Activity 3 Homework-Spring 2018 (1) ut References Mailings Review View Help Tell
ID: 194507 • Letter: A
Question
Activity 3 Homework-Spring 2018 (1) ut References Mailings Review View Help Tell me what you want to do ay. . ..1 tE. O.L. 1 No Spac Heading 7 Subtle EmL Emphasis Intense E." Strong acte Based on the following data from a three point testcross: 1. Determine the gene order and phase for the genes A, B and C. 2. Determine the distance between genes A, B and C 3. Draw a linkage map with the correct gene order, allele combinations on each chromatid and distance between the three genes. For this exercise, phenotypes with capital letters are dominant and phenotypes in lower case letters are recessive. For example, "phenotype A" mean the dominant phenotype for the gene and phenotype a" means the recessive phenotype for that gene. Show all your calculations. Remember that the three genes can be in any order. Remember that the dominant alleles do not necessarily need to all be on the same chromosome in the triple heterozygote parent Phenotypes Atb 305 atB ATB 54 56 61 843 TotalExplanation / Answer
Answer:
a). Gene order= TAB
b). T----------12.1cM--------A-----------14.71cM--------------B
c). TaB / tAb
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is aTB/ Atb
1).
If single crossover occurs between a & T..
Normal combination: aT / At
After crossover: at/AT
at progeny= 54+3=57
AT progeny = 51+4=55
Total this progeny = 102
The recombination frequency between a&T = (number of recombinants/Total progeny) 100
RF = (102/843)100 = 12.1%
2).
If single crossover occurs between T & B..
Normal combination: TB / tb
After crossover: Tb/tB
Tb progeny= 51+61=112
tB progeny = 54+56=110
Total this progeny = 222
The recombination frequency between T&B = (number of recombinants/Total progeny) 100
RF = (22/843)100 = 26.33%
3).
If single crossover occurs between a & B..
Normal combination: aB / Ab
After crossover: ab/AB
ab progeny= 3+61=64
AB progeny = 4+56=60
Total this progeny = 124
The recombination frequency between a&B = (number of recombinants/Total progeny) 100
RF = (124/843)100 = 14.71%
Recombination frequency (%) = Distance between the genes (cM)
T----------12.1cM--------A-----------14.71cM--------------B
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