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Adaptation and Survival - Evolutionary Survival Strategies

Grade 9IB

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

🔑Concepts

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Natural Selection: The mechanism of evolution where organisms with favorable traits are more likely to survive (PsurvivalP_{survival}) and reproduce, passing these traits to the next generation.

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Variation: Differences in DNA sequences among individuals in a population, often caused by mutations or sexual reproduction (2n2^n combinations in meiosis).

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Structural Adaptations: Physical features of an organism's body that contribute to survival, such as the streamlined body of a shark for vmaxv_{max} in water.

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Behavioral Adaptations: Ways an organism acts to survive, such as the migration of birds to warmer climates during winter.

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Physiological Adaptations: Internal systemic responses to environmental stimuli, such as the production of concentrated urine by desert animals to conserve H2OH_2O.

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Selective Pressure: External factors (like predators, climate, or food availability) that affect an organism's ability to survive in a given environment.

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Fitness (ww): A measure of evolutionary success, defined by an individual's genetic contribution to the next generation relative to others.

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Speciation: The formation of new and distinct species in the course of evolution, often occurring when populations become geographically or reproductively isolated.

📐Formulae

Survival Rate (S)=NsurvivingNinitial×100%\text{Survival Rate (S)} = \frac{N_{surviving}}{N_{initial}} \times 100\%

Relative Fitness (w)=Survival/Reproduction rate of a phenotypeSurvival/Reproduction rate of the fittest phenotype\text{Relative Fitness (w)} = \frac{\text{Survival/Reproduction rate of a phenotype}}{\text{Survival/Reproduction rate of the fittest phenotype}}

Selection Coefficient (s)=1−w\text{Selection Coefficient (s)} = 1 - w

P+Q=1 (where P and Q are allele frequencies)P + Q = 1 \text{ (where P and Q are allele frequencies)}

💡Examples

Problem 1:

In a population of 200 rabbits, 40 possess a thick fur trait that allows them to survive extreme cold. After a severe winter, 30 of the thick-furred rabbits survive, while only 20 of the 160 thin-furred rabbits survive. Calculate the survival rate (SS) for both groups and determine which trait has the higher fitness.

Solution:

Sthick=3040×100=75%Sthin=20160×100=12.5%\begin{array}{r} S_{thick} = \frac{30}{40} \times 100 = 75\% \\ S_{thin} = \frac{20}{160} \times 100 = 12.5\% \end{array}

Explanation:

The survival rate of the thick-furred rabbits (75%75\%) is significantly higher than that of the thin-furred rabbits (12.5%12.5\%). This indicates that thick fur is a favorable structural adaptation under cold selective pressure, leading to higher biological fitness.

Problem 2:

If the maximum number of offspring produced by the 'fittest' genotype in a population is 10, and a different genotype produces an average of 4 offspring, calculate the relative fitness (ww) and the selection coefficient (ss) for the second genotype.

Solution:

w=410=0.4w = \frac{4}{10} = 0.4 s=1−0.4=0.6s = 1 - 0.4 = 0.6

Explanation:

The relative fitness (w=0.4w = 0.4) shows the second genotype is only 40%40\% as successful as the fittest. The selection coefficient (s=0.6s = 0.6) represents the intensity of natural selection acting against that genotype.