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Principles of Inheritance and Variation - Polygenic Inheritance

Grade 12CBSEBiology

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

🔑Concepts

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Polygenic inheritance refers to the inheritance of a trait that is controlled by three or more genes, known as polygenes.

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Unlike Mendelian inheritance, which deals with discrete phenotypes (e.g., tall vs. dwarf), polygenic inheritance results in continuous variation in phenotypes.

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The phenotype in polygenic inheritance is determined by the additive effect of each dominant allele. For example, in human skin color, each dominant allele (AA, BB, or CC) contributes to a certain amount of melanin production.

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Polygenic traits are highly influenced by environmental factors, which smooth out the phenotypic differences to create a bell-shaped normal distribution curve.

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Human skin color is a classic example studied by Davenport. It is controlled by three genes: A,B,A, B, and CC. The genotype AABBCCAABBCC represents the darkest skin, while aabbccaabbcc represents the lightest skin.

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The intermediate phenotype, such as AaBbCcAaBbCc, is known as a 'mulatto' and represents the most common phenotype in a population resulting from such a cross.

📐Formulae

Number of Phenotypes=2n+1\text{Number of Phenotypes} = 2n + 1

Number of Genotypes=3n\text{Number of Genotypes} = 3^n

Frequency of Parental Extremes=(14)n\text{Frequency of Parental Extremes} = \left(\frac{1}{4}\right)^n

Where n=number of polygene pairs controlling the trait.\text{Where } n = \text{number of polygene pairs controlling the trait.}

💡Examples

Problem 1:

In a case of polygenic inheritance for human skin color controlled by 33 gene pairs (A,B,A, B, and CC), calculate the total number of possible phenotypes and the number of possible genotypes in the F2F_2 generation.

Solution:

Given n=3n = 3. Number of phenotypes: 2n+1=2(3)+1=72n + 1 = 2(3) + 1 = 7 Number of genotypes: 3n=33=273^n = 3^3 = 27

Explanation:

The number of phenotypes is calculated using the formula 2n+12n + 1 because each additive allele creates a distinct level of pigmentation. The number of genotypes follows 3n3^n as each gene pair can have three possible combinations (AA,Aa,aaAA, Aa, aa).

Problem 2:

Two individuals with the genotype AaBbCcAaBbCc (mulatto) are crossed. What is the probability of obtaining an offspring with the darkest skin color (AABBCCAABBCC)?

Solution:

The probability of an offspring being homozygous dominant for one gene pair (e.g., AAAA) in a dihybrid or trihybrid cross is 14\frac{1}{4}. Since there are 33 independent gene pairs: Probability=(14)×(14)×(14)=164\text{Probability} = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{64}

Explanation:

The darkest phenotype requires all 66 alleles to be dominant. The frequency of the extreme phenotypes in a polygenic cross is given by the formula (14)n(\frac{1}{4})^n, where nn is the number of gene pairs. Here, n=3n = 3, so (14)3=164(\frac{1}{4})^3 = \frac{1}{64}.