Modelling a biological system: rate equations, parameters and fit · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Rate models you can defend with data

Science · Genetics & Evolution · ages 23-24
Name ______________________   Date ____________

  1. In dA/dt = s - kA and dB/dt = kA - hB, concentrations are in µM and time is in hours. Which parameter units make every rate term consistent?

    • s: µM/h; k and h: µM/h
    • s: µM/h; k and h: h^-1
    • s: h^-1; k and h: µM/h

  2. At fixed volume, A converts one-for-one into B at rate kA. B is removed at rate hB. No other process changes B. Which equation balances B?

    • dB/dt = kA + hB
    • dB/dt = hB - kA
    • dB/dt = kA - hB

  3. A perfect fit to training data proves the model's mechanism and makes held-back testing unnecessary.

    Circle one:   True   False

  4. With other settings fixed, raising a by 5% moves a positive prediction from 20 to 26. Separately raising b by 5% moves it from 20 to 20.4. What follows?

    • Greater local sensitivity to a for these perturbations
    • Exclusive dependence on a across all parameter settings
    • Unique estimates of both parameters from the observations

  5. In dP/dt = rP(1 - P/K), all measured P values are far below K. The early data are sparse and noisy. Which parameter is usually less constrained?

    • The carrying capacity K
    • Both parameters must be equally well constrained
    • The growth-rate parameter r

  6. For dX/dt = s - kX, k = 0.2 h^-1 is known. Two steady training measurements are 8 and 10 µM. Fit s from 1.2, 1.8 or 2.4 µM/h by minimising the sum of squared concentration errors. Which candidate fits best?

    • s = 1.2 µM/h
    • s = 2.4 µM/h
    • s = 1.8 µM/h

  7. For dX/dt = s - kX with s,k > 0, exact steady-level observations identify s/k but not s or k separately. Which quantity stays unchanged when both parameters are doubled?

    • The steady concentration X*
    • The initial rate when X starts at zero
    • The rate constant controlling return to the steady level

  8. A full model reduces exactly to dX/dt = -kX when input is zero. With k = 0.2 h^-1 and X(0) = 5 µM, its code predicts increasing X. What should you check first?

    • The noise score for held-back biological measurements
    • The sign, units and zero-input implementation
    • The crowding term of an unrelated logistic growth model
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Answer key

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Rate models you can defend with data W1-mt_Al6sNjUOW5-s1

  1. s: µM/h; k and h: h^-1 · Each rate has units µM/h. Thus s is in µM/h, while k and h must be in h^-1 so that kA and hB are in µM/h.
  2. dB/dt = kA - hB · Conversion adds to B at rate kA. Removal subtracts from B at rate hB. Therefore dB/dt = kA - hB.
  3. False · Training agreement alone proves neither the mechanism nor predictive accuracy on unseen observations.
  4. Greater local sensitivity to a for these perturbations · The prediction changes by 30% for a and 2% for b. Equal fractional parameter changes produce a larger response for a in this comparison.
  5. The carrying capacity K · When P/K is small, growth is close to rP. Changing a large K then has little effect on early predictions, so these data usually constrain K less.
  6. s = 1.8 µM/h · The candidates predict steady concentrations of 6, 9 and 12 µM. Their squared-error sums are 20, 2 and 20 µM². Thus s = 1.8 µM/h fits best among the candidates.
  7. The steady concentration X* · The steady concentration is X* = s/k. Doubling both parameters leaves (2s)/(2k) = s/k unchanged. The initial rate at X = 0 and the return rate constant both double.
  8. The sign, units and zero-input implementation · The initial rate is -0.2 × 5 = -1 µM/h. The solution 5exp(-0.2t), with t in hours, decreases. Increasing X contradicts the stated limit.
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