Genetics ICSE Class 10: Mendel's Laws, Crosses and Sex-linked Inheritance

What Genetics is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Biology. Genetics is in the 1. Basic Biology (iii) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Biology (Science Paper 3) syllabus, examination year 2027

Key concepts

Key terms

• Gene: a unit of heredity, a segment of DNA that controls a character.
• Allele: one of the alternative forms of a gene, found at the same locus on homologous chromosomes (T for tall, t for dwarf).
• Homozygous: both alleles the same (TT or tt); breeds true. Heterozygous: two different alleles (Tt); a hybrid.
• Dominant: the allele that expresses itself in the heterozygote (T). Recessive: the allele that is masked in the heterozygote and shows only when homozygous (t).
• Genotype: the genetic make-up (TT, Tt, tt). Phenotype: the visible character (tall, dwarf).
• Variation: differences among individuals of a species. Mutation: a sudden, heritable change in a gene or chromosome, a source of new variation.

Worked example

A pea plant has genotype Tt. State its phenotype and whether it is homozygous or heterozygous.

Try it first, then show the solutionSolution

1. T (tall) is dominant over t (dwarf).
2. Tt has two different alleles: heterozygous.
3. The dominant allele shows: tall.

Answer: Tall; heterozygous.

Mendel's monohybrid cross and the law of dominance

Mendel crossed pure tall (TT) with pure dwarf (tt) pea plants.

• P: TT × tt; gametes T and t.
• F1: all Tt, all tall. The dwarf character seemed to vanish.
• F1 selfed: Tt × Tt; gametes T, t from each.

	T	t
T	TT	Tt
t	Tt	tt

• F2 phenotype ratio: 3 tall : 1 dwarf.
• F2 genotype ratio: 1 TT : 2 Tt : 1 tt.

Law of dominance: when two contrasting alleles are present together, only one (the dominant) expresses itself; the other (recessive) is masked.

Worked example

In the F2 of a monohybrid cross, 800 plants were raised. How many would be expected to be (a) dwarf, (b) homozygous tall, (c) heterozygous?

Try it first, then show the solutionSolution

1. Genotype ratio 1 TT : 2 Tt : 1 tt (quarters).
2. (a) tt = 800 ÷ 4 = 200.
3. (b) TT = 200.
4. (c) Tt = 400.

Answer: (a) 200 (b) 200 (c) 400

Law of segregation

Law of segregation (purity of gametes): the two alleles of a character, present together in an individual, separate (segregate) during gamete formation, so that each gamete receives only one allele.

This is why the recessive dwarf character, hidden in F1 (Tt), reappears in F2: the t allele was not changed or blended, only masked, and a gamete carrying t can meet another carrying t.

The cell basis is meiosis: homologous chromosomes, carrying the two alleles, go to different gametes.

Worked example

Why does the dwarf character reappear in the F2 generation?

Try it first, then show the solutionSolution

1. F1 plants are Tt; the t allele is present but masked.
2. During gamete formation, T and t separate, giving T and t gametes in equal numbers.
3. When two t gametes fuse, tt dwarf plants form (1 in 4).

Answer: Alleles segregate unchanged into gametes (law of segregation).

Dihybrid cross and the law of independent assortment

Mendel crossed plants differing in two characters: round yellow (RRYY) × wrinkled green (rryy).

• F1: all RrYy, round yellow (both dominant).
• F1 selfed: each F1 plant forms four kinds of gametes in equal numbers: RY, Ry, rY, ry.
• F2 phenotypes (16 boxes): 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Two new combinations (round green and wrinkled yellow) appear that were in neither parent.

Law of independent assortment: the alleles of different characters separate independently of one another during gamete formation, so all combinations are formed. (The syllabus needs only the phenotypes of the dihybrid cross.)

Worked example

In a dihybrid cross, 320 F2 plants were obtained. How many would be wrinkled and yellow?

Try it first, then show the solutionSolution

1. F2 phenotype ratio 9 : 3 : 3 : 1, total 16 parts.
2. Wrinkled yellow = 3 parts.
3. 3/16 × 320 = 60.

Answer: 60

Sex determination in human beings

Humans have 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes.

• Female: XX; every egg carries an X.
• Male: XY; half the sperms carry X, half carry Y.

	X (sperm)	Y (sperm)
X (egg)	XX girl	XY boy

So the sex of the child is decided by the father's sperm: an X-bearing sperm gives a girl, a Y-bearing sperm gives a boy. The chance of either is 1 : 1 (50%) at each birth.

Worked example

A zygote receives a Y chromosome from the father. What will its sex be, and why?

Try it first, then show the solutionSolution

1. The mother always gives an X.
2. X from the mother + Y from the father = XY.
3. XY develops into a male.

Answer: Male (XY).

X-linked inheritance: haemophilia and colour blindness

Some genes lie on the X chromosome and have no partner on the Y. Diseases caused by recessive alleles on the X are X-linked:

• Haemophilia: blood does not clot normally, so a small cut can bleed for a long time.
• Colour blindness (red-green): the person cannot tell red from green.

Write the normal allele as Xᴴ (or Xᶜ) and the disease allele as Xʰ (or Xᶜᵇ).

• A male (XY) has only one X, so one recessive allele makes him affected (XʰY).
• A female needs two (XʰXʰ) to be affected; XᴴXʰ is a carrier: normal herself but can pass the allele on.
• So these diseases are far more common in males, who receive their X from their mother.

Carrier mother (XᴴXʰ) × normal father (XᴴY): daughters XᴴXᴴ or XᴴXʰ (normal, half are carriers); sons XᴴY or XʰY (half are affected).

Worked example

A colour-blind man marries a woman with normal vision who is not a carrier. What proportion of their (a) sons and (b) daughters will be colour blind? What will the daughters be?

Try it first, then show the solutionSolution

1. Father XᶜᵇY; mother XX (normal, non-carrier).
2. Sons get Y from father and normal X from mother: all normal.
3. Daughters get Xᶜᵇ from father and normal X from mother: all carriers, none colour blind.

Answer: (a) None (b) None; all daughters are carriers.

Genetics in real life

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

Topics in this chapter

• Monohybrid Cross and Mendel's Laws
• Dihybrid Cross and Law of Independent Assortment
• Sex Determination and X-linked Diseases

Genetics: key facts

Monohybrid F2 phenotype ratio

$3:1$

Dominant : recessive.

Monohybrid F2 genotype ratio

$1:2:1$

TT : Tt : tt.

Dihybrid F2 phenotype ratio

$9:3:3:1$

Both dominant : one dominant (two kinds) : both recessive.

Sex ratio at fertilisation

$XX:XY=1:1$

Father's X or Y sperm decides.

Gametes of a dihybrid F1

$RrYy\to RY,Ry,rY,ry$

Four kinds in equal numbers.

Genetics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Genetics

The question patterns to expect, and the method that works for each.

Tricks to remember Genetics

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

• D-S-I

  Mendel's laws in order: Dominance, Segregation, Independent assortment ('Do Some Investigation').

  Use it when: Naming the laws.

• 3:1 looks, 1:2:1 genes

  Phenotype ratio is what you see (3 : 1); genotype ratio is what's inside (1 : 2 : 1).

  Use it when: Not swapping the ratios.

• 9-3-3-1 sums to 16

  A dihybrid F2 always has 16 boxes; check that your parts add up to 16.

  Use it when: Dihybrid numbers.

• Mum passes it to her son

  X-linked diseases go from carrier mother to affected son; fathers never pass them to sons.

  Use it when: Haemophilia and colour blindness.

Genetics: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 6-mark question split into 1 + 2 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Genetics

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

• ✗Mistake: Genotype written for a phenotype ratio

  CISCE examiners' comments, 2024: in a seed-shape cross, candidates copied the diagram instead of naming the seed shape, and wrote the genotype instead of the phenotype for the 3 : 1 ratio.

  Wrong working

  3 : 1 = RR, Rr, Rr : rr

  Right working

  3 round : 1 wrinkled

  ✓Do this: Phenotype = words (round, wrinkled); genotype = letters.

  Usually costs: 1 mark.

• ✗Mistake: Vague definition of allele

  CISCE examiners' comments, 2025: candidates defined allele as 'forms of traits' without saying 'alternative forms of a gene at the same locus', or explained dominance instead.

  Wrong working

  Alleles are dominant and recessive traits.

  Right working

  Alleles are alternative forms of a gene at the same locus on homologous chromosomes.

  ✓Do this: Learn the definition word for word.

  Usually costs: 1 mark.

• ✗Mistake: Wrong dihybrid ratio or law

  CISCE examiners' comments, 2025: in a dihybrid tomato question, candidates wrote only 'tall red' without justifying it, gave the monohybrid ratio or reversed the order, and named the law of segregation instead of independent assortment.

  Wrong working

  F2 ratio 3 : 1; law of segregation.

  Right working

  9 tall red : 3 tall yellow : 3 dwarf red : 1 dwarf yellow; law of independent assortment.

  ✓Do this: Two characters → 9 : 3 : 3 : 1 → independent assortment.

  Usually costs: 1-2 marks.

• ✗Mistake: Inheritance in a new context

  CISCE examiners' comments, 2025: in a mouse-tail question, candidates could not decide which characters pass to the next generation; in a tongue-rolling question they did not link phenotype with genotype.

  Wrong working

  Mice with cut tails produce tailless offspring.

  Right working

  Offspring have normal tails: acquired changes do not alter genes in gametes.

  ✓Do this: Only genes in gametes are inherited; write genotypes before choosing.

  Usually costs: 1 mark.

• ✗Mistake: Ignoring the Y chromosome

  CISCE examiners' comments, 2024: for a zygote with a Y from the father, candidates answered 'either boy or girl'.

  Wrong working

  It may become a boy or a girl.

  Right working

  A Y chromosome means XY, a boy.

  ✓Do this: Y present = male.

  Usually costs: 1 mark.

Practice questions on Genetics

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

1. Q1 · 1 mark · MCQBoard level

   The genotypic ratio in the F2 of a monohybrid cross is:
   (a) 3 : 1
   (b) 1 : 2 : 1
   (c) 9 : 3 : 3 : 1
   (d) 1 : 1

   Hint

   TT : Tt : tt.

   Answer

   (b)

2. Q2 · 1 mark · MCQHarder board level

   A heterozygous tall plant is crossed with a dwarf plant. The percentage of dwarf offspring is:
   (a) 0%
   (b) 25%
   (c) 50%
   (d) 75%

   Hint

   Tt × tt.

   Answer

   (c)

3. Q3 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): A carrier woman for haemophilia does not suffer from the disease.
   Reason (R): The normal allele on her other X chromosome is dominant.
   (a) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (b) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
   (c) (A) is true and (R) is false.
   (d) (A) is false and (R) is true.

   Hint

   Recessive allele masked.

   Answer

   (a)

4. Q4 · 3 marks · Very short answerBoard level

   Give the biological term: (a) the visible character of an organism; (b) an individual with two identical alleles; (c) a sudden heritable change in a gene.

   Hint

   One word each.

   Answer

   (a) Phenotype (b) Homozygous (c) Mutation

5. Q5 · 2 marks · Short answerBoard level

   State the law of dominance with an example.

   Hint

   Tt.

   Answer

   When two contrasting alleles are present together, only the dominant one expresses itself. Example: Tt pea plants are tall, though they carry the dwarf allele.

6. Q6 · 3 marks · Short answerHarder board level

   A woman with normal clotting whose father had haemophilia marries a man with normal clotting. What is the chance that their son will have haemophilia? Show your working.

   Hint

   Her X from her father.

   Answer

   She got Xʰ from her father, so she is XᴴXʰ. Husband XᴴY. Sons: XᴴY or XʰY: 1/2 (50%).

7. Q7 · 2 marks · Short answerHarder board level

   In a dihybrid cross between round yellow (RRYY) and wrinkled green (rryy) peas, 480 F2 seeds were obtained. How many of each phenotype are expected?

   Hint

   Sixteenths.

   Answer

   Round yellow 270, round green 90, wrinkled yellow 90, wrinkled green 30.

8. Q8 · 3 marks · Short answerHarder board level

   Explain why the F1 of a cross between pure tall and pure dwarf plants are all tall, yet dwarf plants reappear in F2.

   Hint

   Dominance and segregation.

   Answer

   F1 are Tt; T is dominant so all are tall (law of dominance). The t allele is not lost: at gamete formation T and t separate (law of segregation), so tt plants appear in F2 (1 in 4).

9. Q9 · 2 marks · Short answerHOTS / competency

   Two brothers, one colour blind and one with normal vision, have parents with normal vision. A student says the father must be a carrier. Correct the student and explain.

   Hint

   Males cannot be carriers.

   Answer

   Males have one X, so they are either affected or normal, never carriers. The colour-blind son got the allele on his X from his mother, who must be a carrier.

10. Q10 · 3 marks · Short answerHOTS / competency

    In guinea pigs, black coat (B) is dominant over white (b). A black guinea pig crossed with a white one gives 5 black and 4 white young. Find the genotype of the black parent and explain.

    Hint

    Some white young appear.

    Answer

    The black parent is Bb. White young (bb) need a b allele from each parent; the white parent gives b, so the black parent must also carry b. Bb × bb gives about 1 black : 1 white, as observed.

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. 1.Define phenotype.Show answer

   The observable characters of an organism.

2. 2.Define genotype.Show answer

   The genetic make-up (allele combination) of an organism.

3. 3.Monohybrid F2 phenotypic ratio?Show answer

   3 : 1.

4. 4.Dihybrid F2 phenotypic ratio?Show answer

   9 : 3 : 3 : 1.

5. 5.Which law does the dihybrid cross show?Show answer

   Law of independent assortment.

6. 6.Sex chromosomes of a human male?Show answer

   XY.

7. 7.Define mutation.Show answer

   A sudden, heritable change in a gene or chromosome.

8. 8.Why is haemophilia more common in males?Show answer

   Males have one X; a single recessive allele on it causes the disease.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define allele, homozygous, heterozygous, dominant and recessive?
• ☐Can you draw a monohybrid cross to F2 with both ratios?
• ☐Can you write the four F2 phenotypes of a dihybrid cross in ratio?
• ☐Can you state all three of Mendel's laws?
• ☐Can you show how the father determines the sex of a child?
• ☐Can you predict the children of a carrier mother and a normal father?

Genetics: frequently asked questions