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Vinvika [58]
3 years ago
8

In animals, cell recognition:

Biology
1 answer:
adoni [48]3 years ago
7 0

Answer:

The correct answer is D.  involves proteins in plasma membranes.

Explanation:

Animal cells have several types of proteins embedded in their cell membrane. These proteins play an important role in the transport of molecules, cell recognition, and cell communication.  

Recognition proteins are a type of glycoproteins present in the plasma membrane that allow one cell of the body to recognize the other body cells by making contact with recognition proteins of other cells. Receptors proteins allow cell-cell communication by receiving extracellular proteins.

These proteins are important for proper growth and development of the cell. Therefore the correct answer is D.  involves proteins in plasma membranes.

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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
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

4 0
2 years ago
Plss answer all! Help!
vladimir1956 [14]

Answer:

there are two types of cell one eukaryotic cell that have true nucleus (genetic material enclosed in membrane) and membrane bounded organelles and other is prokaryotic cell that donot have have true nucleus (genetic material dispersed in cytoplasm) and membrane bounded organelles.

similarities between PC and EC is that both contains cell membrane, ribosomes (not a membrane bounded organelle) .

examples of PC are bacteria , paramecium,fungi etc

examples of EC are plant and animal cells

there are two names for cell membrane i.e plasmalemma, cell surface membrane,

selectively permeable or semipermeable means cell only allows those material to pass through its cell membrane which it needs.this cell'ability is important because to protection against the poisonous substance and to maintain cell needs .

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2 years ago
The cell membrane around a cell forms a barrier that protects and regulates the cell. Which best describes those that can pass t
pickupchik [31]
Ok I think its small polar molecules but I could be wrong. 
7 0
3 years ago
Which describes a constant in an experiment?
umka21 [38]
A constant in an experiment is <u>B. </u><span><u>the quantity that must remain the same</u>. If it is constant, it doesn't change, so it cannot be variable. D makes no sense.
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8 0
3 years ago
As part of a blood drive on campus for the american red cross, you and your friends have just donated 500 ml of blood. you are n
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Due to a reduction in blood volume due to blood donation, the stroke volume will decrease. However, to compensate for this, your heart rate will increase so as to maintain your cardiac output. Cardiac output is stroke volume multiplied with heart rate.
Maintaining normal cardiac output is critical so the whole body receives enough nutrients and oxygen.
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