Q.48 The maximum number of genotypes possible for gametes formed from a diploid cell of the
genotype AaBBcCDd is _____.
Maximum Gametes from AaBBcCDd Genotype: 8 Types Explained
The maximum number of genotypes possible for gametes from a diploid cell with genotype AaBBcCDd is 8. This result follows from Mendel’s law of independent assortment, applied only to heterozygous loci.
Gamete Formation Formula
Gametes form through meiosis, where each heterozygous gene pair segregates independently, yielding 2 alleles per pair. Homozygous pairs (AA or aa) produce only 1 allele type. The total gamete genotypes equal 2n, with n as heterozygous loci count.
For AaBBcCDd:
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Aa: heterozygous (2 options: A, a)
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BB: homozygous (1: B)
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cC: heterozygous (2: c, C)
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Dd: heterozygous (2: D, d)
Thus, n=3 (Aa, cC, Dd), so 23=8 gametes.
All Possible Gametes Listed
The 8 unique gamete genotypes combine alleles from varying loci:
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ABC D
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ABC d
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ABc D
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ABc d
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aBC D
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aBC d
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aBc D
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aBc d
Each arises from independent segregation; homozygous BB fixes B in all.
Common Options Explained
Exam questions often list choices like 4, 8, 16, 32. Here’s why:
| Option | Reason | Incorrect/Correct | Heterozygous Count Assumed |
|---|---|---|---|
| 4 | 22 × 22: Ignores 1 heterozygous (e.g., mistakes cC as homozygous) | Incorrect | 2 |
| 8 | 23 × 23: Matches Aa, cC, Dd exactly | Correct | 3 |
| 16 | 24 × 24: Wrongly counts BB as heterozygous | Incorrect | 4 |
| 32 | 25 × 25: Assumes 5 loci heterozygous, but only 4 genes total | Incorrect | 5 |
Correct answer: 8, as BB contributes no variation.
