a geneticist is using a three-point testcross to map three linked Dros Sapling L
ID: 188296 • Letter: A
Question
a geneticist is using a three-point testcross to map three linked Dros Sapling Learning A geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called a b, and c, where a is associated with anomalous associated with curved bristles. She first crosses homozygous anomalous, buckled flies to curved flies. Next, she testcrosses the Fi progeny to anomalous, buckled progeny distributed as the following results. From this data, calculate the map distance between a and d gait; b is associated with buckled wings; and c is e obtains 1000 curved anomalous, buckled wild type anomalous, buckled, curved buckled, curved anomalous anomalous, curved buckled 281 27 m.u 133 127 Incorrect. You entered an incorrect value for the distance between a and c. You may have calculated the distance between b and c instead. Next Exit O Previous ® Give Up & View Solution # Try AgainExplanation / Answer
Answer = 18 map units
Distance between the alleles determines the probability of crossover.
Greater the distance between alleles, the more likely crossover will occur whereas lesser the distance, crossover frequency will decrease.
The highest number of offsprings depict NoCrossOvers (NCO)
In our case it is,
++c and ab+ , where + signifies the wild type.
The lowest number of offsprings signify DoubleCrossOvers (DCO)
In our case it is ,
+++ and abc
In order for this double crossover, the order of alleles must be bca.
Let us calculate the distance now with the following formula :
SCO+DCO/ Total Offspring * 100= number of map units between genes
Distance between a and c
DCO between a and c = 5+5 = 10
SCO between a and c = 86+84 = 170
Total crossovers between a and c = 170 +10 = 180
Number of map units between a and c = 180/1000*100 = 18 map units.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.