How many homozygous recessive individuals by phenotype will turn out as a result of crossing two heterozygous individuals

How many homozygous recessive individuals by phenotype will turn out as a result of crossing two heterozygous individuals (AaBb) (according to Mendel’s third law)?

According to Mendel’s third law, when two heterozygotes are crossed, phenotypic cleavage will follow the 9: 3: 3: 1 scheme. When constructing a Punnett lattice for this example, the father and mother gametes will have the same appearance; AB, AB, AB and AB – horizontally and vertically. After filling in the table and counting, we obtain the following: 9 individuals with the phenotype A_B_ 3 individuals – A_bb 3 individuals – aaB_ 1 individual – aavv Only the “aavv” individual can be classified as homozygous recessive. Answer: when crossing two heterozygotes, 1 out of 16 offspring will be homozygous recessive in phenotype.