EXERCISES

Q1. Mention the advantages of selecting pea plant for experiment by Mendel.
Answer: Mendel chose the pea plant (Pisum sativum) for his experiments because it offered several practical advantages that made it ideal for studying inheritance. 

1. Distinct contrasting traits: Pea plants show easily distinguishable traits such as tall/dwarf plants and yellow/green seeds. Mendel selected 14 true-breeding varieties that differed in one clear character.
2. True-breeding lines: These plants maintained stable traits for several generations through self-pollination, ensuring reliability in experiments.
3. Ease of cross-pollination: Pea flowers normally self-pollinate but can also be easily cross-pollinated manually, allowing controlled hybridisation.
4. Short life cycle: The plants grow, flower, and produce seeds quickly, making it possible to study multiple generations in a short time.
5. Large number of seeds: Each plant produces many seeds, providing a large sample size for accurate statistical analysis of inheritance patterns

Q2. Differentiate between the following:
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid
Answer:

(a) Dominant and Recessive Traits: A dominant trait is one that appears in the F₁ hybrid, even when only one of its alleles is present. In contrast, a recessive trait remains hidden in the F₁ generation and expresses only when both alleles are recessive.
For example, in pea plants, tallness (T) is dominant, whereas dwarfness (t) is recessive. Thus, both TT and Tt plants are tall, while tt plants are dwarf.

(b) Homozygous and Heterozygous: In a homozygous condition, both alleles for a trait are similar (like TT or tt), so the organism produces only one kind of gamete.
On the other hand, in a heterozygous condition (like Tt), the alleles are different, and the organism produces two types of gametes, one carrying T and the other t. Hence, homozygous individuals are pure for a trait, whereas heterozygous individuals are hybrids.

(c) Monohybrid and Dihybrid: A monohybrid is an individual heterozygous for one character, for example, Tt for height. The cross between tall (TT) and dwarf (tt) plants is a monohybrid cross, which gives an F₂ ratio of 3:1.
In contrast, a dihybrid is heterozygous for two characters, like RrYy for seed shape and colour. A dihybrid cross between Round Yellow (RRYY) and Wrinkled Green (rryy) plants produces an F₂ ratio of 9:3:3:1.

So, while monohybrids study inheritance of a single trait, dihybrids study inheritance of two traits simultaneously.

Q3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer: If a diploid organism is heterozygous for n loci, the number of different types of gametes it can produce is given by 2ⁿ.
Here, n = 4, so the number of gametes = 2⁴ = 16 types of gametes.

Q4. Explain the Law of Dominance using a monohybrid cross.
Answer: The Law of Dominance states that when two contrasting alleles for a trait are present together, one allele (dominant) expresses itself, while the other (recessive) remains hidden.

Explanation using a Monohybrid Cross (Tall × Dwarf):

• Parental Generation (P): True-breeding Tall plant (TT) is crossed with a True-breeding Dwarf plant (tt).
• Gametes: The Tall parent produces gametes with T, and the Dwarf parent produces gametes with t.
• F₁ Generation: After fertilisation, all offspring have the genotype Tt and appear Tall.

Since all the F₁ plants were tall like the Tall parent, Mendel concluded that the factor for tallness (T) is dominant over the factor for dwarfness (t), which remains recessive.

Q5. Define and design a test-cross.
Answer: A test cross is a cross between an organism showing a dominant phenotype (unknown genotype) and a homozygous recessive individual to determine the genotype of the test organism.

Design (Example: Flower Colour in Pea):

Q6. Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer: Let Tall (T) be dominant over dwarf (t).
Female Parent (Homozygous): We’ll consider homozygous recessive: tt
Male Parent (Heterozygous): Tt

Cross: tt♀ XTt ♂

F₁ Genotypes: 50% Tt, 50% tt
F₁ Phenotypes: 50% Tall, 50% Dwarf
Phenotypic Ratio: 1 Tall : 1 Dwarf

Q7. When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be:
Answer:  Given cross: TtYy × Ttyy
This is a dihybrid cross involving two traits:

• Height: Tall (T) is dominant over dwarf (t)
• Seed colour: Yellow (Y) is dominant over green (y)

Step 1: Consider each trait separately

(a) Height:
Cross → Tt × Tt
Genotypic ratio → 1 TT : 2 Tt : 1 tt
Phenotypic ratio → 3 Tall : 1 Dwarf

$$P(\text{Tall})=\frac{3}{4},P(\text{Dwarf})=\frac{1}{\mathrm{4<span\; style="font-size:\; inherit;\; font-family:\; -apple-system,\; system-ui,\; BlinkMacSystemFont,\; \text{'}Segoe\; UI\text{'},\; Helvetica,\; Arial,\; sans-serif,\; \text{'}Apple\; Color\; Emoji\text{'},\; \text{'}Segoe\; UI\; Emoji\text{'},\; \text{'}Segoe\; UI\; Symbol\text{'};"></span>}}$$(b) Colour:
Cross → Yy × yy
Genotypic ratio → 1 Yy : 1 yy
Phenotypic ratio → 1 Yellow : 1 Green

$$P(\text{Yellow})=\frac{1}{2},P(\text{Green})=\frac{1}{\mathrm{2<span\; style="font-family:\; -apple-system,\; system-ui,\; BlinkMacSystemFont,\; \text{'}Segoe\; UI\text{'},\; Helvetica,\; Arial,\; sans-serif,\; \text{'}Apple\; Color\; Emoji\text{'},\; \text{'}Segoe\; UI\; Emoji\text{'},\; \text{'}Segoe\; UI\; Symbol\text{'};\; font-size:\; inherit;"></span>}}$$Step 2: Combine probabilities

(a) Tall and Green:

$$P(\text{Tall and Green}) = P(\text{Tall}) \times P(\text{Green})$$ $$P(\text{Tall and Green}) = \left(\frac{3}{4}\right) \times \left(\frac{1}{2}\right) = \frac{3}{8}$$Expected Proportion = 3/8

(b) Dwarf and Green:

$$P(\text{Dwarf and Green}) = P(\text{Dwarf}) \times P(\text{Green})$$ $$P(\text{Dwarf and Green}) = \left(\frac{1}{4}\right) \times \left(\frac{1}{2}\right) = \frac{1}{8}$$Expected Proportion = 1/8

Final Answer:

• Tall and Green: 3/8
• Dwarf and Green: 1/8

(a) Tall and Green: (3/4) × (1/2) = 3/8
(b) Dwarf and Green: (1/4) × (1/2) = 1/8

Q8. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer: If the two loci are linked, they do not assort independently. The proportion of parental combinations will be higher than recombinants.
If tightly linked, phenotypic distribution approaches 3:1, resembling a monohybrid ratio.

Q9. Briefly mention the contribution of T.H. Morgan in genetics.
Answer: T.H. Morgan’s work on Drosophila melanogaster led to:

1. Discovery of linkage and recombination.
2. Verification of the Chromosomal Theory of Inheritance.
3. Gene mapping, with Alfred Sturtevant creating the first genetic map.

Q10. What is pedigree analysis? Suggest how such an analysis can be useful.
Answer: Pedigree analysis is the study of how a trait or genetic disease is inherited across multiple generations of a family, typically visualized through a family tree called a pedigree chart. This analysis is useful for determining how a trait is passed down, predicting future inheritance, and identifying genetic risks for conditions like sickle cell anemia or cystic fibrosis

Such an analysis can be useful in many ways such as-

1. Trace inheritance of traits or diseases.
2. Determine whether traits are dominant or recessive, autosomal or sex-linked.
3. Aid in genetic counselling and predicting disorders in offspring.

Q11. How is sex determined in human beings?
Answer: In humans, sex is determined by the XY type of sex determination. Females have two X chromosomes (XX) and produce eggs that contain only one X chromosome. Males have one X and one Y chromosome (XY) and produce two kinds of sperms, some with X chromosomes and others with Y chromosomes. During fertilisation, if an egg carrying X combines with a sperm carrying X, the baby will be a girl (XX). If the egg carrying X combines with a sperm carrying Y, the baby will be a boy (XY). Thus, there is an equal chance of having a boy or a girl, and the father’s sperm decides the sex of the child.

Q12. A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
Child’s genotype (O) = ii
Father (A) = Iᴬi
Mother (B) = Iᴮi

Cross: Iᴬi × Iᴮi

Possible Offspring Genotypes: IᴬIᴮ, Iᴬi, Iᴮi, ii
Possible Blood Groups: AB, A, B, O

Q13. Explain the following terms with example:
(a) Co-dominance
(b) Incomplete dominance 
Answer:

(a) Co-dominance: In co-dominance, both alleles of a gene are fully expressed in the heterozygous condition, and neither one is dominant or recessive. As a result, both traits appear together.
Example: In human blood groups, individuals with genotype IᴬIᴮ have AB blood group, where both A and B antigens are expressed equally.

(b) Incomplete dominance: In incomplete dominance, neither allele is completely dominant over the other, so the heterozygous individual shows a blending or intermediate phenotype between the two parents.
Example: In Snapdragon (Antirrhinum), when a red-flowered (RR) plant is crossed with a white-flowered (rr) plant, the offspring (Rr) have pink flowers.

Q14. What is point mutation? Give one example.
Answer: A point mutation is a change in a single base pair of DNA.
Example: Sickle-cell anaemia in which a single base changes from GAG to GUG in the gene for haemoglobin replaces glutamic acid with valine, causing red blood cells to become sickle-shaped.

Q15. Who had proposed the chromosomal theory of inheritance?
Answer: Walter Sutton and Theodore Boveri proposed the Chromosomal Theory of Inheritance, linking chromosomal segregation with Mendelian principles.

Q16. Mention any two autosomal genetic disorders with their symptoms.
Answer:

• Down’s Syndrome (Trisomy 21): This disorder occurs due to the presence of an extra copy of chromosome 21. Individuals with Down’s syndrome usually have a short stature, a round face with an enlarged tongue, and broad, short fingers. They also show mental retardation, weak muscle tone, and a delay in physical and mental development.
• Sickle-cell Anaemia: This is caused by a mutation in the gene that codes for the beta chain of haemoglobin. Due to a single base substitution, glutamic acid is replaced by valine, making the red blood cells sickle-shaped under low oxygen conditions. These abnormally shaped cells block blood flow, leading to anaemia, body pain, fatigue, and organ damage.