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Variation and Selection - Variation and mutation

Grade 11IGCSEBiology

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

🔑Concepts

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Variation is defined as the differences between individuals of the same species.

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Discontinuous variation results in a limited number of distinct phenotypes with no intermediates, such as the ABOABO blood group system (AA, BB, ABAB, and OO).

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Continuous variation results in a range of phenotypes between two extremes, such as body length or body mass, and is often represented by a normal distribution curve.

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Phenotype is the result of the interaction between the genotype and the environment, represented as Phenotype=Genotype+EnvironmentPhenotype = Genotype + Environment.

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Mutation is a random change in the base sequence of DNADNA, which is the source of new alleles.

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Ionizing radiation, such as γ\gamma-rays, XX-rays, and ultraviolet (UVUV) light, can increase the rate of mutation.

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Chemical mutagens, such as those found in tobacco smoke, increase the frequency of mutations in DNADNA.

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Sickle-cell anaemia is caused by a specific gene mutation that changes the base sequence of DNADNA, resulting in the production of abnormal haemoglobin (HbSHb^S).

📐Formulae

Phenotype=Genotype+EnvironmentPhenotype = Genotype + Environment

Mutation Rate=Number of mutationsTotal number of gene copies per generation\text{Mutation Rate} = \frac{\text{Number of mutations}}{\text{Total number of gene copies per generation}}

DNA sequence change→mRNA sequence change→Amino acid sequence change→Altered Protein Structure\text{DNA sequence change} \rightarrow \text{mRNA sequence change} \rightarrow \text{Amino acid sequence change} \rightarrow \text{Altered Protein Structure}

💡Examples

Problem 1:

Identify whether the following human characteristics represent continuous or discontinuous variation: 1. ABOABO Blood Groups, 2. Height ranging from 150 cm150\text{ cm} to 200 cm200\text{ cm}.

Solution:

  1. Discontinuous variation. 2. Continuous variation.

Explanation:

ABOABO blood groups fall into discrete categories (no intermediates), while height shows a complete range of values influenced by both genes and environmental factors like nutrition.

Problem 2:

Explain the genetic basis of Sickle-cell anaemia using the alleles HbAHb^A (normal) and HbSHb^S (sickle cell). What is the phenotype of an individual with genotype HbAHbSHb^A Hb^S?

Solution:

The individual is a carrier of the sickle-cell trait and typically shows high resistance to Malaria.

Explanation:

Sickle-cell anaemia is caused by a mutation in the β\beta-globin gene. Homozygous HbSHbSHb^S Hb^S individuals suffer from the disease, while heterozygous HbAHbSHb^A Hb^S individuals have a selective advantage in areas where Malaria is endemic.

Problem 3:

Calculate the percentage increase in mutation rate if the background radiation increases the mutation frequency from 1×10−61 \times 10^{-6} to 5×10−65 \times 10^{-6} per gene per generation.

Solution:

400%400\% increase.

Explanation:

The increase is calculated using the formula: New Value−Old ValueOld Value×100\frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100. Here, 5×10−6−1×10−61×10−6×100=4×100=400%\frac{5\times 10^{-6} - 1\times 10^{-6}}{1\times 10^{-6}} \times 100 = 4 \times 100 = 400\%.

Variation and mutation - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology