A sexually reproducing species of algae has been engineered to produce biofuels with varying yields. A colony with the highest yield is stimulated to reproduce with a colony with the lowest yield and the resulting F¹ progeny are allowed to grow. Yields among the F₁ generation have a variance of 7.8 g². The F₁are allowed to mate and the resulting progeny are sampled and found to have a variance of 9.1 g². Assuming that the F₁ progeny are genetically uniform, determine the broad sense heritability (H²) and interpret if genetic heritability is low or high. A) H²= 0.86, genetic variability is high B) H²= 0.17, genetic variability is low C) H² = 0.14, genetic variability is high D) H² = 0.14, genetic variability is low E) H² = 0.86, genetic variability is low

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Quizplus · Accepted Answer

Broad sense heritability (H²) is calculated as the ratio of genetic variance to phenotypic variance. In this case, the genetic variance is the difference between the variance of the F₁ generation and the variance of the F₂ generation, which is 9.1 g² - 7.8 g² = 1.3 g². The phenotypic variance is the variance of the F₁ generation, which is 7.8 g². Therefore, H² = 1.3 g² / 7.8 g² = 0.17. This indicates that genetic variability is low, as the majority of the phenotypic variance is due to environmental factors rather than genetic factors.

A Sexually Reproducing Species of Algae Has Been Engineered to Produce