krit.club logo

Inheritance - Monohybrid inheritance

Grade 11IGCSEBiology

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

🔑Concepts

•

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

•

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

•

Genotype is the genetic makeup of an organism in terms of the alleles present (e.g., BBBB, BbBb, or bbbb).

•

Phenotype is the observable features of an organism (e.g., blue eyes or brown eyes).

•

Homozygous means having two identical alleles of a particular gene (e.g., AAAA is homozygous dominant, aaaa is homozygous recessive). An organism that is homozygous for a trait is 'pure-breeding'.

•

Heterozygous means having two different alleles of a particular gene (e.g., AaAa). Heterozygous organisms cannot be pure-breeding.

•

A dominant allele (AA) is an allele that is expressed if it is present. A recessive allele (aa) is only expressed when there is no dominant allele of the gene present.

•

A monohybrid cross is the study of the inheritance of one characteristic controlled by a single gene.

•

A test cross is used to identify an unknown genotype of an organism showing a dominant phenotype by crossing it with a homozygous recessive (aaaa) individual.

📐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 Recessive Phenotype from Aa×Aa=14 or 25%\text{Probability of Recessive Phenotype from } Aa \times Aa = \frac{1}{4} \text{ or } 25\%

Test Cross Result (if parent is Aa)=1:1 (Dominant : Recessive)\text{Test Cross Result (if parent is } Aa) = 1 : 1 \text{ (Dominant : Recessive)}

💡Examples

Problem 1:

In pea plants, the allele for tall stems (TT) is dominant over the allele for dwarf stems (tt). If two heterozygous tall plants (TtTt) are crossed, determine the expected genotypic and phenotypic ratios of the offspring.

Solution:

  1. Parents: Tt×TtTt \times Tt.
  2. Gametes: T,tT, t and T,tT, t.
  3. Punnett Square:
  • Top row: T,tT, t
  • Side column: T,tT, t
  • Offspring: TTTT (Tall), TtTt (Tall), TtTt (Tall), tttt (Dwarf).
  1. Genotypic ratio: 1TT:2Tt:1tt1 TT : 2 Tt : 1 tt.
  2. Phenotypic ratio: 3 Tall:1 Dwarf3 \text{ Tall} : 1 \text{ Dwarf}.

Explanation:

Because TT is dominant, both TTTT and TtTt genotypes result in the tall phenotype. Only the homozygous recessive tttt results in the dwarf phenotype.

Problem 2:

A farmer has a black sheep (Black wool is dominant, BB) but does not know if it is homozygous (BBBB) or heterozygous (BbBb). Describe how a test cross would reveal the genotype.

Solution:

  1. Cross the black sheep with a homozygous recessive white sheep (bbbb).
  2. If the black sheep is BBBB, all offspring will be BbBb (all black).
  3. If the black sheep is BbBb, the offspring ratio will be 1Bb:1bb1 Bb : 1 bb (50%50\% black and 50%50\% white).

Explanation:

If even a single white offspring (bbbb) is produced, the unknown parent must carry the recessive allele (bb) and therefore must be heterozygous (BbBb).

Monohybrid inheritance - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology