Can you please help me with this problem:
Homework Answers
a). If the three genes are assorted independently, the trihybrid testcross progeny would be equal proportions. As the numbers are not equal, hence, the three genes are linked.
b).
Hint: Always non-recombinant genotypes are large numbered than the recombinant genotypes.
Hence, the parental (non-recombinant) triple heterozygous genotypes is b+ wx+ cn / b wx cn
c).
Wild type = b+ wx+ cn+ = 5
Black, wavy, cinnabar = b wx cn = 6
waxy, cinnabar = b+ wx cn = 69
Black = b wx+ cn+ = 67
Cinnabar = b+ wx+ cn = 382
Black, waxy = b wx cn+ = 379
Waxy= b+ wx cn+ = 48
Black, cinnabar = b wx+ cn = 44
Total 1000
1).
If single crossover occurs between b & wx..
Normal combination: b+ wx+ / b wx
After crossover: b+ wx / b wx+
b+ wx progeny= 69+48 = 117
b wx+ progeny = 67+44 = 111
Total this progeny = 228
Total progeny = 1000
The recombination frequency between b & wx = (number of recombinants/Total progeny) 100
RF = (228/1000)100 = 22.8%
2).
If single crossover occurs between wx & cn..
Normal combination: wx + cn / wx cn+
After crossover: wx+ cn+ / wx cn
wx+ cn+ progeny= 5+67 = 72
wx cn progeny = 6+69 = 75
Total this progeny = 147
The recombination frequency between wx & cn = (number of recombinants/Total progeny) 100
RF = (147/10000)100 = 14.7%
3).
If single crossover occurs between b & cn..
Normal combination: b+ cn / b cn+
After crossover: b+ cn+ / b cn
b+ cn+ progeny= 5+48 = 53
b cn progeny = 6+44 = 50
Total this progeny = 103
Total progeny = 1000
The recombination frequency between b & cn = (number of recombinants/Total progeny) 100
RF = (103/1000)100 = 10.3%
Recombination frequency (%) = Distance between the genes (mu)
Order of genes = b---cn---wx
Gene map = b----------10.3mu--------cn---------14.7mu--------------wx
d).
Expected double crossover frequency = (RF between b & cn) * (RF between cn & wx)
= 10.3% * 14.7% = 0.015
The observed double crossover frequency = 6+5 / 1000 = 0.011
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.011 / 0.015
= 0.73
Interference = 1-COC
= 1- 0.73 = 0.27
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