Q.41 The total number of genetically different types of gametes that will be produced by a
heterozygous plant carrying the genotypes AABbCc is _________.

The total number of genetically different types of gametes produced by a heterozygous plant with genotype AABbCc is 4. This result follows from Mendel’s law of independent assortment during meiosis. Only heterozygous loci contribute variation to gametes.​

Genotype Analysis

The genotype AABbCc has three gene loci: AA (homozygous, contributes only A), Bb (heterozygous, contributes B or b), and Cc (heterozygous, contributes C or c). Homozygous loci produce one allele type each, while heterozygous loci produce two. Thus, two heterozygous loci (Bb and Cc) determine gamete diversity.​

Gamete Formation Formula

The number of genetically different gametes equals 2n2n, where $n$ is the number of heterozygous gene pairs. Here, $n=2$ (Bb and Cc), so 22=422=4 gametes form. This formula applies assuming no linkage or other complications.​

Possible Gametes

The plant produces these four gamete types through independent assortment:

• ABC (A from AA, B from Bb, C from Cc)

• AbC (A from AA, b from Bb, C from Cc)

• ABc (A from AA, B from Bb, c from Cc)

• Abc (A from AA, b from Bb, c from Cc)​

Introduction to Gamete Diversity in Genetics

Determining the number of genetically different gametes AABbCc plants produce is crucial for CSIR NET Life Sciences aspirants studying Mendelian genetics. This heterozygous plant genotype follows independent assortment, yielding exactly 4 unique gametes via the standard formula.​

Step-by-Step Solution for AABbCc Gametes

Break down AABbCc: AA contributes 1 type (A), Bb contributes 2 types (B, b), Cc contributes 2 types (C, c). Multiply possibilities: $1\times 2\times 2=4$. Use 2n2n where $n=2$ heterozygous pairs for quick calculation.​

Common Exam Options Explained

CSIR NET questions often include distractors:

• 2 gametes: Counts only one heterozygous locus (wrong, ignores Cc).​

• 6 or 8 gametes: Assumes all loci heterozygous like AaBbCc (ignores AA homozygosity).​

• 4 gametes: Correct, matches two heterozygous loci.​

• 1 gamete: For fully homozygous like AABBCC (not applicable).​

Practice for CSIR NET Genetics

Master this for questions on gamete formation: list alleles per locus, count heterozygous pairs, apply 2n2n. For AABbCc, always 4 genetically different gametes ABC, AbC, ABc, Abc.