Question

(A)

Perform a cross between two sickle cell carriers. What ratio is obtained between carrier, disease free and diseased individuals in \( F_1 \) progeny? Name the nitrogenous base substituted, in the haemoglobin molecule in this disease.
Explain the difference in inheritance pattern of flower colour in garden pea plant and snap-dragon plant with the help of monohybrid crosses. \begin{center} OR \end{center} (B) Explain with the help of well-labelled diagrams how lac operon operates in \textit{E. coli} :

In presence of an inducer.
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):

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