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%