Question

For given penthybrid cross for $AaBbCcDdEe$ how many possible different genotype and phenotype is obtained in $F_2$ generation?

Answer 1

In the self-cross of $AaBbCcDdEe$ X $AaBbCcDdEe$, there are 5 heterozygous genes. In order to calculate the number of genotypes in F${}_2$ generation, the formula used is 3${}^n$, where n is the number of heterozygous genes. So, the number of genotypes will be 3${}^5$= 243.

The number of phenotypes will be 2${}^n$, so 2${}^5$=32.

So, there will be 243 genotypes and 32 phenotypes possible.