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: Understanding the Concept:
The question describes a dihybrid test cross. A test cross involves crossing an individual with a dominant phenotype (but unknown genotype, here given as heterozygous RrIi) with a homozygous recessive individual (rrii) to determine the genotype of the former. The number of zygote types produced is equal to the number of different gamete types produced by the heterozygous parent.
Step 2: Key Formula or Approach:
The number of different gamete types produced by an individual is given by the formula \(2^n\), where \(n\) is the number of heterozygous gene pairs. The number of zygote types in a test cross is equal to the number of gamete types from the heterozygous parent.
Step 3: Detailed Explanation:
The cross is: RrIi \(\times\) rrii


Gametes from the F\(_1\) hybrid (RrIi): This individual is heterozygous for two gene pairs (\(n=2\)). According to the law of independent assortment, it will produce \(2^2 = 4\) types of gametes in equal proportions: RI, Ri, rI, and ri.

Gametes from the double recessive parent (rrii): This individual is homozygous for both gene pairs and can only produce one type of gamete: ri.

Formation of zygotes: When the gametes combine, the following zygotes are formed:

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

Step 4: Final Answer:
There are four genetically distinct types of zygotes produced in this cross.