1. Given the following testcross data for maize (corn) in which the genes for fi
ID: 3166015 • Letter: 1
Question
1. Given the following testcross data for maize (corn) in which the genes for fine stripe (, bronze aleurone (bz) and knotted leaf (Kn) are involved: Number 451 134 97 436 18 119 24 86 1,365 Phenotype Kn + Kn f + Kn bz ++ bz Kn f bz Total: Maize geneticists tend to be like Drosophila geneticists utilizing a + to indicate the wild type allele (which also is dominant) and a lower case letter for the mutant allele (which is recessive). a) Determine the sequence (order) of the three genes. (2 points) b) Calculate the map distances between genes and construct a map of them. (5 points) c) Calculate any interference in crossing over in the regions including these genes. (3 points)Explanation / Answer
Answer:
a). Kn - bz - f
b).
Kn-bz= 16.48 cM
bz-f = 19.71 cM
Kn-f = 31.94 cM
Kn----------16.48cM--------bz-----------19.71cM--------------f
c). Interference = 0.05
Explanation:
Hint:
Parental genotypes are more than any type of recombinant progeny. Hence parental genotype is Kn + + / + f bz
1).
If single crossover occurs between Kn & +..
Normal combination: Kn + / + f
After crossover: Kn f / + +
Kn f progeny= 134+86= 220
++ progeny = 97+119 = 216
Total of this progeny = 436
The recombination frequency between Kn&+ = (number of recombinants/Total progeny) 100
RF = (436/1365)100 = 31.94%
2).
If single crossover occurs between f & bz..
Normal combination: ++/f bz
After crossover: + bz / f +
+ bz progeny= 18+119= 137
f + progeny = 24+134= 158
Total this progeny = 269
The recombination frequency between f&bz = (number of recombinants/Total progeny) 100
RF = (269/1365)100 = 19.71%
3).
If single crossover occurs between Kn & bz..
Normal combination: Kn + / + bz
After crossover: Kn bz / + +
Kn bz progeny= 18+86= 104
++ progeny = 97+24= 121
Total this progeny =225
The recombination frequency between Kn&bz = (number of recombinants/Total progeny) 100
RF = (225/1365)100 = 16.48%
Recombination frequency (%) = Distance between the genes (cM)
Kn----------16.48cM--------bz-----------19.71cM--------------f
Expected double crossover frequency = (RF between h & w) * (RF between w & b)
= 0.1648 * 0.1971 = 0.0325
The observed double crossover frequency = 24+18 / 1000 = 0.0308
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.0308 / 0.0325
= 0.95
Interference = 1-COC
= 1-0.95 = 0.05
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