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Classification and Diversity - Variation within Species

Grade 7IB

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

🔑Concepts

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Variation refers to the differences that exist between individuals of the same species. It can be categorized into two main types: genetic variation and environmental variation.

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Genetic variation is caused by differences in DNADNA sequences and alleles inherited from parents. This includes traits like blood type (A,B,AB,OA, B, AB, O) or natural eye color.

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Environmental variation is caused by the conditions in which an organism lives, such as pHpH levels of soil affecting flower color or diet affecting an animal's mass (kgkg).

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Continuous variation features a range of values with no distinct categories. Examples include height in cmcm or hand span. When plotted, these usually form a 'Bell Curve' or normal distribution.

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Discontinuous variation occurs when characteristics fall into distinct, non-overlapping categories. Examples include the ability to roll your tongue or human blood groups. This is typically represented using a bar chart.

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The total variation in a phenotype (VPV_P) can be expressed as a result of the interaction between genotype (GG) and the environment (EE).

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Mutations are random changes in the DNADNA sequence that can introduce new variations into a population.

📐Formulae

V=G+EV = G + E

xˉ=∑xn\bar{x} = \frac{\sum x}{n}

Percentage frequency=fn×100%Percentage\ frequency = \frac{f}{n} \times 100\%

💡Examples

Problem 1:

In a class of 3030 students, 1212 students can roll their tongues. Calculate the percentage of students who show this discontinuous variation trait.

Solution:

1230×100=40%\frac{12}{30} \times 100 = 40\%

Explanation:

To find the percentage frequency, we take the frequency of the trait (f=12f = 12), divide by the total population (n=30n = 30), and multiply by 100100.

Problem 2:

A scientist measures the mass of five sunflower seeds as 0.05g,0.07g,0.06g,0.08g,0.05g, 0.07g, 0.06g, 0.08g, and 0.04g0.04g. Determine the mean mass and identify if this trait is continuous or discontinuous.

Solution:

Mean mass = 0.06g0.06g; Continuous variation.

Explanation:

The mean is calculated as 0.05+0.07+0.06+0.08+0.045=0.06g\frac{0.05 + 0.07 + 0.06 + 0.08 + 0.04}{5} = 0.06g. Mass is continuous variation because it can take any value within a range and is measured on a scale.

Problem 3:

Two plants of the same species are grown in different environments. Plant AA receives H2OH_2O and fertilizer, while Plant BB receives only H2OH_2O. Plant AA grows to 50cm50cm and Plant BB grows to 30cm30cm. Explain this variation.

Solution:

Environmental Variation.

Explanation:

Since the plants are the same species (likely sharing similar genetics), the difference in height (20cm20cm) is primarily due to environmental factors (EE), specifically the availability of nutrients in the fertilizer.