Question

A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not "intelligent," Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are 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.

Answer 1

Answer:

For experiment 1, the recombinant frequecy is 12%

For experiment 2, the recombinant frequency is 5%

Explanation:

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Answer 2

Answer:

For the first experiment, the recombination frequency is shown below. This experiment comprises the cross between the Tall and the Antennae.  

Tall-antennae - 46% - Expected

Dwarf-no antennae - 42% - Expected

Dwarf-antennae - 7% - Recombinant

Tall-no antennae - 5% - Recombinant

Total = 100%

For the recombinants, the recombination frequency is:  

= 7 + 5 = 12%

Thus, the recombination frequency between the T and A genes is 12%.  

For the second experiment, the cross takes place between the heterozygous antennae and the upturned snout. The results are shown below:  

Antennae-upturned snout - 47% - Expected

No antennae-downward snout - 48 % - Expected

Antennae-downward snout - 2% - Recombinant

No antennae-upturned snout - 3% - Recombinant

Total = 100%

The recombinant frequency is = 2 + 3 = 5%

Thus, the recombination frequency between the A and S genes is 5%.