Campbell Problem 6

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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