Question

The genotypes of husband and wife are \( I^A I^B \) and \( I^A I^O \). Among the blood groups of their children, how many different genotypes and phenotypes are possible?

• A. 4 genotypes and 4 phenotypes
• B. 4 genotypes and 3 phenotypes
• C. 3 genotypes and 4 phenotypes
• D. 2 genotypes and 3 phenotypes

Hint:

Always map out the cross completely. Notice that \(I^A I^A\) and \(I^A I^O\) yield the same phenotype (Blood type A), which reduces the phenotype count compared to the genotype count.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: Human ABO blood grouping is controlled by the gene \(I\) which exhibits multiple allelism and codominance through three alleles: \(I^A\), \(I^B\), and \(I^O\) (or \(i\)).
Step 2: Key Formula or Approach: Perform a standard Mendelian genetic cross using a Punnett square between the given parental genotypes: \( I^A I^B \times I^A I^O \).
Step 3: Detailed Explanation:
The possible gametes for the husband (\(I^A I^B\)) are \(I^A\) and \(I^B\).
The possible gametes for the wife (\(I^A I^O\)) are \(I^A\) and \(I^O\).
Crossing these gametes gives the following offspring combinations:
- \(I^A\) from husband + \(I^A\) from wife = \(I^A I^A\) (Genotype 1 $\rightarrow$ Phenotype A)
- \(I^A\) from husband + \(I^O\) from wife = \(I^A I^O\) (Genotype 2 $\rightarrow$ Phenotype A)
- \(I^B\) from husband + \(I^A\) from wife = \(I^A I^B\) (Genotype 3 $\rightarrow$ Phenotype AB)
- \(I^B\) from husband + \(I^O\) from wife = \(I^B I^O\) (Genotype 4 $\rightarrow$ Phenotype B)
Counting the unique combinations, we clearly see 4 distinct genotypes: \(I^A I^A\), \(I^A I^O\), \(I^A I^B\), and \(I^B I^O\).
Looking at the observable traits (phenotypes), we have 3 distinct groups: Blood Group A, Blood Group B, and Blood Group AB.
Step 4: Final Answer: The cross results in 4 genotypes and 3 phenotypes.