Question

Сколько типов гамет формируется у родительского организма с генотипом AaBb в случае сцепленного наследования при отсутствии кроссинговера??

Answer 1

Образуется 2 типа: АВ и ав

Answer 2

Types of Gametes Formed by a Parent Organism with Genotype AaBb in Linked Inheritance without Crossing Over

In the case of linked inheritance without crossing over, the number of different types of gametes formed by a parent organism with a genotype of AaBb can be determined by examining the possible combinations of alleles for each gene.

The genotype AaBb indicates that the organism has two genes, A and B, with two alleles each, A and a for gene A, and B and b for gene B. In linked inheritance, these genes are located close together on the same chromosome, resulting in a lower likelihood of recombination through crossing over.

To determine the number of different types of gametes, we need to consider the possible combinations of alleles for each gene. Since there are two alleles for each gene, there are four possible combinations for gene A (AA, Aa, aA, aa) and four possible combinations for gene B (BB, Bb, bB, bb).

To find the number of different types of gametes, we multiply the number of possible combinations for each gene. Therefore, the total number of different types of gametes formed by a parent organism with genotype AaBb in linked inheritance without crossing over is 4 x 4 = 16.

Therefore, in the case of linked inheritance without crossing over, a parent organism with genotype AaBb can form 16 different types of gametes.