Answers
Answer:
BBCC (true-black) x bbcc(albino)----P1
BbCc (black)-----------------------------F1
BbCc (F1) x bbcc (double recessive)
| bc | |
| BC | BbCc (black) |
| Bc | Bbcc (albino) |
| bC | bbCc (brown) |
| bc | bbcc (albino) |
If the genes are unlinked, the phenotypic proportions of the cats are as follows.
Black = ¼ = ¼ * 200 = 50
Brown = ¼ = ¼ * 200 = 50
Albino = ½ = ½ * 200 = 100
Recombination frequency = (Recombinants / Total progeny)100
RF= (60/ 200)100
= 30%
RF(%) = Distance between the genes (m.u.)
So the distance between the genes = 30 m.u.
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