What Is Biology & Characteristics of Life

Question 1

(a) (4 marks)

• Homeostasis = maintenance of a stable internal environment despite external change (1).
• The passive term $-k(T-T_e)$ is pure physics: heat always flows toward ambient and would drive $T$ to $20\,^\circ\text{C}$ (1).
• The $c(T_s-T)$ term acts in the opposite direction to any deviation from the setpoint $T_s$: if $T$ falls below $37$ it becomes positive (heating), if $T$ rises above it becomes negative (cooling) — this is negative feedback (1).
• It is biological because it requires active energy expenditure (shivering, vasoconstriction, sweating) directed at restoring the setpoint, not just responding to ambient (1).

(b) (6 marks) Set $dT/dt = 0$: $-k(T^\ast - T_e) + c(T_s - T^\ast) = 0$ $-kT^\ast + kT_e + cT_s - cT^\ast = 0$ $T^\ast(k+c) = kT_e + cT_s$ $T^\ast = \frac{kT_e + cT_s}{k+c}$ (3) Substituting: $T^\ast = \dfrac{0.10(20) + 0.40(37)}{0.10+0.40} = \dfrac{2.0 + 14.8}{0.50} = \dfrac{16.8}{0.50} = 33.6\,^\circ\text{C}$ (2).

• Interpretation: because passive cooling never fully vanishes, the setpoint is not perfectly reached — a steady-state offset (proportional-control error). Biologically, real bodies use integral-type control (hormonal/behavioural) to close this gap; the model shows finite gain leaves a residual error (1).

(c) (6 marks) Rewrite: $\dfrac{dT}{dt} = -(k+c)T + (kT_e + cT_s) = -(k+c)(T - T^\ast)$ (2). This is first-order linear; solution: $T(t) = T^\ast + \big(T(0) - T^\ast\big)e^{-(k+c)t}$ (2) With $T(0)=39$, $T^\ast=33.6$, $k+c=0.50$: $T(t) = 33.6 + 5.4\,e^{-0.50\,t}$ Effective time constant $\tau = \dfrac{1}{k+c} = \dfrac{1}{0.50} = 2.0\ \text{min}$ (2).

(d) (4 marks)

• Independent variable: ambient temperature $T_e$ (what experimenter sets) (1).
• Dependent variable: core body temperature $T$ (measured response) (1).
• Controlled variable: any of — species/animal size, activity level, humidity, food intake, time of measurement (1).
• (1) for correctly matching "independent = manipulated, dependent = measured."

Question 2

(a) (4 marks)

• Anabolism: synthesis of complex molecules from simpler ones; consumes energy (endergonic), e.g. protein synthesis (1.5).
• Catabolism: breakdown of complex molecules into simpler ones; releases energy (exergonic), e.g. cellular respiration (1.5).
• During growth, anabolism dominates (net synthesis increases body mass) (1).

(b) (4 marks) Take $\log_{10}$ of $B = aM^{3/4}$: $\log_{10}B = \log_{10}a + \tfrac{3}{4}\log_{10}M$ (2) This is linear in $\log_{10}M$ (form $y = c + m x$) with slope $= 3/4 = 0.75$ and intercept $\log_{10}a$ (2).

(c) (6 marks) Two-point slope on log-log axes: $n = \frac{\log_{10}B_H - \log_{10}B_M}{\log_{10}M_H - \log_{10}M_M}$ $= \frac{\log_{10}(116) - \log_{10}(0.14)}{\log_{10}(70) - \log_{10}(0.020)}$ $= \frac{2.0645 - (-0.8539)}{1.8451 - (-1.6990)} = \frac{2.9184}{3.5441} = 0.8235$ (4) Numerically $n \approx 0.82$, close to but slightly above the theoretical $0.75$; scatter/measurement error and using only two points explains the deviation (2).

(d) (4 marks) Using Human: $a = B/M^{3/4} = 116 / 70^{0.75}$. $70^{0.75} = e^{0.75\ln 70} = e^{0.75(4.2485)} = e^{3.1864} = 24.20$ (2). $a = 116/24.20 = 4.79$ (1). Units: $B$ in W, $M^{3/4}$ in $\text{kg}^{3/4}$, so $a$ has units $\text{W·kg}^{-3/4}$ (1).

(e) (4 marks) Expected algorithm outline (award 1 each):

1. Transform each point: $x_i = \log_{10}M_i$, $y_i = \log_{10}B_i$.
2. Compute least-squares slope $n = \dfrac{\sum(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)^2}$ and intercept $b = \bar y - n\bar x$.
3. Recover $a = 10^{b}$.
4. Report $n$, $a$, and optionally $R^2$ for goodness of fit.

Question 3

(a) (6 marks) Crystal "growth" is accretion (adding identical units), not biological growth. It lacks (any 3, 2 marks each):

• Metabolism — no anabolic/catabolic chemical energy transformations.
• Reproduction — does not produce offspring carrying heritable information (breaking a crystal is fragmentation, not reproduction with inheritance).
• Responsiveness/homeostasis — cannot detect and regulate its internal state against stimuli.
• Organisation into cells — not made of cells, the basic unit of life. (Conclusion: growth alone is insufficient; life requires the full set of characteristics.)

(b) (6 marks)

• Emergent property: a property arising from interactions among components at a given level that is not present in, nor predictable from, the parts alone (2).
• Cell level: life itself / self-replication — individual organelles or molecules (lipids, proteins, DNA) are not alive, but their organised interaction produces a living cell (2).
• Ecosystem level: nutrient cycling / energy flow / community stability — cannot be deduced from any single population; emerges from feeding relationships and abiotic interactions among many species (2).

(c) (6 marks)

• Hypothesis: a testable, falsifiable proposed explanation/prediction for a specific observation (1).
• Theory: a broad, well-substantiated explanation supported by large bodies of evidence, that explains why (1).
• Law: a concise (often mathematical) description of a consistent natural relationship that describes what happens without explaining why (1).
• Classification: (i) = law (quantitative descriptive rule, the $Q_{10}$ relationship) (1); (ii) = theory (cell theory, broad and evidence-backed) (1); (iii) = hypothesis (specific testable if-then prediction) (1).

[
  {"claim":"Steady-state temperature T* = 33.6 C", "code":"k=Rational(1,10); c=Rational(4,10); Te=20; Ts=37; Tstar=(k*Te+c*Ts)/(k+c); result = (Tstar == Rational(336,10))"},
  {"claim":"Time constant tau = 2.0 min", "code":"k=Rational(1,10); c=Rational(4,10); tau=1/(k+c); result = (tau == 2)"},
  {"claim":"Two-point log-log exponent n approx 0.8235", "code":"n=(log(116,10)-log(Rational(14,100),10))/(log(70,10)-log(Rational(2,100),10)); result = abs(float(n)-0.8235) < 0.01"},
  {"claim":"Kleiber constant a approx 4.79 W kg^-3/4 using human point", "code":"a=116/70**Rational(3,4); result = abs(float(a)-4.79) < 0.05"}
]