Question

Prophese tolmeiosis At any time throughout division Question 4 A test cross of AaBb x aabb yields the following babies: 35% babies with AB phenotype 35% babies with ab phenotype 15% babies with Ab phenotype 15% babies with a phenotype Which of the following conclusions is correct? The A and genes are located on different chromosomes The A and B genes are located 70 map units apart on the same chromosome The Aand geves are located 35 map units apart on the same chromosome The Aand penes elected 30 units apart on the same chromosome The A and genes are located 15 apnis part on the same chromosome Question 5 The following pedigree documents an X-linked recessive disorder: I tho,

Answer 1

Q4)

• The A and B genes are located 30 map units apart on the same chromosome.

In the test-cross progeny, the two parental type progenies are AB(35%) and ab(35%). This indicates that the double Heterozygous parent (AaBb) has the two dominant alleles in cis (AB/ab).

Now,

Since the total recombinant progeny are 30% (15% of aB and 15% of Ab),

Map Distance = Recombination Frequency = 30

Thus, the two loci A and B are 30 map units apart on the same chromosome.