B. Females heterozygous for the recessive second chromosome mutations pn. px, an
ID: 197149 • Letter: B
Question
B. Females heterozygous for the recessive second chromosome mutations pn. px, and sp are mated to a male homozygous for all three mutations. The offspring are as follows x sp cn x sp + x + cn 1.461 497 +sp+ t + cn ,482 1539 10,000 What is the genotype of the females that gave rise to these progeny? Using the two point cross method, what are the distances among these three genes? Using the three point cross method, determine which gene is in the middle (which has to be flipped); Which method provides more confidence as to the correct order of the genes? Calculate the coefficient of coincidence and comment on its meaning.Explanation / Answer
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the female (non-recombinant) genotype is px sp + / + + cn.
1).
If single crossover occurs between px & sp..
Normal combination: px sp / ++
After crossover: px +/+ sp
px + progeny= 1+11`=12
+ sp progeny = 9+0=9
Total this progeny = 21
The recombination frequency between px&sp = (number of recombinants/Total progeny) 100
RF = (21/10000)100 = 0.21%
2).
If single crossover occurs between sp & +..
Normal combination: sp + / + cn
After crossover: sp cn/++
sp cn progeny= 9+1461= 1470
+ + progeny = 11+1539= 1550
Total this progeny = 3020
The recombination frequency between sp&+ = (number of recombinants/Total progeny) 100
RF = (3020/10000)100 = 30.2%
3).
If single crossover occurs between px & +..
Normal combination: px + / + cn
After crossover: px cn/++
px cn progeny= 1461+1= 1462
+ + progeny = 0+1539 = 1539
Total this progeny = 3001
The recombination frequency between px&+ = (number of recombinants/Total progeny) 100
RF = (3001/1000)100 = 30.01%
Recombination frequency (%) = Distance between the genes (cM)
sp----------0.21cM--------px-----------30.01cM--------------cn
Expected double crossover frequency = (RF between sp & px) * (RF between px & cn)
= 0.21% * 30.1% = 0.0006
The observed double crossover frequency = 1+0 / 10000 = 0.0001
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.0001 / 0.0006
= 0.17
Observed doublecross over is 17% less than the expected double cross over
Interference = 1-COC
= 1-0.17 = 0.83
The correct oreder of female genotype = sp px + / + + cn
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