Heredity

Given/Concept: Mendel performed monohybrid crosses using contrasting traits in pea plants (e.g., tall vs. short, round seeds vs. wrinkled seeds).

Explanation:

Step 1: Mendel crossed pure-breeding tall plants (TT) with pure-breeding short plants (tt). All plants in the F₁ generation were tall.

$$TT \times tt \rightarrow \text{All } Tt \text{ (Tall)}$$

Step 2: When F₁ plants (Tt) were self-pollinated, the F₂ generation showed both tall and short plants in a ratio of approximately 3:1.

$$Tt \times Tt \rightarrow TT : Tt : tt = 1:2:1$$
$$\text{Phenotype ratio} = 3 \text{ Tall} : 1 \text{ Short}$$

Step 3: The trait that appeared in F₁ (tallness) and in 3/4 of F₂ plants is called the dominant trait. The trait that disappeared in F₁ but reappeared in 1/4 of F₂ plants (shortness) is called the recessive trait.

Conclusion: Since the recessive trait (shortness) was hidden in F₁ but reappeared in F₂, Mendel concluded that traits can be dominant (expressed even in a single copy) or recessive (expressed only when both copies are identical). This shows that both copies of a gene are present in an organism, but only the dominant one is expressed when both are present together.

Given/Concept: Mendel performed dihybrid crosses to study the inheritance of two traits simultaneously.

Explanation:

Step 1: Mendel crossed pure-breeding round yellow seed plants (RRYY) with pure-breeding wrinkled green seed plants (rryy).

$$RRYY \times rryy \rightarrow \text{All } RrYy \text{ (Round Yellow — F}_1\text{)}$$

Step 2: F₁ plants (RrYy) were self-pollinated to get F₂ generation.

$$RrYy \times RrYy$$

Step 3: The F₂ generation showed four phenotypic classes in the ratio:
$$\text{Round Yellow : Round Green : Wrinkled Yellow : Wrinkled Green} = 9:3:3:1$$

Step 4: New combinations — Round Green and Wrinkled Yellow — appeared in F₂ that were not present in either parent. This is only possible if the genes for seed shape and seed colour are inherited independently of each other.

Conclusion: If the two traits were linked and always inherited together, only parental combinations (Round Yellow and Wrinkled Green) would appear. The appearance of new combinations in a 9:3:3:1 ratio proves that traits are inherited independently (Law of Independent Assortment).

Given:
- Father's blood group = A
- Mother's blood group = O
- Daughter's blood group = O

Analysis:

Step 1: The daughter has blood group O. She received one allele from her father (blood group A) and one from her mother (blood group O).

Step 2: Since the daughter's blood group is O, the allele for blood group O must have been expressed. This means the father must have contributed an allele for O (not A) to the daughter.

Step 3: For the father (blood group A) to pass an O allele to his daughter, his genotype must be heterozygous ($I^A i$), i.e., he carries one A allele and one O allele. The daughter received the $i$ (O) allele from the father and $i$ (O) allele from the mother, making her genotype $ii$ (blood group O).

$$\text{Father: } I^A i \times \text{Mother: } ii \rightarrow \text{Daughter: } ii \text{ (Blood group O)}$$

Step 4: Since the father has blood group A with genotype $I^A i$, the A allele is expressed even in the presence of the O allele. This means blood group A is dominant over blood group O.

Conclusion: Yes, this information is enough to conclude that blood group A is dominant over blood group O, because the father expresses blood group A despite carrying an O allele, which means A masks O — confirming A is dominant and O is recessive.

Concept: Sex determination in human beings is based on sex chromosomes.

Step 1 — Sex chromosomes in parents:
- Every human has 23 pairs of chromosomes. Of these, 22 pairs are autosomes and 1 pair are sex chromosomes.
- Females have two X chromosomes → genotype XX
- Males have one X and one Y chromosome → genotype XY

Step 2 — Gamete formation:
- The mother (XX) produces eggs, all of which carry one X chromosome.
- The father (XY) produces two types of sperms:
- 50% sperms carry X chromosome
- 50% sperms carry Y chromosome

Step 3 — Fertilisation:
$$\text{If X sperm + X egg} \rightarrow XX \rightarrow \textbf{Girl}$$
$$\text{If Y sperm + X egg} \rightarrow XY \rightarrow \textbf{Boy}$$

Step 4 — Probability:
Since half the sperms carry X and half carry Y, there is a 50% chance of a boy and 50% chance of a girl.

Conclusion: The sex of the child is determined by the type of sperm (carrying X or Y chromosome) that fertilises the egg. The mother always contributes an X chromosome; it is the father's sperm that determines whether the child will be a boy (Y sperm) or a girl (X sperm). Therefore, the father is responsible for determining the sex of the child.

Correct Answer: (c) TtWW

Justification:

Given observations:
- All progeny have violet flowers → violet is dominant (W); the tall parent must be homozygous dominant for flower colour (WW), since crossing with white (ww) gives all violet (Ww).
- Almost half the progeny are short → shortness reappears, meaning the tall parent must be heterozygous for height (Tt), so that crossing Tt × tt gives 1 Tt (tall) : 1 tt (short), i.e., half tall and half short.

Cross verification:
$$TtWW \times ttWw$$
- For height: $Tt \times tt \rightarrow Tt : tt = 1:1$ (half tall, half short ✓)
- For flower colour: $WW \times ww \rightarrow$ all $Ww$ (all violet ✓)

Conclusion: The genetic make-up of the tall parent is TtWW, i.e., option (c).

Given: Children with light-coloured eyes tend to have parents with light-coloured eyes.

Analysis:

Step 1: If light eye colour were dominant, then a parent with light eyes could be either homozygous (LL) or heterozygous (Ll). A heterozygous parent (Ll) crossed with another light-eyed parent could produce dark-eyed children (ll). But the study says light-eyed parents consistently have light-eyed children.

Step 2: If light eye colour were recessive, then light-eyed parents would be homozygous recessive (ll × ll), and all their children would also be ll (light-eyed). This is consistent with the observation.

Step 3: However, the same observation is also consistent with light eye colour being dominant and homozygous (LL × LL → all LL, all light-eyed).

Conclusion: No, we cannot conclusively determine from this information alone whether light eye colour is dominant or recessive. The observation is consistent with both possibilities:
- Light colour could be recessive (ll parents always give ll children).
- Light colour could be dominant and homozygous (LL parents always give LL children).

To determine dominance, we would need to observe what happens when a light-eyed individual mates with a dark-eyed individual and study the eye colour of the offspring.

Project Title: Determining the Dominant Coat Colour in Dogs

Objective: To identify which coat colour (e.g., black or brown/white) is dominant in dogs.

Materials Required: Dogs of two contrasting pure-breeding coat colours (e.g., pure black-coated dogs and pure white-coated dogs), record-keeping tools.

Procedure:

Step 1 — Selection of pure-breeding parents (P generation):
Select pure-breeding dogs of two contrasting coat colours — for example, pure black-coated dogs and pure white-coated dogs. (Pure-breeding can be confirmed by allowing them to breed among themselves for several generations and observing that the coat colour does not change.)

Step 2 — Cross (F₁ generation):
Cross the pure black-coated dogs with pure white-coated dogs.
$$\text{Black (BB)} \times \text{White (bb)} \rightarrow F_1 \text{ progeny}$$
Observe and record the coat colour of all F₁ offspring.
- The coat colour that appears in all F₁ offspring is the dominant trait.

Step 3 — Self-cross / Inter-cross of F₁ (F₂ generation):
Allow F₁ dogs to mate among themselves.
$$F_1 \times F_1 \rightarrow F_2 \text{ progeny}$$
Observe and record the coat colours of F₂ offspring.

Step 4 — Analysis:
- The coat colour appearing in approximately 3/4 of F₂ offspring is the dominant colour.
- The coat colour appearing in approximately 1/4 of F₂ offspring is the recessive colour.
- The recessive colour reappears in F₂ after being absent in F₁.

Expected Result: If black coat appears in all F₁ and in 3:1 ratio in F₂, then black coat colour is dominant and white coat colour is recessive.

Conclusion: By following Mendel's monohybrid cross approach, we can determine the dominant coat colour in dogs.

Concept: Equal genetic contribution of both parents is ensured through the process of meiosis (gamete formation) and fertilisation.

Step 1 — Chromosome number in body cells:
Every human body cell (somatic cell) contains 46 chromosomes (23 pairs) — one chromosome of each pair inherited from the mother and one from the father.

Step 2 — Gamete formation (Meiosis):
During the formation of gametes (sperm and egg) by meiosis, the chromosome number is halved.
- Each sperm (from father) contains 23 chromosomes (haploid, $n$).
- Each egg/ovum (from mother) contains 23 chromosomes (haploid, $n$).

$$\text{Diploid parent (2n = 46)} \xrightarrow{\text{Meiosis}} \text{Haploid gamete (n = 23)}$$

Step 3 — Fertilisation:
During fertilisation, the sperm fuses with the egg:
$$\text{Sperm (23 chromosomes)} + \text{Egg (23 chromosomes)} \rightarrow \text{Zygote (46 chromosomes)}$$

Step 4 — Equal contribution:
- The zygote (and hence the child) receives exactly 23 chromosomes from the father (via sperm) and 23 chromosomes from the mother (via egg).
- This means 50% of the genetic material comes from the father and 50% from the mother.

Conclusion: The halving of chromosomes during meiosis (gamete formation) and the subsequent fusion of two haploid gametes during fertilisation ensures that each parent contributes equally (50%) to the genetic make-up of the progeny, maintaining the chromosome number constant across generations.