At steady state every entry of S·v is 0.
Review this in the lessonMultiply the lists [−4, −4, −2] and [2, 4, 1] entry by entry, then add the results. What do you get? (This is called a dot product.)
ACKr (ackA) does this: acetyl phosphate + ADP ⇌ acetate + ATP. In syn3A’s stoichiometric matrix, what is the entry in the row for F6P and the column for ACKr?
GLCpts (the PTS (ptsI, ptsH, crr, ptsG)) does this: glucose (outside) + PEP → G6P + pyruvate. How many nonzero entries does GLCpts’s column of syn3A’s stoichiometric matrix have?
PEP’s row in syn3A’s matrix has −1 under GLCpts, +1 under ENO, −1 under PYK, and zeros everywhere else. Suppose the fluxes are: GLCpts 8, ENO 2, PYK 7 (mmol gDW⁻¹ h⁻¹). Multiply the row by the fluxes: how fast is PEP building up (positive) or running down (negative), in mmol gDW⁻¹ h⁻¹?
A toy network has three molecules, A, B and C (the rows, in that order), and three reactions (the columns). Its stoichiometric matrix S and its fluxes v are: What is the entry of S·v for B: how fast B changes?
At steady state, acetyl-CoA must be made exactly as fast as it’s used. In syn3A’s core, it’s made by PDH (3) and used by PTAr (unknown). All fluxes are in mmol gDW⁻¹ h⁻¹. What must the PTAr flux be, in mmol gDW⁻¹ h⁻¹?
RememberLinear programming finds the best value of a straight-line objective inside a region set by straight-line limits; the best is always at a corner. In the two-number syn3A problem, ATP = 2g + a, with g ≤ G = 7.4, a ≤ A = 6.9 and a ≤ 2g: the best is , 21.7 at (7.4, 6.9). A knockout sets its reactions to 0: no ackA means a = 0 (ATP = 2g); no ldh means a = 2g (ATP = 4g). Accuracy = (TP + TN) ÷ all, sensitivity = TP ÷ (TP + FN), specificity = TN ÷ (TN + FP).
Review this in the lessonIn the two-number version of syn3A’s FBA, the cell chooses its glucose uptake g and its acetate output a (both in mmol gDW⁻¹ h⁻¹), and makes ATP = 2g + a. How much ATP does it make at the point g = 3.45, a = 6.9, in mmol gDW⁻¹ h⁻¹?
In the two-number version of syn3A’s FBA, the cell chooses glucose uptake g and acetate output a (both in mmol gDW⁻¹ h⁻¹). The region of allowed choices has corners at (g = 0, a = 0), (g = 7.4, a = 6.9), (g = 3.45, a = 6.9), (g = 7.4, a = 0). Lactate made is 2g − a. What is the most lactate the cell can make, in mmol gDW⁻¹ h⁻¹?
In the two-number version of syn3A’s FBA, the cell chooses its glucose uptake g and its acetate output a (both in mmol gDW⁻¹ h⁻¹), and makes ATP = 2g + a. Three limits apply: g can be at most 10; a can be at most 6.9; and a can’t be more than 2g, because lactate (= 2g − a) can’t be negative. What is the most ATP the cell can make, in mmol gDW⁻¹ h⁻¹?
In the two-number version of syn3A’s FBA, the cell chooses its glucose uptake g and its acetate output a (both in mmol gDW⁻¹ h⁻¹), and makes ATP = 2g + a. Normally g can be at most 10, a at most 4, and a no more than 2g. If gene ackA (JCVISYN3A_0230, acetate kinase) is knocked out, by what percent does the most ATP the cell can make drop?
Breuer et al. (2019) compared their model’s essential-gene calls with transposon data from real syn3A cells, counting quasi-essential genes as essential. Of 153 genes: 123 the model calls essential and the cell needs (true positives, TP); 18 the model calls non-essential but the cell needs (false negatives, FN); 12 the model calls non-essential and the cell can lose (true negatives, TN); 0 the model calls essential but the cell can lose (false positives, FP). What is the model’s specificity, as a percent?
In the two-number version of syn3A’s FBA, the cell chooses its glucose uptake g and its acetate output a (both in mmol gDW⁻¹ h⁻¹), and makes ATP = 2g + a. Normally g can be at most 2, a at most 6.9, and lactate = 2g − a. If ldh (JCVISYN3A_0475, lactate dehydrogenase) is knocked out, lactate must be 0. What is the most ATP the cell can make, in mmol gDW⁻¹ h⁻¹?
8 problems from across the lesson, in random order. Here the topic is hidden until you solve each one, so part of the job is working out which tool you need.
A reaction (made-up numbers) has a half-time of 8 hours on its own and 50 ms with an enzyme. How many times faster is the reaction with the enzyme?
In the two-number version of syn3A’s FBA, the cell chooses its glucose uptake g and its acetate output a (both in mmol gDW⁻¹ h⁻¹), and makes ATP = 2g + a. Normally g can be at most 7.4, a at most 6.9, and a no more than 2g. If gene ackA (JCVISYN3A_0230, acetate kinase) is knocked out, by what percent does the most ATP the cell can make drop?
At steady state, ATP must be made exactly as fast as it’s used. In syn3A’s core, it’s made by PGK (20), PYK (10) and ACKr (5) and used by PFK (10) and ATPM (unknown). All fluxes are in mmol gDW⁻¹ h⁻¹. What must the ATPM flux be, in mmol gDW⁻¹ h⁻¹?
Suppose 75% of syn3A’s pyruvate goes down the acetate route and the rest to lactate. The acetate route (pdhC, lpdA, pta, ackA) earns 1 extra ATP per pyruvate, at acetate kinase. If syn3A takes up 15,000 glucose per second, how many ATP per second does it make, net?
One substrate molecule sits in the active site of syn3A’s lactate dehydrogenase (ldh, JCVISYN3A_0475), which reacts at kcat = 643 per second. What is the half-time, the time by which half of such molecules have reacted, in ms?
Breuer et al. (2019) compared their model’s essential-gene calls with transposon data from real syn3A cells, this time counting quasi-essential genes as non-essential. Of 153 genes: 101 the model calls essential and the cell needs (true positives, TP); 4 the model calls non-essential but the cell needs (false negatives, FN); 26 the model calls non-essential and the cell can lose (true negatives, TN); 22 the model calls essential but the cell can lose (false positives, FP). What is the model’s sensitivity, as a percent?
About 15,000 PEP per second go through pyruvate kinase in syn3A (one per glucose: the other PEP pays for glucose entry). syn3A’s pyruvate kinase (pyk, JCVISYN3A_0221) has kcat = 3,200 per second and 551 copies per cell. These numbers were measured in related enzymes and are used in syn3A’s model. What percent of the enzymes’ top speed does the cell use?
A poison (made-up numbers) binds syn3A’s phosphofructokinase away from the active site and locks it. Without the poison, the cell’s PFK has Vmax = 50,800 per second. It has Ki = 1 µM. What is Vmax with 9 µM of it present, in per second?