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Organisms and their Environment - Population size

Grade 11IGCSEBiology

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

🔑Concepts

A population is defined as a group of organisms of one species, living in the same area at the same time.

Population growth follows a sigmoid (S-shaped) curve consisting of four main phases: the lag phase, the log (exponential) phase, the stationary phase, and the death phase.

During the lag phase, the population size NN increases slowly as organisms adapt to their environment and reach reproductive age.

In the log phase, the growth rate is at its maximum because resources like food and H2OH_2O are abundant and limiting factors are minimal.

The stationary phase occurs when the birth rate equals the death rate, often because the population has reached the carrying capacity (KK) of the environment.

The death phase (or decline phase) occurs when the death rate exceeds the birth rate due to factors like accumulation of toxic waste, depletion of food, or disease.

Limiting factors are environmental factors that restrict population growth, such as light intensity, temperature, predation, and competition for resources.

The human population has been increasing exponentially due to improvements in agriculture, sanitation, and healthcare, which have significantly reduced the death rate (DD).

📐Formulae

ΔN=(B+I)(D+E)\Delta N = (B + I) - (D + E) where BB = Births, II = Immigration, DD = Deaths, and EE = Emigration.

Growth Rate=ΔNΔt\text{Growth Rate} = \frac{\Delta N}{\Delta t}

Percentage Change=New populationOld populationOld population×100\text{Percentage Change} = \frac{\text{New population} - \text{Old population}}{\text{Old population}} \times 100

💡Examples

Problem 1:

A population of N=500N = 500 beetles lives in a garden. Over one year, there are 120120 births, 6060 deaths, 1010 beetles immigrate, and 55 beetles emigrate. Calculate the new population size.

Solution:

ΔN=(120+10)(60+5)=13065=65\Delta N = (120 + 10) - (60 + 5) = 130 - 65 = 65 New Population=500+65=565\text{New Population} = 500 + 65 = 565

Explanation:

To find the change in population, we sum the factors that increase the population (births and immigration) and subtract the factors that decrease it (deaths and emigration). Adding this change to the initial population gives the final count.

Problem 2:

In a laboratory culture, a population of yeast cells increases from 200200 to 10001000 in 44 hours. Calculate the average growth rate per hour.

Solution:

ΔN=1000200=800\Delta N = 1000 - 200 = 800 Growth Rate=800 cells4 hours=200 cells/hour\text{Growth Rate} = \frac{800 \text{ cells}}{4 \text{ hours}} = 200 \text{ cells/hour}

Explanation:

The average growth rate is determined by dividing the total increase in individuals (ΔN\Delta N) by the total time interval (Δt\Delta t).

Population size - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology