Reproductive System & Development

Question 1

(a) (6 marks)

• Low–moderate estradiol from developing follicle exerts negative feedback, suppressing GnRH/LH/FSH secretion (keeps gonadotropins low during early follicular phase). (2)
• When estradiol rises above a threshold and is sustained (~200 pg/mL for ~48 h), feedback switches to positive — high estradiol now stimulates GnRH and sensitises pituitary gonadotropes. (2)
• This triggers the massive LH surge (and smaller FSH surge), which induces meiosis I completion in the oocyte and ovulation ~24–36 h later. (2)

(b) (6 marks)

• $320 = 40\,e^{4k} \Rightarrow e^{4k}=8 \Rightarrow k = \tfrac{\ln 8}{4} = \tfrac{3\ln 2}{4}$. (2)
• Numeric: $k = 0.5199\approx 0.520\ \text{day}^{-1}$. (2)
• Threshold: $200 = 40 e^{k t^*} \Rightarrow e^{k t^*}=5 \Rightarrow t^* = \dfrac{\ln 5}{k} = \dfrac{\ln 5}{0.5199} = 3.096\approx 3.10$ days. (2)

(c) (8 marks)

• $\dfrac{dF}{dt} = a - bE_0 e^{kt}$. (1)
• Integrate: $F(t) = a t - \dfrac{bE_0}{k}e^{kt} + C$. (2)
• Apply $F(0)=12$: $12 = -\dfrac{bE_0}{k} + C \Rightarrow C = 12 + \dfrac{bE_0}{k}$. With $\dfrac{bE_0}{k}=\dfrac{0.05\cdot40}{0.5199}=3.847$, so $C = 15.85$. (2)
• Thus $F(t) = 10t - 3.847\,e^{0.5199 t} + 15.85$. (1)
• Slope at $t=2$: $\dfrac{dF}{dt}\big|_2 = 10 - 0.05\cdot40\,e^{2k} = 10 - 2\cdot e^{1.0397} = 10 - 2(2.828) = 10 - 5.657 = 4.34 >0$ ⇒ FSH still rising at $t=2$ but slowing. (1)
• Biological interpretation: as estradiol climbs, its inhibitory term grows; FSH growth decelerates and will turn to decline — the mid-follicular FSH dip produced by negative feedback selects the dominant follicle. (1)

Question 2

(a) (8 marks)

• Drag: $F = 6\pi\eta r v = 6\pi (10^{-2})(2.5\times10^{-6})(50\times10^{-6})$. $= 6\pi\times1.25\times10^{-12} = 2.356\times10^{-11}\ \text{N} \approx 2.4\times10^{-11}$ N. (3)
• Reynolds: $Re = \dfrac{\rho v r}{\eta} = \dfrac{1000\cdot50\times10^{-6}\cdot2.5\times10^{-6}}{10^{-2}} = \dfrac{1.25\times10^{-7}}{10^{-2}} = 1.25\times10^{-5}$. (3)
• $Re \ll 1$: viscous forces dominate inertia; there is no coasting — if flagellar beating stops, the sperm halts almost instantly (motion is time-reversible / "scallop theorem"), so continuous non-reciprocal beating is needed. (2)

(b) (6 marks)

• Effective speed $v_{\text{eff}} = v/5 = 10\ \mu\text{m/s} = 10\times10^{-6}$ m/s. (1)
• Time $t = L/v_{\text{eff}} = 0.15 / (10\times10^{-6}) = 1.5\times10^{4}$ s. (2)
• $=1.5\times10^4/60 = 250$ min. (1)
• This is far longer than ~30 min observed; the gap is closed by uterine/oviductal muscular contractions and ciliary currents actively transporting sperm, not sperm swimming alone. (2)

(c) (6 marks)

• Acrosome reaction: on contact with zona pellucida (ZP3 binding), the sperm acrosome releases hydrolytic enzymes (acrosin) digesting a path through the zona pellucida. (2)
• Cortical reaction: fusion triggers $Ca^{2+}$ release; cortical granules exocytose enzymes that harden/modify the zona (zona reaction), creating a slow block to polyspermy (fast block = membrane depolarisation). (2)
• Motility is energetically costly (ATP for flagellum) and the tract is long/hostile (acidic vagina, immune cells, mucus); attrition means only ~hundreds reach the egg from ~$3\times10^8$. (2)

Question 3

(a) (8 marks)

• Recurrence: $N_n = 2 N_{n-1},\ N_0 = 1$; closed form $N_n = 2^{n}$. (2)
• Volume per cell: $v_n = \dfrac{V_{\text{tot}}}{2^{n}}$ (total volume conserved). (2)
• Cell radius: $r_n = r_0 2^{-n/3}$ since $v\propto r^3$. Surface per cell $\propto r_n^2 = r_0^2 2^{-2n/3}$. (2)
• Total surface $S_n = 2^{n}\cdot(\text{area per cell}) = S_0\, 2^{n}\,2^{-2n/3} = S_0\,2^{n/3}$. Ratio $S_n/S_0 = 2^{n/3}$. (2)

(b) (6 marks)

def cleavage_cells(n):
    return 2**n
 
def surface_ratio(n):
    return 2**(n/3)   # total surface / original single-cell surface

• $n=0$: cells $=1$, ratio $=1$. (2)
• $n=1$: cells $=2$, ratio $=2^{1/3}\approx1.26$. (2)
• $n=4$: cells $=16$, ratio $=2^{4/3}\approx2.52$. (2) (Increasing surface-to-volume aids nutrient/gas exchange as blastula forms.)

(c) (6 marks)

• No failure in one cycle: $1-p$. Over $n$ independent cycles: $(1-p)^n$. (1)
• At least one failure: $P = 1-(1-p)^{n}$. (1)
• Perfect use $p=0.01$, $n=12$: $P = 1 - 0.99^{12} = 1 - 0.8864 = 0.1136 \approx 11.4\%$. (2)
• Typical use $p=0.09$: $P = 1 - 0.91^{12} = 1 - 0.3225 = 0.6775 \approx 67.7\%$ — dramatically higher, showing adherence matters enormously. (1)
• Mechanism: combined pills deliver estrogen+progestin that negatively feed back on the hypothalamus/pituitary, suppressing FSH/LH and preventing the LH surge → no ovulation (also thicken cervical mucus, thin endometrium). (1)

[
  {"claim":"Estradiol growth constant k = 3 ln2 /4 ≈ 0.5199",
   "code":"k = ln(8)/4; result = abs(float(k) - 0.5199) < 1e-3"},
  {"claim":"Threshold time t* = ln5/k ≈ 3.096 days",
   "code":"k = ln(8)/4; t = ln(5)/k; result = abs(float(t) - 3.096) < 1e-2"},
  {"claim":"dF/dt at t=2 is about 4.34 (>0, FSH rising)",
   "code":"k = ln(8)/4; dF = 10 - 0.05*40*exp(2*k); result = abs(float(dF) - 4.343) < 1e-2"},
  {"claim":"Stokes drag F ~ 2.356e-11 N",
   "code":"F = 6*pi*Rational(1,100)*Rational(25,10)*1e-6*50*1e-6; result = abs(float(F) - 2.356e-11) < 1e-13"},
  {"claim":"Reynolds number = 1.25e-5",
   "code":"Re = (1000*50e-6*2.5e-6)/1e-2; result = abs(float(Re) - 1.25e-5) < 1e-8"},
  {"claim":"Surface ratio at n=4 is 2^(4/3) ≈ 2.52",
   "code":"s = 2**(Rational(4,3)); result = abs(float(s) - 2.5198) < 1e-3"},
  {"claim":"Pill perfect-use annual failure prob ≈ 0.1136",
   "code":"P = 1 - 0.99**12; result = abs(float(P) - 0.1136) < 1e-3"},
  {"claim":"Typical-use annual failure prob ≈ 0.6775",
   "code":"P = 1 - 0.91**12; result = abs(float(P) - 0.6775) < 1e-3"}
]