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Inheritance - Monohybrid inheritance

Grade 12IGCSEBiology

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

🔑Concepts

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Inheritance is the transmission of genetic information from generation to generation via gametes.

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A gene is a length of DNADNA that codes for a specific protein, while an allele is an alternative form of a gene (e.g., TT for tall and tt for short).

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The Genotype is the genetic makeup of an organism in terms of the alleles present (e.g., AAAA, AaAa, or aaaa).

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The Phenotype is the observable physical or physiological feature of an organism (e.g., blue eyes or tall height).

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Homozygous means having two identical alleles of a particular gene (e.g., TTTT or tttt). Pure-breeding individuals are homozygous.

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Heterozygous means having two different alleles of a particular gene (e.g., TtTt), where the dominant allele masks the expression of the recessive allele.

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A Monohybrid cross involves the study of the inheritance of a single characteristic controlled by one gene locus.

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A Test Cross is performed by crossing an individual with a dominant phenotype but unknown genotype with a homozygous recessive (aaaa) individual to determine if the unknown is AAAA or AaAa.

📐Formulae

Phenotypic Ratio (Heterozygous Cross)=3:1\text{Phenotypic Ratio (Heterozygous Cross)} = 3 : 1

Genotypic Ratio (Heterozygous Cross)=1:2:1\text{Genotypic Ratio (Heterozygous Cross)} = 1 : 2 : 1

Probability of an allele in a gamete=12\text{Probability of an allele in a gamete} = \frac{1}{2}

Percentage of Recessive Phenotype in F2=(14)×100=25%\text{Percentage of Recessive Phenotype in } F_2 = \left( \frac{1}{4} \right) \times 100 = 25\%

💡Examples

Problem 1:

In pea plants, the allele for smooth seeds (SS) is dominant over the allele for wrinkled seeds (ss). If two heterozygous plants (SsSs) are crossed, calculate the probability of the offspring having wrinkled seeds.

Solution:

The cross is Ss×SsSs \times Ss. The resulting genotypes are: 1×SS1 \times SS (Smooth), 2×Ss2 \times Ss (Smooth), and 1×ss1 \times ss (Wrinkled). The probability of wrinkled seeds is 25%25\% or 14\frac{1}{4}.

Explanation:

Using a Punnett square, the gametes from both parents are SS and ss. The possible combinations are SSSS, SsSs, sSsS, and ssss. Since ss is recessive, only the ssss genotype results in the wrinkled phenotype.

Problem 2:

A farmer wants to know if a black sheep has a homozygous dominant (BBBB) or heterozygous (BbBb) genotype. Black (BB) is dominant over white (bb). Describe the cross required.

Solution:

Perform a test cross by mating the black sheep with a homozygous recessive white sheep (bbbb). If any offspring are white (bbbb), the black sheep must be BbBb.

Explanation:

If the parent is BBBB, all offspring will be BbBb (Black) because BB×bb→100%BbBB \times bb \rightarrow 100\% Bb. If the parent is BbBb, there is a 50%50\% chance for white offspring because Bb×bb→50%BbBb \times bb \rightarrow 50\% Bb (Black) and 50%bb50\% bb (White).

Monohybrid inheritance - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology