
AABbCc genotype forms how many types of gametes
(a) 4
(b) 8
(c) 2
(d) 6
Answer
522.6k+ views
Hint: Genotype is considered a genetic combination of traits in an organism. The Mendel’s law of inheritance is followed during the formation of the gametes. The idea of the Punnett square transformed the knowledge of genotype and phenotype.
Complete answer
The plant having the genotype AaBbCc is a triploid plant. To calculate the total number of gametes that are produced by a particular genotype, a specific formula ${ 2 }^{ n }$ is used, where n= number of heterogeneous alleles that are found in the genotype. Here, the given genotype consists of two heterogeneous alleles Bb, and Cc while one homozygous allele is AA. So, it results in the production of ${ 2 }^{ 2 }$= 4 types of gametes. These types of genotypes that are ABC, ABc, AbC, and Abc will be found in the gametes.
Thus the number of gametes produced by the plant having the genotype AABbCc is four.
Additional Information:
-The genotype consists of a complete set of all the heritable genes that can be passed from one generation to the other. In 1903, the Danish botanist Wilhelm Johannsen coined the term ‘genotype’ for the first time.
-The phenotype and the genotype of the offspring of a particular cross can be determined with the help of a square known as ‘Punnett square’. It is the pictorial representation of Mendelian inheritance.
-The Mendel's law of segregation states that from either parent one allele of each gene is found in each gamete.
-The Mendel's law of independent assortment states that each pair of alleles are separated or segregated independently.
So, the correct answer is ‘Four’.
Note:
To calculate the phenotype present in the cross, ${2}^{n}$ formula is used, while for calculating the genotypes present in the cross the formula 3n must be used. The phenotypic ratio of a cross can be calculated by the formula ${(3:1)}^{n}$ where n= number of genes present. The phenotypic ratio for example of a trihybrid cross can be written and calculated as (3:1)3 = (27:9:9:9:3:3:3:1).
Complete answer
The plant having the genotype AaBbCc is a triploid plant. To calculate the total number of gametes that are produced by a particular genotype, a specific formula ${ 2 }^{ n }$ is used, where n= number of heterogeneous alleles that are found in the genotype. Here, the given genotype consists of two heterogeneous alleles Bb, and Cc while one homozygous allele is AA. So, it results in the production of ${ 2 }^{ 2 }$= 4 types of gametes. These types of genotypes that are ABC, ABc, AbC, and Abc will be found in the gametes.
Thus the number of gametes produced by the plant having the genotype AABbCc is four.
Additional Information:
-The genotype consists of a complete set of all the heritable genes that can be passed from one generation to the other. In 1903, the Danish botanist Wilhelm Johannsen coined the term ‘genotype’ for the first time.
-The phenotype and the genotype of the offspring of a particular cross can be determined with the help of a square known as ‘Punnett square’. It is the pictorial representation of Mendelian inheritance.
-The Mendel's law of segregation states that from either parent one allele of each gene is found in each gamete.
-The Mendel's law of independent assortment states that each pair of alleles are separated or segregated independently.
So, the correct answer is ‘Four’.
Note:
To calculate the phenotype present in the cross, ${2}^{n}$ formula is used, while for calculating the genotypes present in the cross the formula 3n must be used. The phenotypic ratio of a cross can be calculated by the formula ${(3:1)}^{n}$ where n= number of genes present. The phenotypic ratio for example of a trihybrid cross can be written and calculated as (3:1)3 = (27:9:9:9:3:3:3:1).
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