Q.1) Mention the advantages of selecting pea plant for experiment by Mendel.
Ans: Mendel selected pea plants for his genetics experiments because of the following advantages:
- Pea plants have many easily observable contrasting traits such as tall/dwarf plants, purple/white flowers and yellow/green pods.
- They are naturally self-pollinating, which helps in maintaining pure breeding lines.
- Cross-pollination can be easily carried out by emasculation and artificial pollination.
- Pea plants have a short life cycle and produce a large number of seeds, allowing statistical analysis of results.
Q.2) Differentiate between the following:
(a) Dominant and Recessive
Ans: A dominant allele expresses itself even in the presence of a recessive allele, whereas a recessive allele expresses itself only in the absence of the dominant allele.
(b) Homozygous and Heterozygous
Ans: A homozygous genotype has identical alleles for a trait (TT or tt), while a heterozygous genotype has different alleles (Tt).
(c) Monohybrid and Dihybrid
Ans: A monohybrid cross involves one pair of contrasting traits, whereas a dihybrid cross involves two pairs of contrasting traits.
Q.3) A diploid organism is heterozygous for 4 loci. How many types of gametes can be produced?
Ans: If an organism is heterozygous for four loci, it can produce 16 different types of gametes (2⁴), assuming independent assortment.
Q.4) Explain the Law of Dominance using a monohybrid cross.
Ans: Mendel’s Law of Dominance states that when two contrasting alleles are present, only one (the dominant allele) expresses itself, while the other (recessive allele) remains hidden.
In a cross between tall (TT) and dwarf (tt) pea plants, all F₁ progeny are tall (Tt). In the F₂ generation, the dwarf trait reappears with a phenotypic ratio of 3:1.
Q.5) Define and design a test cross.
Ans: A test cross is performed by crossing an individual showing a dominant phenotype with a homozygous recessive individual to determine its genotype.
If all offspring show the dominant trait, the parent is homozygous dominant. If both dominant and recessive traits appear, the parent is heterozygous.
Q.6) Using a Punnett square, work out the phenotypic distribution in F₁ generation after crossing a homozygous female and heterozygous male.
Ans: When a homozygous female (BB) is crossed with a heterozygous male (AB), the progeny show a phenotypic ratio of 1:1.
Q.7) When a cross is made between TtYy and Ttyy, find the proportion of:
(a) Tall and green
(b) Dwarf and green
Ans:
- Tall and green – 3
- Dwarf and green – 1
Q.8) What will be the phenotypic distribution if two heterozygous parents are crossed and genes are linked?
Ans: If genes are linked, they do not assort independently. The phenotypic ratio deviates from 9:3:3:1 and may resemble a monohybrid ratio, depending on the degree of linkage.
Q.9) Briefly mention the contribution of T.H. Morgan in genetics.
Ans: T.H. Morgan proposed the chromosomal theory of inheritance and demonstrated linkage and recombination using Drosophila. He showed that genes are located on chromosomes and inherited together if linked.
Q.10) What is pedigree analysis? How is it useful?
Ans: Pedigree analysis is the study of inheritance of traits across generations using family trees. It helps determine whether a trait is dominant, recessive or sex-linked and predicts genetic disorders.
Q.11) How is sex determined in human beings?
Ans: Sex determination in humans depends on sex chromosomes. Females are XX and males are XY. The sperm determines the sex of the child by contributing either X or Y chromosome.
Q.12) A child has blood group O. Father has blood group A and mother has blood group B. Find the genotypes.
Ans: Both parents must be heterozygous (IAi and IBi). Possible offspring genotypes are IAIB, IAi, IBi and ii.
Q.13) Explain:
(a) Co-dominance
Ans: In co-dominance, both alleles express equally in a heterozygous condition, e.g. AB blood group.
(b) Incomplete dominance
Ans: In incomplete dominance, neither allele is completely dominant, resulting in an intermediate phenotype, e.g. pink flowers in snapdragon.
Q.14) What is point mutation? Give an example.
Ans: A point mutation is a change in a single nucleotide base in DNA. Sickle-cell anaemia is caused by a point mutation that substitutes valine for glutamic acid.
Q.15) Who proposed the chromosomal theory of inheritance?
Ans: The chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Boveri in 1902.
Q.16) Mention any two autosomal genetic disorders with symptoms.
Ans:
- Sickle cell anaemia: Causes sickle-shaped RBCs, leading to breathlessness, fatigue and pain.
- Phenylketonuria: Leads to mental retardation due to accumulation of phenylalanine.