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Biology - Human Physiology (Systems: Circulatory, Respiratory, and Digestive)

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Mechanical vs. Chemical Digestion: Digestion involves the breakdown of food. Chemical digestion is mediated by enzymes such as Amylase (converting starch to maltose), Protease (proteins to amino acids), and Lipase (lipids to fatty acids and glycerol). The reaction for glucose oxidation during respiration is C6H12O6+6O26CO2+6H2OC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O.

Absorption and Surface Area: The small intestine contains villi and microvilli to increase the surface area (AA) for nutrient absorption. The rate of diffusion is influenced by Fick's Law, where RateSurface Area×Concentration GradientDiffusion DistanceRate \propto \frac{Surface\ Area \times Concentration\ Gradient}{Diffusion\ Distance}.

Circulatory System Structure: The heart functions as a double pump. The systemic circuit carries oxygenated blood to the body, while the pulmonary circuit carries deoxygenated blood to the lungs. Blood pressure is highest in the Aorta and lowest in the Vena Cava, following the gradient ΔP\Delta P.

Gas Exchange in Alveoli: Gas exchange occurs via simple diffusion. O2O_2 moves from the alveoli (high PO2P_{O_2}) to the capillaries (low PO2P_{O_2}), while CO2CO_2 moves in the opposite direction following its own partial pressure gradient.

Hemoglobin and Oxygen Transport: Oxygen binds to hemoglobin in red blood cells to form oxyhemoglobin: Hb+4O2Hb(O2)4Hb + 4O_2 \rightleftharpoons Hb(O_2)_4. The affinity of hemoglobin for oxygen changes based on the pHpH and CO2CO_2 concentration (the Bohr Shift).

Breathing Mechanics: Ventilation is driven by volume and pressure changes. According to Boyle's Law, P1VP \propto \frac{1}{V}. When the diaphragm contracts, thoracic volume (VV) increases, causing pressure (PP) to decrease below atmospheric pressure (PatmP_{atm}), forcing air into the lungs.

📐Formulae

C6H12O6+6O26CO2+6H2O+3638 ATPC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + 36\text{--}38\ ATP

C6H12O62C3H6O3+2 ATPC_6H_{12}O_6 \rightarrow 2C_3H_6O_3 + 2\ ATP

CO=HR×SVCO = HR \times SV

VE=VT×fV_E = V_T \times f

P1V1=P2V2P_1V_1 = P_2V_2

💡Examples

Problem 1:

Calculate the Cardiac Output (COCO) for an athlete whose heart rate (HRHR) is 60 bpm60\ bpm and whose stroke volume (SVSV) is 80 mL80\ mL.

Solution:

CO=60 bpm×80 mL=4800 mL/minCO = 60\ bpm \times 80\ mL = 4800\ mL/min or 4.8 L/min4.8\ L/min.

Explanation:

Cardiac Output is defined as the volume of blood pumped by the heart per minute, calculated by the product of the heart rate and the volume of blood pumped per beat (stroke volume).

Problem 2:

A student measures their ventilation rate. If they breathe 1515 times per minute (frequency ff) and each breath has a tidal volume (VTV_T) of 0.5 L0.5\ L, what is their total ventilation (VEV_E)?

Solution:

VE=0.5 L×15=7.5 L/minV_E = 0.5\ L \times 15 = 7.5\ L/min.

Explanation:

The total ventilation rate is the total volume of gas entering the lungs per minute, found by multiplying the volume per breath by the number of breaths per minute.

Problem 3:

Identify the chemical change when starch is acted upon by salivary amylase in a solution at pH7pH \approx 7.

Solution:

Starch+H2OAmylaseMaltoseStarch + H_2O \xrightarrow{Amylase} Maltose

Explanation:

Amylase is a carbohydrate enzyme that catalyzes the hydrolysis of the glycosidic bonds in starch to produce simpler disaccharides like maltose.