A homozygous recessive with cd genes was crossed with dominant (+ +). If their hybrid is test crossed and the following result was obtained+ +/cd-900cd/cd-880+d/cd-115c+/cd-105Than what will be the distance between genes c and d. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
5.75 unit
B
11 unit
C
44 unit
D
90 unit

Solution:

Recombination frequency = Number of recombinants/ total number of offsprings X 100 %. In the above example, number of recombinants =220; total number of offsprings = 2000. So 220/2000 X 100 = 11 map units. Hence distance between gene c and d is 11 map units.

A homozygous recessive with cd genes was crossed with dominant (+ +). If their hybrid is test crossed and the following result was obtained
+ +/cd -900
cd/cd -880
+d/cd -115
c+/cd -105
Than what will be the distance between genes c and d.

Solution:

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