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In mendelian dihybrid cross how many zygotes are homozygous recessive for one of the character only in F 2 generation. Answer is 6/16 How?
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In mendelian dihybrid cross how many zygotes are homozygous recessive ...
Mendelian Dihybrid Cross

A Mendelian dihybrid cross is a genetic experiment where two traits are studied together. The traits are controlled by two different genes, each occupying a different locus on a chromosome.

F2 Generation

The F2 generation is the second filial generation that results from the cross between two F1 individuals. In a Mendelian dihybrid cross, the F2 generation produces 16 different possible genotypes.

Homozygous Recessive Zygotes

Homozygous recessive zygotes are individuals with two copies of a recessive allele for a particular gene. In a Mendelian dihybrid cross, there are four possible homozygous recessive zygotes for each of the two genes being studied.

Calculating the Probability

To calculate the probability of getting homozygous recessive zygotes for one of the two genes only, we need to use a Punnett square. The Punnett square shows all the possible combinations of alleles that can result from the cross between two individuals.

For example, if we cross two individuals that are heterozygous for both genes (AaBb x AaBb), the Punnett square would look like this:

| | A | a |
|---|---|---|
| B | AB| aB|
| b | Ab| ab|

In this case, we can see that there are six possible genotypes that result in homozygous recessive zygotes for one of the two genes only: aabb, Aabb, aaBb, AaBb, Aabb, and aaBB.

Therefore, the probability of getting homozygous recessive zygotes for one of the two genes only is 6/16 or 0.375.
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In mendelian dihybrid cross how many zygotes are homozygous recessive for one of the character only in F 2 generation. Answer is 6/16 How?
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