Anatomy & Physiology Practice Exam — Free

Trace the reflex arc from receptor to effector. The receptor is the muscle spindle in the quadriceps, which detects the sudden change in muscle length. Its sensory neuron (a group Ia afferent) enters the spinal cord through the dorsal root and makes an excitatory, monosynaptic connection onto the alpha motor neurons of the quadriceps, producing the contraction that extends the knee.

The same afferent gives off collateral branches that synapse on a small inhibitory interneuron in the ventral horn. That interneuron releases an inhibitory transmitter (glycine) onto the motor neurons of the hamstrings, hyperpolarizing them and reducing their output. Because the agonist is excited and the antagonist inhibited by one and the same sensory input, the circuit is called reciprocal inhibition; it stops the two muscles from working against each other across the joint.

Hamstring relaxation is produced by reciprocal inhibition through an inhibitory spinal interneuron (option A).

Option C is the most tempting wrong answer, but primary sensory neurons entering the spinal cord are excitatory (they release glutamate); inhibition always requires an interneuron, which is why the antagonist pathway is disynaptic rather than monosynaptic. Option B describes the wrong receptor: Golgi tendon organs respond to tension, not length, and their reflex inhibits the muscle they sit in, not the antagonist. Option D is wrong because the stretch reflex is a spinal reflex that is complete before any cortical processing occurs; the brain can adjust the reflex's gain but is not needed for the antagonist to relax.

(a) Stroke volume is the blood ejected by one ventricle in one beat. The ventricle fills to the end-diastolic volume (EDV) and empties down to the end-systolic volume (ESV), so the difference is what left:

$$SV = EDV - ESV = 130\ \text{mL} - 60\ \text{mL} = 70\ \text{mL}$$

(b) Cardiac output is the volume pumped per minute, so multiply the volume per beat by the number of beats per minute:

$$CO = HR \times SV = 72\ \text{beats/min} \times 70\ \text{mL/beat} = 5040\ \text{mL/min} = 5.04\ \text{L/min}$$

This is a typical resting value, which is a useful sanity check: an answer of 504 L/min would mean a unit slip.

(c) Ejection fraction expresses the stroke volume as a fraction of the volume that was available to eject, which is the EDV, not the ESV:

$$EF = \frac{SV}{EDV} \times 100\% = \frac{70}{130} \times 100\% = 53.8\%$$

Note that each quantity answers a different question: SV is per beat, CO is per minute, and EF describes how completely the ventricle empties regardless of its size.

(a) SV = 70 mL; (b) CO = 5.04 L/min; (c) EF = 53.8 %.

(a) Clearance is the volume of plasma cleared of a substance per minute, $C = \dfrac{U \times V}{P}$. Because inulin is filtered freely and then neither reabsorbed nor secreted, every milligram that appears in the urine got there by filtration alone, so its clearance equals the GFR:

$$GFR = C_{in} = \frac{(25\ \text{mg/mL})(2.0\ \text{mL/min})}{0.50\ \text{mg/mL}} = \frac{50\ \text{mg/min}}{0.50\ \text{mg/mL}} = 100\ \text{mL/min}$$

(b) The filtration fraction is the portion of the plasma entering the kidneys that is filtered into Bowman's capsule:

$$FF = \frac{GFR}{RPF} = \frac{100\ \text{mL/min}}{600\ \text{mL/min}} = 0.167 \approx 16.7\%$$

(c) Apply the same clearance formula to X:

$$C_X = \frac{(6.0\ \text{mg/mL})(2.0\ \text{mL/min})}{0.20\ \text{mg/mL}} = 60\ \text{mL/min}$$

Compare with the GFR. A freely filtered substance whose clearance is below the GFR is being removed from the filtrate somewhere along the tubule, so X is net reabsorbed. To quantify it, compare the filtered load with the excretion rate:

$$\text{Filtered load} = GFR \times P_X = 100\ \text{mL/min} \times 0.20\ \text{mg/mL} = 20\ \text{mg/min}$$ $$\text{Excretion rate} = U_X \times V = 6.0\ \text{mg/mL} \times 2.0\ \text{mL/min} = 12\ \text{mg/min}$$ $$\text{Net reabsorption} = 20 - 12 = 8\ \text{mg/min}$$

Had the clearance of X exceeded 100 mL/min, the tubules would have been adding X to the filtrate, which is net secretion.

(a) GFR = 100 mL/min; (b) FF = 0.167 (16.7 %); (c) clearance of X = 60 mL/min, so X is net reabsorbed at 8 mg/min.