Monohybrid Cross

What a monohybrid cross studies

A monohybrid cross follows one gene with two alleles, one dominant and one recessive. It is used to predict the genotypes and phenotypes of offspring and to confirm patterns of inheritance.

Generations: P, F1, F2

• P (parental generation): the two organisms first crossed, often true-breeding (homozygous), e.g. TT x tt.
• F1 (first filial generation): the offspring of the P cross; if P is TT x tt, all F1 are Tt.
• F2 (second filial generation): the offspring from crossing two F1 individuals (Tt x Tt), which shows the classic 3:1 phenotype ratio.

Worked example: Tt x Tt

Take tallness in pea plants, where T (tall) is dominant over t (short). Two heterozygous tall plants, genotype Tt, are crossed.

Parental genotypes: Tt x Tt.

Gametes: each Tt parent produces gametes T and t.

The gametes combine as shown in the Punnett square below.

Reading the result

Offspring genotypes: 1 TT : 2 Tt : 1 tt. Since TT and Tt are both tall, the phenotype ratio is 3 tall : 1 short.

This 3:1 ratio is the signature result of crossing two heterozygotes for one gene, and appeared repeatedly in Mendel's own pea experiments.

How it is examined

SPM questions typically give a scenario (e.g. two heterozygous parents), and ask you to state the genotypes, draw the Punnett square with gametes in circles, and state the phenotype ratio of the offspring.

Worked problem 1 (easy): a true-breeding cross

In pea plants, purple flower colour (P) is dominant over white flower colour (p). A true-breeding purple plant, genotype PP, is crossed with a white plant, genotype pp.

Predict the genotype and phenotype of the F1 offspring.

Each PP parent forms only P gametes and each pp parent forms only p gametes, so every fertilisation joins a P gamete with a p gamete. All offspring are Pp and flower purple; there is no ratio, because every offspring shares one genotype.

Worked problem 2 (medium): two heterozygotes

Two purple F1 plants from the cross above, each Pp, are crossed together. Predict the genotype ratio and the phenotype ratio of their offspring.

Each Pp parent forms two kinds of gamete, P and p, in equal numbers. Combining them gives offspring genotypes of 1 PP : 2 Pp : 1 pp.

Because PP and Pp both flower purple, the phenotype ratio is 3 purple : 1 white.

Worked problem 3 (SPM-level): deduce the parents

A gardener crosses two purple-flowered pea plants and finds that some of the offspring flower white. State the genotypes of the two parents and give the expected phenotype ratio, explaining your reasoning.

A white plant is pp, so it carries two recessive alleles and must have received one p allele from each parent. Both parents flower purple, so neither can be pp, and each must carry a hidden p allele; both parents are therefore heterozygous, Pp.

The cross Pp x Pp gives offspring of 1 PP : 2 Pp : 1 pp, an expected phenotype ratio of 3 purple : 1 white. The white offspring are the clue that reveals both purple parents as carriers of the recessive allele.

Practice problems

1. In cattle, the polled (hornless) allele H is dominant over the horned allele h. A homozygous polled bull is crossed with a horned cow. State the genotype and phenotype of the F1 calves.
2. Two heterozygous polled cattle, each Hh, are mated. State the genotype ratio and the phenotype ratio of their offspring.
3. A polled cow mated repeatedly with a horned bull produces polled and horned calves in roughly equal numbers. Deduce the genotype of the cow and explain your reasoning.

Common mistakes

Source:SRC-DSKP-EN

Frequently asked questions