Two genes ' $A$ ' and ' $B$ ' are linked. In a dihybrid cross involving these two genes, the $F_1$ heterozygote is crossed with homozygous recessive parental type (aa bb ). What would be the ratio of offspring in the next generation?
$1: 1: 1: 1$
$9: 3: 3: 1$
$3: 1$
$1: 1$
In the $F_2$-generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are
phenotypes-4, genotypes-16
phenotypes-9, genotypes-4
phenotypes-4, genotypes-8
phenotypes-4, genotypes-9
Mother and father of a person with ' 0 ' blood group have ' $A$ ' and ' $B$ ' blood group respectively. What would be the genotype of both mother and father?
Mother is homozygous for ' A ' blood group and father is heterozygous for ' B '
Mother is heterozygous for ' A ' blood group and father is homozygous for ' B '
Both mother and father are heterozygous for ' A ' and ' B ' blood group respectively
Both mother and father are homozygous for ' A ' and ' B ' blood group respectively
What is the cross between the progeny of $F_1$ and the homozygous recessive parent called? How is it useful?
When a progeny of $F_1$ is crossed with the homozygous recessive parent, it is called test cross.
Test cross between pure dominant $(\mathrm{A})$ and hybrid dominant $(\mathrm{B})$ individuals with recessive parent is shown below
Such a cross is useful to determine the genotype of an unknown trait, i.e., whether it is heterozygous or homozygous dominant for the trait.
Do you think Mendel's Laws of inheritance would have been different in the characters that he chose were located on the same chromosome.
If the characters are present on the same chromosome they would not assort independently as they are linked on the same chromosome. Percentage of linkage depends on the distance between the genes. With linkage no conclusive laws can be drown.