Question

The test cross ratio for a trait governed by two genes with masking gene action will be-

• A. 1:1:1:1
• B. 2:1:1
• C. 3:1
• D. 1:1

Hint:

To quickly find a test cross ratio from an $F_2$ ratio, group the 16 parts of the $F_2$ ratio ($12:3:1$) into their corresponding four gametic classes, which simplifies directly to the test cross ratio $2:1:1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Masking gene action is a form of gene interaction known as dominant epistasis. It occurs when a dominant allele at one gene locus (e.g., \( A \)) masks the phenotypic expression of alleles at a second gene locus (e.g., \( B \) or \( b \)).

Step 2: Detailed Explanation:
Let us designate the two gene loci as \( A/a \) and \( B/b \).
Let allele \( A \) be epistatic (masking) over alleles \( B \) and \( b \).
To find the test cross ratio, we cross a dihybrid (\( AaBb \)) with a homozygous double recessive tester (\( aabb \)): \[ AaBb \times aabb \]
The cross produces four genotypes in equal frequencies (\( 1:1:1:1 \)):
1. \( AaBb \) (Frequency: 1)
2. \( Aabb \) (Frequency: 1)
3. \( aaBb \) (Frequency: 1)
4. \( aabb \) (Frequency: 1)
Let us determine the phenotypic expression for each genotype under dominant epistasis:
- Genotypes with \( A \): Both \( AaBb \) and \( Aabb \) contain the dominant masking allele \( A \). Thus, both will express the same "masked" phenotype (Phenotype 1). Their frequencies pool together: \[ 1 + 1 = 2 \]
- Genotype \( aaBb \): Lacks the dominant \( A \) allele, so the effect of dominant allele \( B \) is expressed (Phenotype 2). Frequency is 1.
- Genotype \( aabb \): Lacks both dominant alleles, expressing the double recessive phenotype (Phenotype 3). Frequency is 1.
Combining these phenotypic classes yields a ratio of: \[ 2 : 1 : 1 \]

Step 3: Final Answer:
Therefore, the test cross ratio under masking gene action is 2:1:1.