You are trying to map the linked recessive genes black body
Solution
Answer:
(F1 wild female)b+b vg+vg cn cn X (homozygous recessive male) bb vgvg cncn
Phenotypes
Short hand genotypes
Observed Number
Wild-type
b+ vg+ cn+
4150
Black, vestigial, cinnabar
b vg cn
4075
vestigial
b+ vg cn+
425
Cinnabar, black
b vg+ cn
450
Cinnabar, vestigial
b+ vg cn
381
black
b vg+ cn+
437
Cinnabar
b+ vg+ cn
50
Black, vestigial
b vg cn+
25
9993
Imaginary order is b+ vg+ cn+ / b vg cn
1. If single cross over (SCO) occurs between b+ & vg+
Normal order= b+---------vg+ & b-----vg
After cross over= b+-----vg & b------vg+
b+-----vg recombinants are 425 + 381 = 806
b------vg+ recombinants are 450+ 437 = 887
Total recombinants = 1693
RF = (1693/9993)*100 =16.94%
2. If single cross over (SCO) occurs between vg+ & cn+
Normal order= vg+---------cn+ & vg----cn
After cross over= vg+-----cn & vg------cn+
vg+-----cn recombinants are 450+50 = 500
vg------cn+ recombinants are 425+25 = 450
Total recombinants = 950
RF = (950/9993)*100 = 9.5%
3. If single cross over (SCO) occurs between b+ & cn+
Normal order= b+---------cn+ & b------cn
After cross over= b+-----cn & b------cn+
b+-----cn recombinants are 381+ 50 = 431
b------cn+ recombinants are 437+ 25= 462
Total recombinants = 893
RF = (893/9993)*100 =8.94%
% RF = distance between the genes
The order of genes is -----
b+---8.94 cM---cn+----9.5 cM.----vg+
| Phenotypes | Short hand genotypes | Observed Number |
| Wild-type | b+ vg+ cn+ | 4150 |
| Black, vestigial, cinnabar | b vg cn | 4075 |
| vestigial | b+ vg cn+ | 425 |
| Cinnabar, black | b vg+ cn | 450 |
| Cinnabar, vestigial | b+ vg cn | 381 |
| black | b vg+ cn+ | 437 |
| Cinnabar | b+ vg+ cn | 50 |
| Black, vestigial | b vg cn+ | 25 |
| 9993 |


