You perform a trihybrid test cross for an individual of genotype A/a . B/b. D/d
ID: 208835 • Letter: Y
Question
You perform a trihybrid test cross for an individual of genotype A/a . B/b. D/d You obtain the following phenotypes of offspring (in numerical order): ABd: 422
abD: 406 aBD: 53 ABD: 40 Abd: 39 abd: 32 aBd: 5 AbD: 3
A. Give the correct genotype of the trihybrid parent, using correct notation. Explain how you know.(2 points) B. Determine the gene order and show how you know. (2 points)
C. If necessary, rewrite the phenotypes in the correct order and ensure that reciprocal classes are correctly identified. Determine the map distances of the two smaller distances. Draw a map showing their arrangement and the correct distances. To save time you don’t need to calculate the larger distance but tell me what you would have to be careful about if you were going to make that calculation
Explanation / Answer
Answer:
A). Bad / bAD
B). Order of genes = BAD
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotype is ABd/abD
1).
If single crossover occurs between A&B..
Normal combination: AB/ab
After crossover: Ab/aB
Ab progeny= 39+3=42
aB progeny = 53+5=58
Total of this progeny = 100
Total progeny = 1000
The recombination frequency between A&B = (number of recombinants/Total progeny) 100
RF = (100/1000)100 = 10%
2).
If single crossover occurs between B&d..
Normal combination: Bd/bD
After crossover: BD/bd
BD progeny= 53+40=93
bd ct progeny = 39+32=71
Total this progeny = 164
The recombination frequency between B&d = (number of recombinants/Total progeny) 100
RF =(164/1000)100 = 16.4%
3).
If single crossover occurs between A&d..
Normal combination: Ad/aD
After crossover: AD/ad
AD progeny= 40+3=43
ad progeny = 32+5=37
Total this progeny = 80
The recombination frequency between A&d = (number of recombinants/Total progeny) 100
RF = (80/1000)100 = 8%
Recombination frequency (%) = Distance between the genes (cM)
b---------10cM---------a-------8cM--------d
C).
Distance between the gene b & a = 10 map units or centi Morgans
Distance between the gene a & d = 8 map units or centi Morgans
D).
b---------10cM---------a-------8cM--------d
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