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REY [17]
3 years ago
7

Team of scientists is studying several plant species that they think could become new farm crops. The scientists want to apply s

elective breeding to develop one of the species into useful crop plants. Which property of a plant species would be MOST USEFUL for meeting the scientists' goal? A. very low genetic variation B. very high genetic variation C. producing only a few fruits and seeds each growing season
Biology
1 answer:
algol [13]3 years ago
3 0

Answer:

A

Explanation:

because the scientists want to breed a certain type of the crop, therefore they are trying to single out one type instead of creating a bunch of new ones

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Answer:

The phenotypic variance= 2113

The eenvironmenal variance= 644

The genetic variance= 1469

The broad-sense heritability = 0.695

Explanation:

To determine the phenotypic variance, we need to know the calculate the environmental variance and the genetic variance first;

because Phenotypic variance = V(p) such that;

V(p) = V(e) + V(g)                  

where V(e)  =  environmental variance and V(g)   =  genetic variance

Therefore, Environmental variance can be defined as a portion of phenotypic variance due to differences in the environments to which the individuals in a population have been exposed. The total amount of variance observed is dependent on the genetic component, determined by the variation that is inherited.

This implies that V(e)= initial variance which was there in the parents ( i.e  Variance of inbred line 1 = 311  and Variance of inbred line 2 = 333 )

V(e) = (311 + 333) = 644

Genetic Variance simply refer to the variance that occurs or preferrably , that is observed after inbreeding (i.e Variance of F1 = 316  and Variance of F2 = 1153)

V(g)  = (316 + 1153) = 1469

Now, back to phenotypic variance = V(e) + V(g)

substituting the parameters we have;  1469 + 644 = 2113

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= V(g)/ V(p)

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