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Genetics: Some Basic FundamentalsICSE Class 10 Biology Important Questions

13 hand-picked ICSE Class 10 Biology important questions for Genetics: Some Basic Fundamentals, each with a full model answer — the formats and topics most likely to appear in your board exam.

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High-yield ICSE Genetics questions cover Mendel's laws of dominance and segregation, monohybrid and dihybrid crosses worked with Punnett squares, key terms (gene, allele, genotype, phenotype, dominant, recessive, homozygous, heterozygous), and sex determination in humans. Predicting ratios (3:13:13:1 and 9:3:3:19:3:3:19:3:3:1) and drawing a Punnett square appear almost every year.

About Genetics: Some Basic Fundamentals

In the ICSE Class 10 Biology chapter Genetics: Some Basic Fundamentals you study Mendel's experiments on the garden pea, his laws of inheritance, monohybrid and dihybrid crosses using Punnett squares, the meanings of basic genetic terms, and the mechanism of sex determination in human beings. The chapter needs precise definitions and neat cross-diagrams.

Mendel's experiments and terminologyLaw of dominance and law of segregationMonohybrid cross and $3:1$ ratioDihybrid cross and $9:3:3:1$ ratioSex determination in humans

Key concepts & formulas

Basic terms

A gene is a unit of heredity; alleles are alternative forms of a gene. Genotype is the genetic make-up; phenotype is the observable trait. Homozygous (TTTTTT or tttttt) has like alleles; heterozygous (TtTtTt) has unlike alleles.

Law of dominance

In a pair of contrasting characters (alleles), only one — the dominant allele — expresses itself in the F1 hybrid while the other (recessive) remains hidden.

Law of segregation

The two alleles of a character separate (segregate) during gamete formation so that each gamete receives only one allele of a pair; they reunite randomly at fertilisation.

Monohybrid and dihybrid ratios

A monohybrid cross (Tt×TtTt \times TtTt × Tt) gives an F2 phenotypic ratio of 3:13:13:1; a dihybrid cross gives 9:3:3:19:3:3:19:3:3:1.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The observable characteristics of an organism constitute its:

  1. (a)

    Genotype

  2. (b)

    Phenotype

  3. (c)

    Karyotype

  4. (d)

    Allele

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Answer: (b) Phenotype.

Phenotype is the set of visible or expressed traits; genotype refers to the underlying genetic constitution.

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Q2MCQEasy1 mark

Mendel carried out his classic experiments on the:

  1. (a)

    Sweet pea

  2. (b)

    Garden pea (Pisum sativum)

  3. (c)

    Maize

  4. (d)

    Snapdragon

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Answer: (b) Garden pea (Pisum sativum).

Mendel chose the garden pea for its contrasting characters, ease of cultivation and self-pollinating nature.

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Q3MCQModerate1 mark

The phenotypic ratio obtained in the F2 generation of a typical dihybrid cross is:

  1. (a)

    3:13:13:1

  2. (b)

    1:2:11:2:11:2:1

  3. (c)

    9:3:3:19:3:3:19:3:3:1

  4. (d)

    1:11:11:1

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Answer: (c) 9:3:3:19:3:3:19:3:3:1.

A dihybrid cross between two heterozygotes (RrYy×RrYyRrYy \times RrYyRrYy × RrYy) gives an F2 phenotypic ratio of 9:3:3:19:3:3:19:3:3:1.

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Q4MCQHOTS1 mark

A tall pea plant is crossed with a dwarf plant and all offspring are tall. When these tall offspring are self-pollinated, the fraction of dwarf plants expected is:

  1. (a)

    12\dfrac{1}{2}1/2

  2. (b)

    14\dfrac{1}{4}1/4

  3. (c)

    34\dfrac{3}{4}3/4

  4. (d)

    000

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Answer: (b) 14\dfrac{1}{4}1/4.

The F1 are all TtTtTt. Selfing Tt×TtTt \times TtTt × Tt gives 1TT:2Tt:1tt1\,TT : 2\,Tt : 1\,tt1\,TT : 2\,Tt : 1\,tt, so dwarf (tttttt) plants are 14\dfrac{1}{4}1/4 of the F2.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In humans the father determines the sex of the child.

Reason (R): The mother produces two types of eggs, one carrying an X chromosome and the other a Y chromosome.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (c) A is true but R is false. The father is heterogametic (XYXYXY) and produces X- and Y-bearing sperms, so he determines the sex; the mother (XXXXXX) produces only X-bearing eggs, so R is incorrect.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Define (i) allele and (ii) heterozygous.

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(i) Allele: One of the two or more alternative forms of a gene that occupy the same position (locus) on homologous chromosomes and control contrasting forms of a character, e.g. TTT for tallness and ttt for dwarfness.

(ii) Heterozygous: An organism having two different (unlike) alleles of a gene for a character, e.g. TtTtTt.

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Q7Very ShortModerate2 marks

Differentiate between dominant and recessive traits with one example each.

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A dominant trait is one that expresses itself in the F1 hybrid even when only one allele for it is present, e.g. tallness (TTT) in pea plants. A recessive trait is one that is masked in the hybrid and appears only when both alleles for it are present (homozygous condition), e.g. dwarfness (ttt) in pea plants.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Work out a monohybrid cross between a homozygous tall pea plant (TTTTTT) and a homozygous dwarf pea plant (tttttt) up to the F2 generation, giving the phenotypic ratio.

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P generation: TTTTTT (tall) ×\times× tttttt (dwarf).

F1: All plants are TtTtTttall (tallness is dominant).

F1 selfed: Tt×TtTt \times TtTt × Tt. Gametes: TTT and ttt from each parent.

ICSE Class 10 Biology — Genetics: Some Basic Fundamentals: Work out a monohybrid cross between a homozygous tall pea plant (TT) and a homozygous dwarf pea plant (tt) up to the F2 g

F2 genotypes: 1TT:2Tt:1tt1\,TT : 2\,Tt : 1\,tt1\,TT : 2\,Tt : 1\,tt.

F2 phenotypes: 333 tall :1: 1: 1 dwarf, i.e. the phenotypic ratio is 3:1\mathbf{3:1}3:1.

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Q9Short AnswerEasy3 marks

Explain the mechanism of sex determination in human beings with the help of a cross.

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In humans, of the 23 pairs of chromosomes, one pair is the sex chromosomes. Females are XXXXXX (homogametic) and males are XYXYXY (heterogametic).

The mother produces only X-bearing eggs, while the father produces two kinds of sperm — X-bearing and Y-bearing.

  • If an X sperm fertilises the egg XX\rightarrow XX \rightarrow→ XX → girl.
  • If a Y sperm fertilises the egg XY\rightarrow XY \rightarrow→ XY → boy.

XX (mother)×XY (father)XX:XY=1:1XX\ (\text{mother}) \times XY\ (\text{father}) \rightarrow XX : XY = 1:1XX (mother) × XY (father) → XX : XY = 1:1

Thus there is a 50%50\%50\% chance of a boy and 50%50\%50\% of a girl, and the father determines the sex of the child.

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Q10Short AnswerModerate3 marks

State Mendel's law of segregation and law of dominance.

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Law of dominance: When two organisms differing in a pair of contrasting characters (alleles) are crossed, only one character — the dominant one — appears in the F1 generation, while the other (recessive) remains hidden.

Law of segregation (law of purity of gametes): The two alleles of a character present in an organism separate during the formation of gametes, so that each gamete carries only one allele of the pair. The alleles remain pure and reunite at random during fertilisation. This is why the recessive character can reappear in the F2 generation.

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Long answer questions (5 marks)

Q11Long AnswerHOTS5 marks

Work out a dihybrid cross between a pure round-yellow seeded pea plant (RRYYRRYYRRYY) and a pure wrinkled-green seeded pea plant (rryyrryyrryy), where round (RRR) and yellow (YYY) are dominant. Show the F1, the gametes of the F1, the Punnett square, and the F2 phenotypic ratio.

Show model answer

P: RRYYRRYYRRYY (round, yellow) ×\times× rryyrryyrryy (wrinkled, green).

F1: All RrYyRrYyRrYyround and yellow seeds (both dominants expressed).

Gametes of F1 (RrYyRrYyRrYy): RY, Ry, rY, ryRY,\ Ry,\ rY,\ ryRY, Ry, rY, ry.

Punnett square (F1 ×\times× F1):

ICSE Class 10 Biology — Genetics: Some Basic Fundamentals: Work out a dihybrid cross between a pure round-yellow seeded pea plant (RRYY) and a pure wrinkled-green seeded pea plant

F2 phenotypes:

  • 999 round, yellow
  • 333 round, green
  • 333 wrinkled, yellow
  • 111 wrinkled, green

F2 phenotypic ratio =9:3:3:1= 9:3:3:1= 9:3:3:1. This illustrates Mendel's law of independent assortment — the two characters (seed shape and colour) are inherited independently of each other.

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Q12Long AnswerModerate5 marks

(a) Why did Mendel choose the garden pea for his experiments? Give three reasons. (b) Define the terms genotype, phenotype and homozygous. (c) What is a test cross?

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(a) Reasons for choosing the garden pea:
(i) It shows a number of clear, contrasting characters (e.g. tall/dwarf, round/wrinkled seeds).
(ii) It is naturally self-pollinating, giving pure lines, yet can be easily cross-pollinated by the experimenter.
(iii) It has a short life cycle and produces many offspring in one generation, allowing quick results over many generations.

(b) Definitions:

  • Genotype: The genetic constitution (allelic make-up) of an organism for a character, e.g. TtTtTt.
  • Phenotype: The observable or expressed characteristic of an organism, e.g. tall.
  • Homozygous: An organism having two identical alleles of a gene for a character, e.g. TTTTTT or tttttt (also called pure).

(c) Test cross: A cross between an organism showing the dominant phenotype (whose genotype is unknown) and a homozygous recessive individual, used to find out whether the dominant organism is homozygous or heterozygous. If any recessive offspring appear, the tested parent is heterozygous.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

In humans, the ability to roll the tongue (RRR) is dominant over the inability to roll it (rrr). A man heterozygous for tongue-rolling marries a woman who cannot roll her tongue.

(i) Write the genotypes of the man and the woman.
(ii) Work out the cross and give the genotypes of the children.
(iii) What percentage of the children are expected to be tongue-rollers?
(iv) State whether tongue-rolling is a dominant or recessive character in this case.

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(i) Man (heterozygous roller) =Rr= Rr= Rr; woman (non-roller) =rr= rr= rr.

(ii) Cross Rr×rrRr \times rrRr × rr. Gametes: man R,rR, rR, r; woman r,rr, rr, r.

rrrrrr
RRRRrRrRrRrRrRr
rrrrrrrrrrrrrrr

Children: 2Rr2\,Rr2\,Rr (rollers) :2rr: 2\,rr: 2\,rr (non-rollers), i.e. genotypes RrRrRr and rrrrrr in a 1:11:11:1 ratio.

(iii) Tongue-rollers (RrRrRr) are 24=50%\dfrac{2}{4} = \mathbf{50\%}2/4 = 50\% of the children.

(iv) Tongue-rolling is a dominant character (it is expressed in the heterozygous RrRrRr children).

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