BiologyICSE Class 10 · Genetics

Monohybrid Cross, Genetic Terms and Mendel's Laws of Dominance and Segregation (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A monohybrid cross follows one pair of contrasting characters. Crossing pure tall (TT) with pure dwarf (tt) peas gives all tall F1 hybrids (Tt). Selfing F1 gives F2 in the phenotypic ratio 3 tall : 1 dwarf and genotypic ratio 1 TT : 2 Tt : 1 tt, explained by the laws of dominance and segregation.

Notice that all F1 are Tt and tall, and in the F2 square three boxes are tall (TT, Tt, Tt) while one is dwarf (tt).

Illustration drawn for QuestionBuddy.

Understand Monohybrid Cross and Mendel's Laws

Mendel and his pea plants. Gregor Johann Mendel worked for seven years (1856–1863) on the garden pea. Peas suit genetic work: they have clear contrasting characters, they normally self-pollinate (so pure, true-breeding lines exist), they can be cross-pollinated by hand, they have a short life cycle and give many seeds. Mendel studied seven pairs of contrasting characters:

Character	Dominant	Recessive
Stem height	tall	dwarf
Flower colour	violet	white
Flower position	axial	terminal
Pod shape	inflated	constricted
Pod colour	green	yellow
Seed shape	round	wrinkled
Seed colour	yellow	green

Key terms (all in the syllabus).

Gene: the unit of heredity; a segment of DNA that controls a character.
Allele: one of the alternative forms of a gene occupying the same locus on homologous chromosomes (T and t).
Dominant: the allele that expresses itself even in the presence of the other (T). Recessive: the allele that is masked when the dominant one is present and shows only when homozygous (t).
Homozygous: both alleles alike (TT or tt), a pure line. Heterozygous: two different alleles (Tt), a hybrid.
Genotype: the genetic make-up (TT, Tt, tt). Phenotype: the observable character (tall, dwarf).
Variation: differences between individuals of the same species. Mutation: a sudden heritable change in a gene or chromosome; it creates new alleles and so new variation.

The monohybrid cross step by step.
1. P (parental) generation: pure tall TT × pure dwarf tt.
2. Gametes: each parent gives one allele per gamete: T from the tall parent, t from the dwarf.
3. F1 (first filial) generation: all Tt, all tall. The dwarf character has not been lost, only hidden.
4. F1 × F1 (selfing): Tt × Tt. Each parent forms two kinds of gametes, T and t, in equal numbers.
5. F2 by Punnett square:

Gametes	T	t
T	TT (tall)	Tt (tall)
t	Tt (tall)	tt (dwarf)

6. F2 phenotypic ratio 3 tall : 1 dwarf. F2 genotypic ratio 1 TT : 2 Tt : 1 tt.

Mendel's first two laws.

Law of dominance: when a pair of contrasting alleles is present in an individual, only one (the dominant) expresses itself; the other (recessive) remains masked. Evidence: all F1 plants are tall.
Law of segregation (law of purity of gametes): the two alleles of a pair separate during gamete formation, so each gamete receives only one allele and is 'pure' for it. Evidence: dwarf plants, absent in F1, reappear in F2 unchanged, in one quarter of the plants. The cell basis is meiosis, in which homologous chromosomes carrying the two alleles go into different gametes.

Working backwards from offspring (useful for HOTS).

Offspring seen	Parents must be
all dominant	at least one parent homozygous dominant
3 dominant : 1 recessive	both heterozygous
1 dominant : 1 recessive	heterozygous × homozygous recessive
all recessive	both homozygous recessive

If two parents showing the dominant trait have a child showing the recessive trait, both parents must be heterozygous. Crossing an unknown dominant-looking individual with a homozygous recessive one (a test cross) reveals its genotype: all dominant offspring → homozygous; half recessive → heterozygous.

Human example: tongue rolling (dominant R, non-rolling r). Two rollers who have a non-rolling child are both Rr.

Flow chart: monohybrid cross. P: TT (tall) × tt (dwarf) → gametes T, t → F1: Tt (all tall) → selfed: Tt × Tt → gametes T, t and T, t → F2: 1 TT : 2 Tt : 1 tt → 3 tall : 1 dwarf.

Draw-and-label checklist: a monohybrid cross chart. Write P with the phenotypes in words and genotypes below; circle the gametes; draw arrows to F1; write 'selfing' or ⊗; draw a 2 × 2 Punnett square with gametes on both sides; write the phenotype under every box; finish with both ratios in words: 'Phenotypic ratio 3 tall : 1 dwarf; genotypic ratio 1 TT : 2 Tt : 1 tt'.

Monohybrid Cross and Mendel's Laws: key facts

Monohybrid ratios: F1 (TT × tt) all Tt, all tall
F2 phenotype 3 tall : 1 dwarf
F2 genotype 1 TT : 2 Tt : 1 tt
Test cross (Tt × tt) 1 tall : 1 dwarf; Always say which ratio you are giving.
Mendel's seven characters (dominant first): Stem height tall over dwarf
Flower colour violet over white
Flower position axial over terminal
Pod shape inflated over constricted
Pod colour green over yellow
Seed shape round over wrinkled
Seed colour yellow over green; Pod colour green is dominant but seed colour yellow is dominant: a common trap.
Laws illustrated by the monohybrid cross: Law of dominance: F1 shows only the dominant character | Law of segregation: alleles separate in gamete formation, recessive reappears in F2; Independent assortment needs a dihybrid cross.
Genetic terms in one line each: Allele alternative form of a gene at the same locus
Homozygous identical alleles (TT, tt)
Heterozygous different alleles (Tt)
Genotype genetic make-up
Phenotype observable character
Mutation sudden heritable change in a gene; Every one has been a 1-mark question.

F1 (TT × tt)	all Tt, all tall
F2 phenotype	3 tall : 1 dwarf
F2 genotype	1 TT : 2 Tt : 1 tt
Test cross (Tt × tt)	1 tall : 1 dwarf

Stem height	tall over dwarf
Flower colour	violet over white
Flower position	axial over terminal
Pod shape	inflated over constricted
Pod colour	green over yellow
Seed shape	round over wrinkled
Seed colour	yellow over green

Allele	alternative form of a gene at the same locus
Homozygous	identical alleles (TT, tt)
Heterozygous	different alleles (Tt)
Genotype	genetic make-up
Phenotype	observable character
Mutation	sudden heritable change in a gene

How to answer Monohybrid Cross and Mendel's Laws questions

ICSE genetics questions give a cross (often as a chart) and ask for phenotypes, genotypes, ratios, the law shown, or parents' genotypes from offspring.

1Choose letters: a capital for the dominant allele, the same small letter for the recessive (T/t, R/r). State them: 'Let T = tall, t = dwarf'.
2Write the parents' genotypes, then their gametes (one allele each, circled).
3Draw the Punnett square and fill every box; write the phenotype under each.
4Count and give the ratio asked: phenotypic in words (3 tall : 1 dwarf), genotypic with symbols (1 TT : 2 Tt : 1 tt).
5For numbers out of a total, multiply the total by the fraction: one quarter of 800 = 200.
6For parents from offspring, start from the recessive offspring: it must be homozygous recessive, so each parent gave one recessive allele.

Monohybrid Cross and Mendel's Laws 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.

Recessive trait reappearing

Suppose two tongue-rolling parents, Mr and Mrs Iyer, are surprised that their younger son Karthik cannot roll his tongue, while his sister can. Tongue rolling (R) is dominant to non-rolling (r).

Find the genotypes of the parents and the chance that their next child is a non-roller.

Try it first, then show the solutionSolution

Karthik shows the recessive trait, so he is rr: he received one r from each parent.
Both parents roll their tongues, so each has at least one R; with the r they passed on, each is Rr.
Rr × Rr gives 1 RR : 2 Rr : 1 rr.
The chance of rr (non-roller) for each new child is 1/4.

Answer: Both parents are Rr; each child has a 1/4 (25%) chance of being a non-roller.

Memory hook: A recessive child exposes two hidden carriers.

Finding a pure line

Suppose a farmer has a tall pea plant and wants to sell its seeds as 'pure tall'. Her neighbour says that a tall plant may be hiding a dwarf allele.

How can she find out, and what result would show the plant is pure?

Try it first, then show the solutionSolution

A tall plant may be TT or Tt; its phenotype cannot tell them apart.
Cross it with a dwarf plant (tt), which can give only t gametes.
If the tall plant is TT, all offspring are Tt, tall.
If it is Tt, half the offspring are tt, dwarf.

Answer: Cross it with a dwarf plant: all-tall offspring show it is pure (TT); any dwarf offspring show it is Tt.

Memory hook: Cross with the recessive to reveal the hidden.

Tricks to remember Monohybrid Cross and Mendel's Laws

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

3 : 1 looks, 1 : 2 : 1 genes
Phenotype (what you see) has two numbers; genotype (what you carry) has three.
Use it when: Never giving the wrong ratio for the wrong question.
Segregation = separation at gamete time
Alleles live together in the plant but travel alone in a gamete.
Use it when: Stating the law of segregation.
Green pods, yellow seeds
On the pod, green wins; inside it, yellow wins.
Use it when: Mendel's seven characters.

Monohybrid Cross and Mendel's Laws: solved examples

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

Example 1: Two terms · 2 marks · Short answerBoard level

Define: (a) allele, (b) phenotype. (Allele was asked in 2025.)

Try it first, then show the solutionSolution

(a) Alleles are the alternative forms of a gene that occupy the same locus on homologous chromosomes and control contrasting expressions of one character (1 mark).
(b) Phenotype is the observable or physical expression of a character, such as tall or dwarf (1 mark).

Answer: (a) Alternative forms of a gene at the same locus. (b) The observable character.

Example 2: Full monohybrid cross · 3 marks · Short answerBoard level

In peas, round seed (R) is dominant over wrinkled seed (r). A pure round-seeded plant is crossed with a wrinkled-seeded plant. Show the cross up to F2 and give (a) the F1 phenotype, (b) the F2 phenotypic ratio, (c) the F2 genotypic ratio.

Try it first, then show the solutionSolution

P: RR (round) × rr (wrinkled); gametes R and r; F1: Rr, all round (1 mark).
F1 selfed: Rr × Rr; gametes R, r from each; Punnett square gives RR, Rr, Rr, rr (1 mark).
(b) 3 round : 1 wrinkled; (c) 1 RR : 2 Rr : 1 rr (1 mark).

Answer: (a) All round (b) 3 round : 1 wrinkled (c) 1 RR : 2 Rr : 1 rr.

Example 3: Numbers from ratios · 3 marks · Short answerHarder board level

An F2 population from a cross of pure tall and pure dwarf pea plants has 800 plants. How many are expected to be (a) dwarf, (b) heterozygous tall, (c) homozygous tall? (d) If you pick a tall F2 plant at random, what is the chance that it is pure-breeding?

Try it first, then show the solutionSolution

(a) One quarter dwarf: 800 ÷ 4 = 200 (½ mark).
(b) Half Tt: 400 (½ mark); (c) one quarter TT: 200 (1 mark).
(d) Tall plants are TT (200) and Tt (400); pure ones are 200 of 600 = 1/3 (1 mark).

Answer: (a) 200 (b) 400 (c) 200 (d) 1/3.

Example 4: Which law and why · 3 marks · Short answerHarder board level

In a monohybrid cross, the dwarf character that disappeared in F1 reappears in F2. Name the law this illustrates, state it, and explain why the F2 dwarfs are as short as the original dwarf parent.

Try it first, then show the solutionSolution

Law of segregation (purity of gametes) (½ mark).
Statement: the two alleles of a character separate during gamete formation, so each gamete carries only one allele (1 mark).
The t allele was masked in Tt but not changed or blended; F2 dwarfs (tt) receive two unchanged t alleles, so they are fully dwarf (1½ marks).

Answer: Law of segregation: alleles separate into gametes unchanged, so tt plants in F2 are fully dwarf.

Example 5: Parents from children · 3 marks · Case-basedHOTS / competency

Suppose two brown-eyed parents (brown B dominant over blue b) have four children: three brown-eyed and one blue-eyed. (a) Give the genotypes of the parents and of the blue-eyed child. (b) What are the possible genotypes of the brown-eyed children? (c) The brown-eyed daughter marries a blue-eyed man. Can they have a blue-eyed child? Explain.

Try it first, then show the solutionSolution

(a) Blue-eyed child bb; each parent gave b and is brown, so both are Bb (1 mark).
(b) BB or Bb (1 mark).
(c) Yes, if she is Bb: Bb × bb gives half bb (blue); if she is BB, all children are Bb (brown). There is a 2/3 chance she is Bb (1 mark).

Answer: (a) Bb × Bb; child bb. (b) BB or Bb. (c) Yes, if she is Bb (then half the children are expected to be blue-eyed).

Example 6: A new character · 3 marks · Short answerHOTS / competency

In rabbits black fur (B) is dominant to white fur (b). Over two litters, a black doe mated with a white buck has 9 black and 8 white young. (a) What is the genotype of the black parent? (b) What ratio do these numbers approach, and why? (c) What result would show that a black rabbit is homozygous?

Try it first, then show the solutionSolution

(a) White young (bb) must get b from the black parent, so it is Bb (1 mark).
(b) Bb × bb gives 1 Bb : 1 bb; 9 : 8 is close to 1 : 1 (actual numbers vary by chance) (1 mark).
(c) Crossed with white, a homozygous black (BB) gives only black young (1 mark).

Answer: (a) Bb (b) about 1 : 1, from Bb × bb (c) All black young when crossed with a white rabbit.

Where students lose marks in Monohybrid Cross and Mendel's Laws

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

Mistake: Giving genotype when phenotype is asked
CISCE examiners' comments (2024): in a seed-shape cross, candidates copied the diagram instead of naming the seed shape in words, wrote the genotype instead of the phenotype for the 3 : 1 ratio, and wrote about dominant and recessive traits instead of a pair of contrasting characters.
Wrong working
Phenotypic ratio: 1 RR : 2 Rr : 1 rr.
Right working
Phenotypic ratio: 3 round : 1 wrinkled.
Do this: Phenotype in words; genotype in letters.
Usually costs: 1 mark.
Mistake: Vague definition of allele
CISCE examiners' comments (2025): candidates wrote 'forms of traits or characters' without 'alternative forms of a gene at the same locus', or described dominance instead.
Do this: Learn: 'alternative forms of a gene occupying the same locus on homologous chromosomes'.
Usually costs: 1 mark.
Mistake: Not linking phenotype to genotype
CISCE examiners' comments (2025): in a tongue-rolling sibling question, candidates chose randomly and failed to link phenotype with genotype; examiners suggested Punnett squares to deduce parents' genotypes.
Do this: Start from the recessive child (rr) and work back to the parents.
Usually costs: 1 mark.
Mistake: Thinking acquired characters are inherited
CISCE examiners' comments (2025): in a question on mice whose tails were cut, candidates could not decide which characters pass to the next generation.
Do this: Only changes in the genes of gametes are inherited; injuries to body cells are not.
Usually costs: 1 mark.

Exam tips

Always define your symbols before a cross.
Show gametes separately (circled); the examiner gives a mark for them.
Write both ratios with labels: 'phenotypic' and 'genotypic'.
Name the law and state it; one without the other gets half marks.
Use 'F1' and 'F2', not 'first children'.

Practice questions on Monohybrid Cross and Mendel's Laws

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 · MCQBoard level
F1 tall pea plants (Tt) are self-pollinated. Among the tall plants of the F2, the ratio of homozygous to heterozygous plants is:
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1
Hint
Tall F2 plants are TT or Tt; count each in the Punnett square.
Answer
(b) 1 : 2 (1 TT to 2 Tt).
Q2 · 1 mark · Very short answerBoard level
Name the type of individual that has two identical alleles for a character.
Hint
Homo = same.
Answer
Homozygous (a pure line).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): All the F1 plants of a cross between pure tall and pure dwarf pea plants are tall.
Reason (R): The allele for tallness is dominant over that for dwarfness.
(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
Law of dominance.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
A heterozygous tall plant (Tt) is crossed with a dwarf plant (tt). Give the phenotypic and genotypic ratios of the offspring. If 120 seeds are produced, how many will grow into dwarf plants?
Hint
Two gametes × one gamete.
Answer
Tt × tt gives 1 Tt : 1 tt; phenotypes 1 tall : 1 dwarf. Dwarf: 120 ÷ 2 = 60.
Q5 · 2 marks · Short answerHarder board level
Distinguish between dominant and recessive alleles, with an example from Mendel's peas.
Hint
Expression in the heterozygote.
Answer
A dominant allele expresses itself even when paired with a different allele (T for tall shows in Tt). A recessive allele expresses itself only when both alleles are recessive (t for dwarf shows only in tt).
Q6 · 2 marks · Short answerHarder board level
Why did Mendel get only tall plants in F1 but both tall and dwarf in F2? Use the words gamete and allele.
Hint
Dominance, then segregation.
Answer
F1 plants received allele T from the tall parent's gamete and t from the dwarf parent's gamete; T is dominant, so all were tall. In F1 the alleles segregated into separate gametes (T or t), and a t gamete meeting a t gamete gave tt dwarfs in F2.
Q7 · 2 marks · Short answerHOTS / competency
Suppose a scientist cuts off the tails of mice for several generations and then breeds them. Will the young be born with tails? Use the idea of genes to explain. (A question of this kind was asked in 2025.)
Hint
Is a cut tail a change in DNA?
Answer
Yes, they will be born with normal tails. Cutting the tail changes only body cells, not the genes in the gametes; only characters controlled by genes in the gametes are inherited.
Q8 · 2 marks · Short answerHOTS / competency
Suppose in a plant red flower colour (R) is dominant to white (r). Two red-flowered plants are crossed and produce 300 seedlings, of which about 75 bear white flowers. (a) What are the genotypes of the parents? (b) How many of the red-flowered seedlings are expected to be heterozygous?
Hint
75 of 300 is one quarter.
Answer
(a) Both Rr (only Rr × Rr gives about one quarter white). (b) Half of 300 = 150 are Rr (the remaining 75 red are RR).
Q9 · 3 marks · Case-basedHarder board level
Read and answer.
Suppose a pure violet-flowered pea plant is crossed with a white-flowered one. All F1 plants are violet. The F1 plants are self-pollinated and give 640 plants in F2.
(a) Which colour is dominant? (1)
(b) How many F2 plants are expected to be white, and how many violet? (1)
(c) Which of Mendel's laws explains the reappearance of white flowers? (1)
Hint
3 : 1.
Answer
(a) Violet. (b) White 640 ÷ 4 = 160; violet 480. (c) Law of segregation.

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.What is the F2 phenotypic ratio of a monohybrid cross?Show answer
3 : 1.
2.What is Tt called: homozygous or heterozygous?Show answer
Heterozygous.
3.Which law is also called the law of purity of gametes?Show answer
The law of segregation.
4.Which pod colour is dominant in peas?Show answer
Green.
5.Of 400 F2 plants of a monohybrid cross, how many show the recessive trait?Show answer
100.
6.What is a sudden heritable change in a gene called?Show answer
Mutation.

Monohybrid Cross and Mendel's Laws: frequently asked questions

What is a monohybrid cross?

A cross between parents that differ in one pair of contrasting characters, such as tall and dwarf, followed to the F2 generation.

What is the difference between genotype and phenotype?

Genotype is the set of alleles an organism carries (TT, Tt, tt); phenotype is the character it shows (tall or dwarf).

Why did the dwarf trait reappear in F2?

Because the alleles segregate during gamete formation; the recessive t allele, hidden in Tt plants, passed unchanged into some gametes, and two t gametes gave tt dwarfs.

Why is the pea plant good for genetics?

It has clear contrasting characters, self-pollinates to give pure lines, can be cross-pollinated by hand, grows quickly and produces many seeds.