Question

What is the probability of obtaining pure homozygous individuals, if a cross is made between $AaBbCcDd$ X $aaBBCCdd$ individuals?

• A. $\frac{1}{16}$
• B. $\frac{2}{16}$
• C. $\frac{16}{248}$
• D. $\frac{1}{64}$

Answer 1

The correct option is A $\frac{1}{16}$

Number of gametes produced = 2^n, where n is numbers of Heterozygous Gene pairs

Genotype AaBbCcDd: Has 4 numbers of Heterozygous Gene pairs

So the number of gametes are 4^4=16

The Genotype aaBBCCdd: Homozygous for all the Gene pairs

So the gamete produced will be only 1 aBCd

Total 16 male gametes and 1 female gamete produced will result in 16 offsprings.

Out of 16, there will be only one genotype aaBBCCdd Homozygous for all genes

So the offsprings Homozygous for all gene pairs= 1/16