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Biology - Transport in Animals

Grade 10IGCSE

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

🔑Concepts

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Circulatory Systems: Mammals have a double circulatory system, meaning blood passes through the heart twice for every complete circuit. This consists of the pulmonary circulation (to the lungs) and systemic circulation (to the rest of the body).

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Heart Structure: The heart consists of four chambers: the right atrium, right ventricle, left atrium, and left ventricle. The muscular wall of the left ventricle is significantly thicker than the right because it must pump blood at high pressure to the entire body.

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Blood Vessels: Arteries carry blood away from the heart at high pressure (thick elastic walls); Veins carry blood towards the heart at low pressure and contain valves to prevent backflow; Capillaries are one-cell thick to allow for the diffusion of substances like O2O_2 and CO2CO_2.

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Red Blood Cells (Erythrocytes): Specialized for oxygen transport. They contain hemoglobin, which binds to oxygen to form oxyhemoglobin: Hb+4O2ightleftharpoonsHb(O2)4Hb + 4O_2 ightleftharpoons Hb(O_2)_4. They lack a nucleus to provide more space for hemoglobin and have a biconcave shape to increase the surface area to volume ratio (SA:VSA:V).

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White Blood Cells: Phagocytes ingest pathogens via phagocytosis, while Lymphocytes produce antibodies to neutralize specific pathogens.

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Tissue Fluid: Formed when plasma is forced out of capillaries due to high hydrostatic pressure. It bathes cells, allowing for the exchange of nutrients and waste products like glucose and CO2CO_2.

📐Formulae

Cardiac Output=Stroke Volume×Heart RateCardiac\ Output = Stroke\ Volume \times Heart\ Rate

SA:V=Surface AreaVolumeSA:V = \frac{Surface\ Area}{Volume}

Pulse Rate=Number of beatsTime (minutes)Pulse\ Rate = \frac{Number\ of\ beats}{Time\ (minutes)}

💡Examples

Problem 1:

A student measures their pulse and counts 1818 beats in 1515 seconds. Calculate the heart rate in beats per minute (bpm).

Solution:

Heart Rate=18 beats15 s×60 s/min=72 bpmHeart\ Rate = \frac{18\ beats}{15\ s} \times 60\ s/min = 72\ bpm

Explanation:

To find the beats per minute, the number of beats in a shorter interval is multiplied by the factor required to reach 6060 seconds. Here, 60/15=460 / 15 = 4, so 18×4=7218 \times 4 = 72.

Problem 2:

Explain why a large multicellular organism needs a transport system compared to a unicellular organism using the concept of SA:VSA:V.

Solution:

As an organism increases in size, its volume (L3L^3) increases much faster than its surface area (L2L^2). For a cube of side 1 cm1\ cm, SA:V=6(1)213=6:1SA:V = \frac{6(1)^2}{1^3} = 6:1. For a cube of side 10 cm10\ cm, SA:V=6(10)2103=0.6:1SA:V = \frac{6(10)^2}{10^3} = 0.6:1.

Explanation:

A low SA:VSA:V ratio means the surface area is insufficient to supply the large volume of the organism with nutrients via simple diffusion alone. Therefore, a specialized transport system (the circulatory system) is required to move substances internally.