Solve this:
Q.136. individual of the cross AAbb aaBB was test crossed. If both non-allelic genes are 24 map units apart then calculate the percentage of aaBb :
(1) 24 %
(2) 12 %
(3) 76 %
(4) 38 %
Dear student,
Option D is correct answer.
Map unit (m.u) is defined as the measure of distance between two genes corresponding to a recombination frequency of one percent. It is simply used to express distance between genes on chromosome. Since distance between two non-allelic genes is 24 map units, the recombination frequency (crossing over frequency) between these genes will be 24% i.e. out of 100 gametes, total 24% gametes will be recombinant gametes and 100-24= 76% will be parental types. So, frequency of parental type gametes "Ab" and "aB" will be 38% (76/2= 38) each. This gives 38% "aaBb" genotype in test cross progeny.
Regards.
Option D is correct answer.
Map unit (m.u) is defined as the measure of distance between two genes corresponding to a recombination frequency of one percent. It is simply used to express distance between genes on chromosome. Since distance between two non-allelic genes is 24 map units, the recombination frequency (crossing over frequency) between these genes will be 24% i.e. out of 100 gametes, total 24% gametes will be recombinant gametes and 100-24= 76% will be parental types. So, frequency of parental type gametes "Ab" and "aB" will be 38% (76/2= 38) each. This gives 38% "aaBb" genotype in test cross progeny.
Regards.