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Vinil7 [7]
4 years ago
5

The ______ system senses changes in the internal and external environment so that the body can adapt to the changes.

Biology
2 answers:
Ivenika [448]4 years ago
7 0

The Answer is A. Nervous

meriva4 years ago
4 0
The answer is either endorcrine or reporudctive, 
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atroni [7]
False, because a niche is how one animal uses its habitat- many animals use many different niches
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4 years ago
When corn plants are too tall, they lodge (topple over) prior to harvest. This is bad since the grain is often lost when it cann
Shalnov [3]

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

Broad-sense heritability can be calculated as the ratio of total genetic variance to total phenotypic variance.

= V(g)/ V(p)

= 1469/ 2113

= 0.695

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3 years ago
Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.
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You know the genotype of the parents, and for the cross, you need to separate into two alleles, both genes.

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Now you can cross them, and all probabilities will be:

AABbCc, AaBbCc, AABbcc and AaBbcc

So, you could say 25% of the offspring will be AaBbCc, 25% AaBbcc, 25% AABbCc, and 25% AABbcc

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I need help on question 16 for quiz 3.2
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Answer:

what

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The layers and structures the needle passed through are- epidermis, dermis, subcutaneous layer or hypodermis, connective tissue layer and muscle.

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