Homework Answers
Answer:
a&b).
c).
Hint: Non-recombinant (parental) combinations are produced more in number than recombinants. The parental (non-recombinant) genotypes is b+ cn+ wx / b cn wx+ (brown body, red eyes, non-waxy wings)
1).
If single crossover occurs between b&cn
Normal combination: b+ cn+ / b cn
After crossover b+ cn / b cn+
b+ cn progeny=112+5 = 117
v+ cv+ progeny = 109+3=112
Total this progeny = 229
Total progeny = 1163
The recombination frequency between b&cn = (number of recombinants/Total progeny) 100
RF = (229/1163)100 = 19.69%
2).
If single crossover occurs between cn&wx
Normal combination: cn+ wx / cn wx+
After crossover: cn+ wx+ / cn wx
cn+ wx+ progeny= 109+52= 161
cn wx progeny = 112+48= 160
Total this progeny = 321
The recombination frequency between cn&wx = (number of recombinants/Total progeny) 100
RF = (321/1163)100 = 27.60%
3).
If single crossover occurs between cn&wx
Normal combination: b+ wx / b wx+
After crossover: b+ wx+ / b wx
b+ wx+ progeny= 5+52= 57
b wx progeny = 3+48= 51
Total this progeny = 108
The recombination frequency between b&wx = (number of recombinants/Total progeny) 100
RF = (108/1163)100 = 9.29%
Recombination frequency (%) = Distance between the genes (mu)
cn--------19.69mu------b----9.29mu----wx
The order of genes = cn –b--wx
d). cn--------19.69mu------b----9.29mu----wx
e). Trihybrid genotype = cn+ -----b+-----wx / cn ----b-----wx+
f). b & cn are in cis linkage
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