Question:

How may types of gametes will be formed by a parent with genotype ‘AaBbCc’?

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Gamete types = $2^n$, where $n$ is the number of heterozygous gene pairs.
Updated On: Apr 18, 2025
  • 8
  • 12
  • 6
  • 4
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The Correct Option is A

Solution and Explanation

  • Gamete Formation: The number of gamete types is determined by the formula $2^n$, where $n$ is the number of heterozygous gene pairs.
  • Genotype Analysis: The genotype ‘AaBbCc’ has three heterozygous pairs (Aa, Bb, Cc), so $n = 3$.
  • Calculation: $2^3 = 8$ possible gamete combinations (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc).
  • Option Analysis:
    • (1) 8: Matches the calculation. Correct.
    • (2) 12: Incorrect, exceeds the possible combinations.
    • (3) 6: Incorrect, does not fit $2^n$ for 3 pairs.
    • (4) 4: Incorrect, corresponds to $2^2$ (two heterozygous pairs).
  • Conclusion: The correct answer is (1) 8, as a parent with genotype AaBbCc can form 8 types of gametes.
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