Question

1. Perform a cross between two sickle cell carriers. What ratio is obtained between carrier, disease free and diseased individuals in F1 progeny? Name the nitrogenous base substituted, in the haemoglobin molecule in this disease.
2. Explain the difference in inheritance pattern of flower colour in garden pea plant and snap-dragon plant with the help of monohybrid crosses.
OR,
Explain with the help of well-labelled diagrams how lac operon operates in E. coli :
1. In presence of an inducer.
2. In absence of an inducer.

Hint:

Remember — Lac operon is a classic example of an inducible operon, and incomplete dominance results in an intermediate phenotype in \( F_1 \) generation.

Answer 1

Step 1 (A): (A) (B) Cross between two sickle cell carriers (Hb\( ^A \)Hb\( ^S \) × Hb\( ^A \)Hb\( ^S \)) results in: (C) 1 Normal (Hb\( ^A \)Hb\( ^A \)) (D) 2 Carriers (Hb\( ^A \)Hb\( ^S \)) (E) 1 Diseased (Hb\( ^S \)Hb\( ^S \)) Ratio = 1:2:1 (F) Substitution: Adenine is replaced by Thymine in the sixth codon of the beta globin gene, leading to substitution of glutamic acid by valine. (G) (H) Garden Pea Plant: Shows complete dominance (e.g., red × white flower yields all red in \( F_1 \)). (I) Snap-Dragon Plant: Shows incomplete dominance (e.g., red × white yields pink in \( F_1 \)). (J) Monohybrid cross in pea plant gives a phenotypic ratio of 3:1, while in snap-dragon it is 1:2:1. Step 2 (B):
In presence of an inducer (e.g., lactose):
Inducer binds to repressor.
Repressor becomes inactive.
RNA polymerase transcribes structural genes (lac Z, Y, A).
Enzymes for lactose metabolism produced.
In absence of an inducer:
Active repressor binds to operator.
RNA polymerase is blocked.
No transcription.
No enzyme production.