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lubasha [3.4K]
3 years ago
5

What stage of mitosis is depicted by the image above?

Biology
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

telophase

Explanation:

because the chromatids are pulled to the opposite sides

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Explain the role of decomposers in the carbon cycle
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In the carbon cycle, decomposers break down dead material from plants and other organisms and release carbon dioxide into the atmosphere, where it's available to plants for photosynthesis.

Explanation:

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Which statement BEST supports the author's analysis that unshelled fish are more challenging game? A) And, fish or no fish, ther
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B; An unshelled fish is lively and elusive past the power of words to portray, and in this, undoubtedly, lies its desirability.
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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
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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

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
Part A - Modification of chromatin structure Which statements about the modification of chromatin structure in eukaryotes are tr
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Some forms of chromatin modification can be passed on to future generation of cells

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methylation of histone tails can promote condensation of the chromatin

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chromatin modifications that can be passed on includes epigenetic modifications that are heritable changes made to the chromatin structure that does not involve the DNA sequences. Some epigenetic modifications include DNA methylation and Histone modifications. examples of histone modification include acetylation, methylation, phosphorylation, ubiquintylation etc. All these function either in allowing the DNA become more accessible to transcritional factors or vice versa. for exmple, histone tail acetylation encourages unwounding of nucleosomes allowing transcriptional factors to have access to the DNa while histone tails methylation further tightens the nucleosomes promoting condensation of the chromatin.

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