BiologyCBSE Class 12 · Principles of Inheritance and Variation

Monohybrid Cross and Test Cross

Updated 15 October 2026 · Fact-checked

A monohybrid cross follows one character, such as tall versus dwarf pea plants. F1 shows only the dominant trait; selfing F1 (Tt × Tt) gives F2 in a 3:1 phenotypic and 1:2:1 genotypic ratio, explained by dominance and segregation. A test cross mates a dominant-looking plant with the recessive parent: 1:1 offspring means heterozygous, all dominant means homozygous.

Understand Monohybrid Cross and Test Cross

Mendel's set-up. Mendel worked on garden pea (*Pisum sativum*) for seven years (1856-1863), using large samples and counting offspring, which gave his results statistical weight. He chose true-breeding lines (which show the same trait over generations of self-pollination) for 7 pairs of contrasting traits: stem height (tall/dwarf), flower colour (violet/white), flower position (axial/terminal), pod shape (inflated/constricted), pod colour (green/yellow), seed shape (round/wrinkled) and seed colour (yellow/green).

The monohybrid cross. Tall (TT) × dwarf (tt):

F1: all tall (Tt). No intermediate heights; the dwarf trait is hidden.
F2 (F1 selfed, Tt × Tt): 3 tall : 1 dwarf. The dwarf trait reappears unchanged in 1/4 of plants.

Punnett square for Tt × Tt (gametes T and t from each parent):

	T	t
T	TT	Tt
t	Tt	tt

Genotypic ratio 1 TT : 2 Tt : 1 tt.
Phenotypic ratio 3 tall : 1 dwarf.

Terms: Mendel's 'factors' are genes; alternative forms of a gene are alleles (T, t). TT and tt are homozygous; Tt is heterozygous. The physical trait is the phenotype; the allele combination is the genotype. The checkerboard was devised by Reginald C. Punnett.

Law of dominance. (i) Each character is governed by discrete units, the factors; (ii) the factors come in pairs; (iii) when the two factors of a pair differ, one (the dominant) masks the other (the recessive). This explains why F1 shows only one parental trait and why F2 shows 3:1.

Law of segregation. Alleles do not blend. During gamete formation, the two alleles of a pair separate so that each gamete receives only one. A homozygote forms one kind of gamete; a heterozygote forms two kinds (T and t) in equal proportion. This is why the recessive trait comes back unchanged in F2.

Test cross. A tall plant may be TT or Tt; they look the same. Cross it with the recessive parent (tt) instead of selfing:

TT × tt → all Tt: all tall → the plant was homozygous.
Tt × tt → Tt and tt: 1 tall : 1 dwarf → the plant was heterozygous.

The recessive parent contributes only recessive alleles, so the offspring directly reveal the gametes of the unknown parent.

Selfing as an alternative. Self a dominant-looking plant: TT gives all tall; Tt gives 3 tall : 1 dwarf.

When a monohybrid F2 is not 3:1.

Incomplete dominance: in snapdragon (*Antirrhinum*), red (RR) × white (rr) → pink F1 (Rr); F2 is 1 red : 2 pink : 1 white. The phenotypic ratio equals the genotypic ratio because the heterozygote is intermediate.
Dominance depends on the level of phenotype: in pea, a gene for starch synthesis has alleles B and b. BB makes large starch grains (round seeds); bb makes small grains (wrinkled seeds); Bb seeds are round (B dominant for shape) but have intermediate-sized starch grains (incomplete dominance for grain size).
Co-dominance: both alleles show fully in the heterozygote. ABO blood groups are controlled by gene I with three alleles (multiple alleles): $I^{A}$ and $I^{B}$ each make a slightly different sugar on red blood cells; i makes none. $I^{A}$ and $I^{B}$ are dominant over i but co-dominant with each other ( $I^{A}$ $I^{B}$ = group AB). There are 6 genotypes and 4 phenotypes.

Why one allele is dominant. Usually the unmodified allele makes a working enzyme; a modified allele may make a normal, a less efficient or a non-functional enzyme (or none). If one working copy is enough, the phenotype looks normal, so the working allele is dominant.

Flow chart: true-breeding parents (TT × tt) → gametes T, t → F1 Tt (tall) → selfing → gametes T, t from each (segregation) → F2 TT, Tt, Tt, tt → 3:1 phenotype, 1:2:1 genotype.

Monohybrid Cross and Test Cross: key facts

F2 of a monohybrid cross: $T t \times T t \to \frac{1}{4} T T + \frac{1}{2} T t + \frac{1}{4} tt$; Phenotype 3 dominant : 1 recessive.
Test cross: $T T \times tt \to all dominant; T t \times tt \to 1 : 1$; Always cross with the homozygous recessive.
Fraction of dominant F2 plants that are heterozygous: $P (T t ∣ tall) = \frac{2/4}{3/4} = \frac{2}{3}$; Only 1/3 of tall F2 plants breed true.
Incomplete dominance F2: $R r \times R r \to 1 red : 2 pink : 1 white$; Phenotypic ratio = genotypic ratio.
ABO blood groups: $I^{A} I^{A}, I^{A} i \to A; I^{B} I^{B}, I^{B} i \to B; I^{A} I^{B} \to A B; ii \to O$; 3 alleles, 6 genotypes, 4 phenotypes.

How to answer Monohybrid Cross and Test Cross questions

Every one-gene problem follows the same seven steps. Write each step; each carries part of the marks.

1Choose symbols: capital for dominant, small of the same letter for recessive (T/t). For incomplete or co-dominance use superscripts or two capitals (R/r with pink Rr; $I^{A}$ , $I^{B}$ , i).
2Write the parents' genotypes with their phenotypes (e.g. TT tall × tt dwarf).
3Write the gametes: one allele per gamete (segregation); a homozygote gives one type, a heterozygote two.
4Fill a Punnett square for F1, then for F2 (or for the test cross).
5Count genotypes and convert to phenotypes; state both ratios with labels.
6For numbers, multiply the total by each fraction and check that the parts add up to the total.
7Conclude with the law shown: dominance (F1 uniform), segregation (recessive reappears in F2), or the deviation (incomplete dominance or co-dominance).

Monohybrid Cross and Test Cross in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Test cross to check a seed stock

Suppose a seed company buys 'pure violet-flowered' pea seeds and wants to be sure they are truly homozygous before selling them. It test-crosses one plant grown from the stock with a white-flowered plant and gets 61 violet and 59 white offspring.

What was the genotype of the plant tested? Was the seller's claim correct? What result would have supported the claim?

Try it first, then show the solutionSolution

Let V = violet (dominant), v = white. The white tester is vv and gives only v gametes.
Offspring 61 violet : 59 white ≈ 1 : 1, which arises only if the tested plant gave V and v gametes equally: it was Vv (heterozygous).
So the stock is not pure. A pure VV plant would have given all violet (Vv) offspring in the test cross.

Answer: The tested plant was Vv; the claim was wrong. All-violet offspring would have shown VV.

Memory hook: Half and half means hidden recessive; all dominant means pure.

Tricks to remember Monohybrid Cross and Test Cross

Short memory aids. Say each one aloud once, then use it on a question the same day.

3:1 hides 1:2:1
TT and Tt both look tall, so 1 + 2 = 3 tall against 1 dwarf.
Use it when: Moving between phenotypic and genotypic ratios.
Test with the weakest
Test cross = cross with the homozygous recessive, the 'weakest' genotype, which hides nothing.
Use it when: Remembering which parent to use.
Pink = in-between, AB = both
Incomplete dominance blends to pink; co-dominance shows both A and B.
Use it when: Telling the two deviations apart.

Monohybrid Cross and Test Cross: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Rr × rr · 1 mark · MCQBoard level

In pea, round seed (R) is dominant over wrinkled seed (r). What fraction of the offspring of Rr × rr will have wrinkled seeds?
(a) 0 (b) 1/4 (c) 1/2 (d) 3/4

Try it first, then show the solutionSolution

Rr gives R and r gametes equally; rr gives only r.
Offspring: 1 Rr (round) : 1 rr (wrinkled), so half are wrinkled.

Answer: (c) 1/2.

Example 2: Test cross with a Punnett square · 2 marks · Very short answerBoard level

Green pod (G) is dominant over yellow pod (g) in pea. Using Punnett squares, show the test crosses of a green-podded plant of unknown genotype and interpret the results.

Try it first, then show the solutionSolution

If GG × gg: gametes G and g; all offspring Gg, green: the plant is homozygous (1 mark).
If Gg × gg: gametes G, g and g; offspring Gg (green) and gg (yellow) in 1:1: the plant is heterozygous (1 mark).

Answer: All green → GG; 1 green : 1 yellow → Gg.

Example 3: Monohybrid cross to F2 and the laws · 5 marks · Long answerHarder board level

Work out a monohybrid cross between true-breeding tall and dwarf pea plants up to F2. Give the phenotypic and genotypic ratios and explain which law each generation demonstrates.

Try it first, then show the solutionSolution

Parents TT (tall) × tt (dwarf); gametes T and t (1 mark).
F1: all Tt, tall. Only one parental trait appears: law of dominance (1 mark).
F1 selfed: Tt × Tt; each forms T and t gametes in equal numbers (1 mark).
Punnett square: TT, Tt, Tt, tt; phenotypic ratio 3 tall : 1 dwarf; genotypic 1 TT : 2 Tt : 1 tt (1 mark).
The dwarf trait reappears unchanged in F2 because alleles separate into different gametes without blending: law of segregation (1 mark).

Answer: F1 all Tt (tall) → dominance; F2 3:1 and 1:2:1 → segregation.

Example 4: Snapdragon F2 counts · 3 marks · Short answerHarder board level

A red-flowered snapdragon (RR) is crossed with a white-flowered one (rr). F1 plants are selfed and 480 F2 plants are raised. How many red, pink and white plants are expected? Why does F2 not show 3:1?

Try it first, then show the solutionSolution

F1: Rr, pink (incomplete dominance) (1 mark).
F2: 1 RR : 2 Rr : 1 rr = red 480/4 = 120, pink 480/2 = 240, white 120 (1 mark).
Neither allele fully masks the other, so the heterozygote is intermediate and is counted separately: phenotypic ratio equals the genotypic ratio 1:2:1 (1 mark).

Answer: 120 red, 240 pink, 120 white; incomplete dominance makes Rr pink.

Example 5: Round seeds, intermediate grains · 3 marks · Short answerHOTS / competency

In pea, BB seeds are round with large starch grains, bb seeds are wrinkled with small grains, and Bb seeds are round with intermediate grains. Is B dominant or incompletely dominant? Explain how the answer depends on the phenotype chosen.

Try it first, then show the solutionSolution

If seed shape is the phenotype, Bb is round like BB, so B is completely dominant over b (1 mark).
If starch grain size is the phenotype, Bb has intermediate grains, so B shows incomplete dominance (1 mark).
Dominance is not a fixed property of an allele; it depends on which aspect of the gene's product (phenotype) we observe (1 mark).

Answer: Dominant for seed shape; incompletely dominant for grain size; dominance depends on the phenotype considered.

Example 6: Unknown tall violet plant by selfing · 3 marks · Short answerHarder board level

A tall pea plant with violet flowers has an unknown genotype for height. Using only selfing, show how you would find whether it is TT or Tt, giving the ratios expected in each case.

Try it first, then show the solutionSolution

Self the plant and raise the offspring (1 mark).
If TT: all gametes T, all offspring TT, all tall; genotypic and phenotypic ratio 1 (all tall) (1 mark).
If Tt: offspring 1 TT : 2 Tt : 1 tt; phenotype 3 tall : 1 dwarf; appearance of dwarf plants proves Tt (1 mark).

Answer: All tall on selfing → TT; 3 tall : 1 dwarf → Tt.

Where students lose marks in Monohybrid Cross and Test Cross

The slips examiners see most often, and what to write instead.

Mistake: Writing two alleles in a gamete
Students copy the parent genotype into the gamete row.
Wrong working
Gametes of Tt: Tt.
Right working
Gametes of Tt: T and t (segregation).
Do this: One allele per gamete, always.
Usually costs: 1-2 marks in crosses.
Mistake: Unlabelled ratios
Students write '1:2:1' without saying which ratio.
Wrong working
F2 ratio is 1:2:1.
Right working
Genotypic 1:2:1; phenotypic 3:1.
Do this: Label every ratio as genotypic or phenotypic.
Usually costs: 1 mark.
Mistake: Test-crossing with the dominant parent
Back cross and test cross are confused.
Wrong working
Test cross: F1 Tt × TT.
Right working
Test cross: Tt × tt (recessive parent).
Do this: A test cross is a back cross with the recessive parent.
Usually costs: 1-2 marks.
Mistake: Calling the AB group a case of incomplete dominance
Any heterozygote that differs from parents is called incomplete dominance.
Wrong working
$I^{A}$ $I^{B}$ shows incomplete dominance.
Right working
$I^{A}$ $I^{B}$ shows co-dominance: both sugars are fully present.
Do this: Blend → incomplete; both full → co-dominance.
Usually costs: 1 mark.

Exam tips

Show every step of a cross: parents, gametes, F1, F1 selfed, gametes, Punnett square, ratios, conclusion.
Use the same letter for both alleles (T/t), never two different letters.
For 5-mark questions, end with the law shown by each generation.
For probability questions, write the fraction first and multiply for independent events.
Name the example for each deviation: snapdragon (incomplete dominance), ABO blood groups (co-dominance and multiple alleles), pea starch grains (dominance depends on phenotype).

Practice questions on Monohybrid Cross and Test Cross

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): In a monohybrid cross, the F2 genotypic and phenotypic ratios are the same when the inheritance shows incomplete dominance.
Reason (R): In incomplete dominance the heterozygote has a phenotype different from both homozygotes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Pink is a third phenotype.
Answer
(a).
Q2 · 1 mark · MCQHarder board level
The first child of a couple, both with normal skin pigmentation, is albino (an autosomal recessive trait). What is the probability that their second child is also albino?
(a) 0 (b) 1/4 (c) 1/2 (d) 3/4
Hint
What must both parents be?
Answer
(b) 1/4. Both parents must be carriers (Aa); each child independently has a 1/4 chance of aa.
Q3 · 2 marks · Very short answerHOTS / competency
Two tall F2 plants from a Tt × Tt cross are chosen at random and crossed with each other. What is the probability that their first offspring is dwarf?
Hint
Each tall F2 plant is Tt with probability 2/3.
Answer
A dwarf offspring needs both parents to be Tt: (2/3) × (2/3) = 4/9; then Tt × Tt gives tt with 1/4. Probability = 4/9 × 1/4 = 1/9.
Q4 · 3 marks · Short answerHarder board level
In shorthorn cattle, a calf that gets the red-coat allele from one parent and the white-coat allele from the other has a roan coat, with red and white hairs both present. (a) Name this pattern of inheritance. (b) Give the offspring ratio of roan × roan. (c) How does this differ from pink snapdragons?
Hint
Both colours are fully expressed.
Answer
(a) Co-dominance: both alleles are fully expressed in the heterozygote. (b) 1 red : 2 roan : 1 white. (c) In snapdragon (incomplete dominance) the heterozygote shows a blended intermediate colour (pink); in roan cattle each colour is expressed separately (red hairs and white hairs together).
Q5 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a student test-crosses three tall pea plants P, Q and R with dwarf plants and records: P → 98 tall; Q → 51 tall, 49 dwarf; R → 102 tall, 0 dwarf.
(a) Give the genotypes of P, Q and R. (1)
(b) Why is a dwarf plant used as the tester? (1)
(c) If Q were selfed instead and gave 400 offspring, how many would be expected to breed true for tallness? Show working. (2)
Hint
Breed true = homozygous TT.
Answer
(a) P = TT, Q = Tt, R = TT. (b) The dwarf (tt) gives only t gametes, so each offspring's phenotype shows which allele came from the plant being tested. (c) Tt selfed gives 1/4 TT: 400 × 1/4 = 100 plants breed true for tallness (the 100 tt dwarfs also breed true, but for dwarfness).
Q6 · 2 marks · Very short answerHarder board level
A man of blood group A marries a woman of blood group B; their first child is group O. Give the genotypes of the parents and the probability that their next child has group AB.
Hint
An O child is ii.
Answer
The O child is ii, so each parent carries i: man $I^{A}$ i, woman $I^{B}$ i. Children: $I^{A}$ $I^{B}$ (AB), $I^{A}$ i (A), $I^{B}$ i (B), ii (O), each 1/4. P(AB) = 1/4.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.F2 phenotypic ratio of a monohybrid cross?Show answer
3 : 1.
2.Test cross result if the plant is heterozygous?Show answer
1 dominant : 1 recessive.
3.F1 colour of red × white snapdragon?Show answer
Pink.
4.How many genotypes and phenotypes in the ABO system?Show answer
6 genotypes, 4 phenotypes.
5.What fraction of tall F2 plants are heterozygous?Show answer
2/3.
6.Which law explains why the recessive trait reappears in F2?Show answer
The law of segregation.

Monohybrid Cross and Test Cross: frequently asked questions

What is a monohybrid cross?

A cross between parents differing in one character, such as tall and dwarf, followed through F1 and F2.

What is a test cross and why is it done?

Crossing a dominant-looking individual with a homozygous recessive one to find out whether it is homozygous (all dominant offspring) or heterozygous (1:1 offspring).

What is the law of segregation?

The two alleles of a gene separate during gamete formation, so each gamete gets only one allele; they do not blend.

How is incomplete dominance different from co-dominance?

In incomplete dominance the heterozygote is intermediate (pink snapdragon). In co-dominance both alleles are fully expressed (AB blood group).

Why is the F2 genotypic ratio 1:2:1?

Each F1 parent gives T and t gametes equally, so random fusion gives TT, Tt, tT and tt in equal chances: 1 TT : 2 Tt : 1 tt.