How many kinds of gametes can be produced by an individual of genotype `(A)/(a)(B)/(b)(C)/(c )(D)/(d)(e )/(e ) (F)/(f)`

Options

A. 32
B. 16
C. 8
D. 4

Answer

To determine how many kinds of gametes can be produced by an individual with the genotype \( (A)/(a)(B)/(b)(C)/(c)(D)/(d)(e)/(e)(F)/(f) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotype**: Write down the complete genotype.
- The genotype is: \( A/a, B/b, C/c, D/d, e/e, F/f \).

2. **Classify Alleles**: Determine which alleles are heterozygous and which are homozygous.
- Heterozygous alleles: \( A/a, B/b, C/c, D/d, F/f \)
- Homozygous alleles: \( e/e \) (this pair will not contribute to gamete diversity).

3. **Count the Heterozygous Alleles**: Count the number of heterozygous pairs.
- There are 5 heterozygous pairs: \( A/a, B/b, C/c, D/d, F/f \).

4. **Apply the Formula for Gametes**: Use the formula \( 2^n \), where \( n \) is the number of heterozygous pairs.
- Here, \( n = 5 \).
- Therefore, the number of different gametes is \( 2^5 \).

5. **Calculate the Result**: Compute \( 2^5 \).
- \( 2^5 = 32 \).

### Final Answer:
The individual can produce **32 different kinds of gametes**.

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