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Principles of Inheritance and Variation - Inheritance of One Gene (Monohybrid Cross)

Grade 12CBSEBiology

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

🔑Concepts

A Monohybrid Cross is a genetic cross between two individuals involving a single pair of contrasting characters (e.g., stem height: Tall vs Dwarf).

Mendel's Law of Dominance states that in a heterozygote (TtTt), one allele (dominant) masks the expression of the other allele (recessive). Thus, the F1F_1 generation always expresses the dominant trait.

The Law of Segregation (Purity of Gametes) states that the two alleles for a trait separate during gamete formation, such that each gamete receives only one allele. This is a universal law with no exceptions in diploid organisms.

The Punnett Square, developed by Reginald C. Punnett, is a graphical representation used to calculate the probability of all possible genotypes of offspring in a genetic cross.

A Test Cross is performed to identify the genotype of an organism showing a dominant phenotype. The organism is crossed with a homozygous recessive parent (tttt). If the offspring are all tall, the parent was TTTT; if the ratio is 1:11:1, the parent was TtTt.

In the F2F_2 generation of a monohybrid cross, the phenotypic expression follows a 3:13:1 ratio, while the underlying genetic makeup follows a 1:2:11:2:1 ratio.

📐Formulae

Phenotypic Ratio (F2)=3:1\text{Phenotypic Ratio (} F_2 \text{)} = 3 : 1

Genotypic Ratio (F2)=1(TT):2(Tt):1(tt)\text{Genotypic Ratio (} F_2 \text{)} = 1 (TT) : 2 (Tt) : 1 (tt)

Test Cross Ratio (Heterozygous)=1(Tt):1(tt)\text{Test Cross Ratio (Heterozygous)} = 1 (Tt) : 1 (tt)

Probability of homozygous dominant in F2=14\text{Probability of homozygous dominant in } F_2 = \frac{1}{4}

Probability of heterozygous in F2=12\text{Probability of heterozygous in } F_2 = \frac{1}{2}

💡Examples

Problem 1:

In a monohybrid cross between a pure tall pea plant (TTTT) and a pure dwarf pea plant (tttt), calculate the number of dwarf plants expected in the F2F_2 generation if the total number of offspring produced is 800800.

Solution:

The phenotypic ratio in the F2F_2 generation is 3:13:1 (Tall:Dwarf). The fraction of dwarf plants is 14\frac{1}{4}. Total plants = 800800. Number of dwarf plants = 14×800=200\frac{1}{4} \times 800 = 200.

Explanation:

According to the Law of Segregation, the F1F_1 generation (TtTt) produces two types of gametes, TT and tt, in equal proportion. Selfing Tt×TtTt \times Tt results in 25%25\% homozygous tall (TTTT), 50%50\% heterozygous tall (TtTt), and 25%25\% homozygous dwarf (tttt).

Problem 2:

A pea plant with violet flowers (dominant) is crossed with a plant with white flowers (recessive). If the resulting progeny shows 50%50\% violet and 50%50\% white flowers, determine the genotype of the parent violet plant.

Solution:

The ratio 1:11:1 indicates a Test Cross. Let VV be the allele for violet and vv be the allele for white. Since white flowers are vvvv, and some offspring are white (vvvv), the violet parent must be heterozygous (VvVv).

Explanation:

If the parent were homozygous dominant (VVVV), all offspring would be VvVv (violet). The appearance of the recessive trait in 50%50\% of the offspring proves the parent is a heterozygote: Vv×vv1Vv:1vvVv \times vv \rightarrow 1 Vv : 1 vv.

Inheritance of One Gene (Monohybrid Cross) Revision - Class 12 Biology CBSE