Questions within 8.1
Q1. If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B is likely to have arisen earlier. In asexual reproduction, variations arise from minor inaccuracies during DNA copying. A trait that is present in a larger percentage of the population (60%) has had more time to spread through successive generations, indicating it appeared earlier. Trait A (10%) is more recent and has not yet become as common.
Q2. How does the creation of variations in a species promote survival?
Answer:
Variations provide a range of traits within a species. When environmental conditions change (e.g., a heat wave, new predator, or disease), individuals with certain variations may be better adapted to survive and reproduce. For example, bacteria that can withstand heat will survive better in a heat wave. This natural selection of advantageous variations increases the chances of the species surviving as a whole.
Activity 8.1
QActivity 8.1. Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and calculate the percentage of students having each (Fig. 8.2). Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.
Answer:
Procedure: (1) Observe each student's earlobes and classify them as free (detached) or attached. (2) Record data and calculate percentages: e.g., (Number with free earlobes / total students) × 100%. (3) Ask students about their parents' earlobe types. (4) Correlate: For a student with free earlobes, check if at least one parent has free earlobes; for attached earlobes, check if both parents have attached earlobes or if one parent has free and one has attached. Possible rule: Free earlobes are likely dominant over attached earlobes. If a child has free earlobes, at least one parent must have free earlobes (assuming simple dominance). If both parents have attached earlobes, all children have attached earlobes.
Activity 8.2
QActivity 8.2. In Fig. 8.3, what experiment would we do to confirm that the F2 generation did in fact have a 1:2:1 ratio of TT, Tt and tt trait combinations?
Answer:
To confirm the 1:2:1 ratio, we could perform a test cross. Take the F2 tall plants (which could be TT or Tt) and cross them with a short (tt) plant. If the tall plant is TT, all offspring will be tall (Tt). If the tall plant is Tt, half the offspring will be tall (Tt) and half short (tt). By counting the progeny from many such crosses, we can deduce the actual genotype ratio among the F2 tall plants. Alternatively, allow F2 plants to self-pollinate and observe the trait ratios in the next generation: TT plants will produce all tall, Tt plants will produce 3 tall : 1 short, and tt plants will produce all short. The proportion of each pattern will match the expected 1:2:1 ratio.
Questions within 8.2
Q1. How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel crossed a tall pea plant with a short pea plant. The F1 progeny were all tall, showing that only the tall trait was expressed. When these F1 plants self-pollinated, the F2 generation had both tall and short plants in a 3:1 ratio. The reappearance of the short trait in F2 shows that the trait for shortness was present but hidden in F1. Thus, tallness is dominant (expressed in F1) and shortness is recessive (only expressed when both copies are present).
Q2. How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel performed dihybrid crosses, e.g., crossing plants with round, yellow seeds (RRYY) and wrinkled, green seeds (rryy). The F1 progeny all had round, yellow seeds (RrYy). When F1 plants self-pollinated, F2 showed four types in a 9:3:3:1 ratio: round yellow, round green, wrinkled yellow, wrinkled green. The appearance of new combinations (round green and wrinkled yellow) shows that the inheritance of seed shape is independent of the inheritance of seed colour. The factors (genes) for different traits assort independently during gamete formation.
Q3. A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No, this information alone is not enough to determine dominance. Blood group inheritance involves three alleles: IA, IB, and i. IA and IB are dominant over i. Blood group A can be IAIA or IAi; blood group O is ii. The daughter with blood group O (ii) inherited one i from each parent. So the man must have genotype IAi (since he has A and can pass i). The woman is ii. Here, A is dominant over O because IAi shows A. However, without knowing the genotype of the man (whether IAIA or IAi), we cannot conclude dominance from the given data alone—but in this case, the daughter's O blood type shows the man carries an i, so A is dominant.
Q4. How is the sex of the child determined in human beings?
Answer:
In humans, sex is determined by sex chromosomes. Females have two X chromosomes (XX), males have one X and one Y (XY). During gamete formation, females produce eggs with only X chromosomes. Males produce sperm with either X or Y chromosomes. If an X-bearing sperm fertilises the egg, the child is XX (female). If a Y-bearing sperm fertilises the egg, the child is XY (male). Thus, the father's sperm determines the sex of the child.
Exercises
Q1. A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as (a) TTWW (b) TTww (c) TtWW (d) TtWw
Answer:
The correct option is (c) TtWW. Explanation: All progeny have violet flowers, so violet is dominant over white. Since all progeny have violet, the tall parent must be homozygous for violet (WW) because the short parent is white (ww). For height: half the progeny are short, so the tall parent must be heterozygous for height (Tt) to produce both tall and short offspring when crossed with the short (tt) parent. Thus, tall parent is TtWW.
Q2. A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
No, we cannot determine dominance from this observation alone. Light-coloured eyes could be dominant (if at least one parent has light eyes, children get light eyes) or recessive (if both parents have light eyes, children get light eyes). The study only shows a correlation, not the inheritance pattern. For example, if light eyes are recessive, both parents must be homozygous recessive to have a light-eyed child. If light eyes are dominant, at least one parent must have the dominant allele. More data (e.g., children of parents with dark eyes) is needed to decide.
Q3. Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Project outline: (1) Choose a specific breed of dogs with known coat colours (e.g., black and white). (2) Obtain a pure-breeding black dog and a pure-breeding white dog (homozygous for the trait). (3) Cross them and observe the coat colour of all F1 offspring. The colour that appears in all F1 is dominant. (4) Allow F1 dogs to mate among themselves (or backcross with recessive parent) and observe F2 colours. (5) Record the ratio (e.g., 3:1 if one dominant). (6) Repeat with other colour pairs to confirm. (7) Conclusion: The colour that appears in F1 and in approximately 3/4 of F2 is dominant.
Q4. How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Equal genetic contribution is ensured by the process of meiosis and fertilisation. Each parent produces haploid germ cells (sperm and egg) that contain one set of chromosomes (n). The sperm contributes 23 chromosomes and the egg contributes 23 chromosomes. During fertilisation, they fuse to form a diploid zygote (2n = 46), with half the chromosomes from the father and half from the mother. Thus, the progeny receives equal genetic material from both parents. Additionally, for each gene, the child inherits one allele from the mother and one from the father.