Step 1: Identifying the number of heterozygous loci.
The plant's genotype is AABbCc. In this case: - AA is homozygous, so it can only produce one type of allele (A). - Bb is heterozygous, so it can produce two types of alleles (B and b). - Cc is heterozygous, so it can produce two types of alleles (C and c).
Step 2: Calculating the number of genetically different gametes.
The number of different types of gametes produced depends on the number of heterozygous loci, because each heterozygous locus can produce two types of alleles.
- For gene A (AA): Only one type of allele, A, is possible.
- For gene B (Bb): Two types of alleles, B and b, are possible.
- For gene C (Cc): Two types of alleles, C and c, are possible.
The total number of genetically different gametes can be calculated by multiplying the number of possibilities for each locus: \[ \text{Total number of gametes} = 1 \times 2 \times 2 = 4 \]
Step 3: Conclusion.
Thus, the total number of genetically different types of gametes produced is 4. Therefore, the correct answer is 4.
If a heterozygous tall plant (Tt) is crossed with a dwarf plant (tt), what will be the phenotypic ratio of the offspring?
The vector, shown in the figure, has promoter and RBS sequences in the 300 bp region between the restriction sites for enzymes X and Y. There are no other sites for X and Y in the vector. The promoter is directed towards the Y site. The insert containing only an ORF provides 3 fragments after digestion with both enzymes X and Y. The ORF is cloned in the correct orientation in the vector using the single restriction enzyme Y. The size of the largest fragment of the recombinant plasmid expressing the ORF upon digestion with enzyme X is ........... bp. (answer in integer) 