You are a tropical fish enthusiast who loves the dwarf cichlid Apistrogramma cac
ID: 215901 • Letter: Y
Question
You are a tropical fish enthusiast who loves the dwarf cichlid Apistrogramma cacatuoides. On a recent trip to the Amazon, you collected a spectacular specimen that bred true for long fins (Lf), neon colors (Nc) and extreme aggressivness (Ea). Your existing stock breeds true for short fins (lf), drab colors (nc) and docile behavior (ea). When you mate them, the progeny produced are all long finned, neon colored and aggressive. You backcross the F1 again to your existing stock and obtain the following phenotypic classes. (It might help to draw out the crosses!!)
Phenotype # progeny
Long fins, neon colors, aggressive 290
Short fins, drab colors, docile 280
Long fins, neon colors, docile 8
Short fins, drab colors, aggressive 12
Long fins, drab colors, docile 111
Short fins, neon colors, aggressive 119
Long fins, drab colors, aggressive 86
Short fins, neon colors, docile 94
A. (2 points) Determine the gene order on the chromosome. Explain.
B. (2 points) Determine the map distance between each gene.
Explanation / Answer
Answer:
A).
Gene order on the chromosome: Lf—Ea----Nc / lf ea Nc
B).
Lf----------25cM--------Ea------20cM-------Nc
Explanation:
Phenotype # progeny
Long fins, neon colors, aggressive(Lf Nc Ea) 290
Short fins, drab colors, docile (lf nc ea) 280
Long fins, neon colors, docile (Lf Nc ea) 8
Short fins, drab colors, aggressive (lf nc Ea) 12
Long fins, drab colors, docile(Lf nc ea) 111
Short fins, neon colors, aggressive(lf Nc Ea) 119
Long fins, drab colors, aggressive(Lf nc Ea) 86
Short fins, neon colors, docile (lf Nc ea) 94
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is Lf Nc Ea / lf nc ea
1).
If single crossover occurs between Lf & Nc..
Normal combination: Lf Nc / Lf nc
After crossover: Lf nc / lf Nc
Lf nc progeny= 111+86=197
lf Nc progeny = 119+94=213
Total this progeny = 410
Total progeny = 1000
The recombination frequency between Lf&Nc = (number of recombinants/Total progeny) 100
RF = (410/1000)100 = 41%
2).
If single crossover occurs between Nc/Ea
Normal combination: Nc Ea / nc ez
After crossover: Nc ea / nc Ea
Nc ea progeny= 8+94=102
nc Ea progeny = 12+86=98
Total this progeny = 200
The recombination frequency between Nc & Ea = (number of recombinants/Total progeny) 100
RF = (200/1000)100 = 20%
3).
If single crossover occurs between Lf & Ea..
Normal combination: Lf Ea / lf ea
After crossover: Lf ea / lf Ea
Lf ea progeny= 8+111= 119
lf Ea progeny = 12+119= 131
Total this progeny = 250
The recombination frequency between Lf&Ea = (number of recombinants/Total progeny) 100
RF = (250/1000)100 =25%
Recombination frequency (%) = Distance between the genes (cM)
Lf----------25cM--------Ea------20cM-------Nc
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.