Question

The F\(_1\) hybrid (Rr Ii) is crossed with a variety double recessive for both the traits. How many types of zygotes will be produced in the cross?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In a test cross, the number of resulting phenotypes and genotypes is determined solely by the number of different gametes the heterozygous parent can produce. For a dihybrid cross (2 heterozygous pairs), it's always 4. For a trihybrid, it would be \(2^3 = 8\).

Answer 1

The Correct Option is D

Step 1: Concept Overview:
This problem involves a dihybrid test cross. A test cross is performed to determine the genotype of a dominant phenotype individual (here, RrIi) by crossing it with a homozygous recessive individual (rrii). The number of different zygotes produced equals the number of distinct gamete types generated by the heterozygous parent.
Step 2: Core Principle:
The formula \(2^n\) determines the number of different gamete types an individual produces, where \(n\) represents the number of heterozygous gene pairs. In a test cross, the number of zygote types matches the number of gamete types from the heterozygous parent.
Step 3: Detailed Breakdown:
The cross is: RrIi \(\times\) rrii


Gametes from the F\(_1\) hybrid (RrIi): With two heterozygous gene pairs (\(n=2\)), this individual produces \(2^2 = 4\) gamete types in equal proportions, following the law of independent assortment: RI, Ri, rI, and ri.

Gametes from the double recessive parent (rrii): Being homozygous for both gene pairs, this individual produces only one gamete type: ri.

Zygote Formation: The combination of gametes results in the following zygotes:

RI + ri \(\rightarrow\) RrIi
Ri + ri \(\rightarrow\) Rrii
rI + ri \(\rightarrow\) rrIi
ri + ri \(\rightarrow\) rrii

Step 4: Conclusion:
The test cross produces four genetically distinct zygote types.