Molecular Genetics Problem 6 6. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not “intelligent” Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For a tall heterozygote with antennae, the offspring were tall-antennae, 46; dwarf-antennae 7; dwarf-no antennae 42; tall-no antennae 5. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47; antennae-downturned snout, 2; no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.
Experiment 1 (Frequency/Distance between T and A).
Determine the recombination frequency for the genes controlling Tallness and Antennae:
| 46 tall-antennae |
= 46% |
expected |
| 42 dwarf-no antennae |
= 42% |
expected |
| 7 dwarf-antennae |
= 7% |
recombinant |
| 5 tall-no antennae |
= 5% |
recombinant |
Total = 100
Therefore this recombination frequency between genes T and A is 12%
Experiment 2. (Frequency/Distance between A and S)
Determine the recombination frequency for the genes controlling Antennae and Snout:
| 47 antennae-upturned snout |
= 47% |
expected |
| 48 no antennae-downturned snout |
= 48% |
expected |
| 2 antennae-downturned snout |
= 2% |
recombinant |
| 3 no antennae-upturned snout |
= 3% |
recombinant |
Total = 100
Therefore this recombination frequency between genes A and S is 5%
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