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Monohybrid crosses: Class 10 practice

Each parent passes one allele of a gene. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Dominant and recessive traits

The idea behind monohybrid crosses

Each parent passes one allele of a gene. A Punnett square lists all gamete combinations: Tt × Tt gives TT : Tt : tt = 1 : 2 : 1 (genotype) and 3 : 1 (phenotype) when T is dominant.

Step-by-step method

  1. Write each parent's gametes.
  2. Fill the Punnett square.
  3. Count genotypes and phenotypes.

Worked example 1 (Basic)

Cross TT × Tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 200.

  1. Offspring: TT, Tt, TT, Tt
  2. TT : Tt : tt = 2 : 2 : 0
  3. Tall : short = 4 : 0
  4. Tall among 200: 200
  5. Answer: 200 tall; genotype 2 : 2 : 0; tall : short = 4 : 0

Worked example 2 (Standard)

Cross Tt × tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 800.

  1. Offspring: Tt, Tt, tt, tt
  2. TT : Tt : tt = 0 : 2 : 2
  3. Tall : short = 2 : 2
  4. Tall among 800: 400
  5. Answer: 400 tall; genotype 0 : 2 : 2; tall : short = 2 : 2

Worked example 3 (Challenge)

TT × Tt (T tall, dominant). If 400 seeds grow, how many are expected to be short?

  1. Short (tt) = 0/4
  2. 400 × 0/4 = 0
  3. Answer: 0

Common mistakes and misconceptions

  • Writing TT × tt offspring as 3 : 1
  • Confusing genotype and phenotype ratios

Practice questions

  1. (Basic) Cross TT × tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 100.
  2. (Basic) Cross Tt × Tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 200.
  3. (Standard) Cross Tt × Tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 400.
  4. (Standard) Cross Tt × tt (T = tall, dominant over t). Give the genotype and phenotype ratios, and the expected number of tall plants among 1200.
  5. (Challenge) Tt × Tt (T tall, dominant). If 400 seeds grow, how many are expected to be short?
  6. (Challenge) Tt × tt (T tall, dominant). If 400 seeds grow, how many are expected to be short?

Answer key with full working

  • Q1: Offspring: Tt, Tt, Tt, Tt → TT : Tt : tt = 0 : 4 : 0 → Tall : short = 4 : 0 → Tall among 100: 100 → Answer: 100 tall; genotype all Tt; tall : short = 4 : 0
  • Q2: Offspring: TT, Tt, Tt, tt → TT : Tt : tt = 1 : 2 : 1 → Tall : short = 3 : 1 → Tall among 200: 150 → Answer: 150 tall; genotype 1 : 2 : 1; tall : short = 3 : 1
  • Q3: Offspring: TT, Tt, Tt, tt → TT : Tt : tt = 1 : 2 : 1 → Tall : short = 3 : 1 → Tall among 400: 300 → Answer: 300 tall; genotype 1 : 2 : 1; tall : short = 3 : 1
  • Q4: Offspring: Tt, Tt, tt, tt → TT : Tt : tt = 0 : 2 : 2 → Tall : short = 2 : 2 → Tall among 1200: 600 → Answer: 600 tall; genotype 0 : 2 : 2; tall : short = 2 : 2
  • Q5: Short (tt) = 1/4 → 400 × 1/4 = 100 → Answer: 100
  • Q6: Short (tt) = 2/4 → 400 × 2/4 = 200 → Answer: 200

Syllabus note

This skill is listed under "Heredity and Evolution" for Class 10 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is monohybrid crosses in?

"Heredity and Evolution", Class 10 Biology (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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