A dihybrid cross is a genetic cross between two individuals that involves the inheritance of two traits. The two traits are controlled by two different genes, each with two alleles (dominant and recessive). This cross follows the principles of Mendelian inheritance. In order to predict the genetic outcome of a dihybrid cross, we use a Punnett square or checkerboard. Here's a step-by-step explanation:
Step 1: Assigning Alleles and Parental Genotypes
Consider two traits:
- Trait 1: Seed color (Yellow - Y, Green - y)
- Trait 2: Seed shape (Round - R, Wrinkled - r)
Assume both parents are heterozygous for both traits (YyRr).
Step 2: Gamete Formation
Each parent can produce four different types of gametes, formed by the independent assortment of alleles:
- Parent 1 (YyRr) can produce the following gametes: YR, Yr, yR, yr
- Parent 2 (YyRr) can produce the same gametes: YR, Yr, yR, yr
Step 3: Setting up the Punnett Square (Checkerboard)
Now, set up a 4x4 Punnett square, where each gamete from one parent is placed along the top and the other parent's gametes along the side. The results are as follows:
\[ \begin{array}{c|c c c c} & YR & Yr & yR & yr \\ \hline YR & YYRR & YYRr & YyRR & YyRr \\ \hline Yr & YYRr & YYrr & YyRr & Yyrr \\ \hline yR & YyRR & YyRr & yyRR & yyRr \\ \hline yr & YyRr & Yyrr & yyRr & yyrr \\ \hline \end{array} \]
Step 4: Genotypic and Phenotypic Ratios
- Genotypic ratio: 1 YYRR : 2 YyRR : 2 YYRr : 4 YyRr : 1 YYrr : 2 Yyrr : 1 yyRR : 2 yyRr : 1 yyrr.
- Phenotypic ratio: 9 Yellow and Round : 3 Yellow and Wrinkled : 3 Green and Round : 1 Green and Wrinkled.
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.
Study the pedigree chart given below, showing the inheritance pattern of blood group in a family:

Answer the following questions:
(a) Give the possible genotypes of individual 1 and 2.
(b) Which antigen or antigens will be present on the plasma membranes of the R.B.Cs of individuals ‘5’ and ‘8’?
The sequence of nitrogenous bases in a segment of a coding strand of DNA is
5' – AATGCTAGGCAC – 3'. Choose the option that shows the correct sequence of nitrogenous bases in the mRNA transcribed by the DNA.