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3 years ago

How does the transfersomes work?

Biology

1 answer:

igomit [66]3 years ago

3 0

Answer:

Different from other types of liposomes, transfersomes are mainly developed for transdermal drug delivery systems. Transfersomes are composed of self-assembled phospholipids (eg. phosphatidylcholine) and edge activators (such as a biocompatible surfactant or an amphiphilic drug) which increases lipid bilayer flexibility and permeability. Therefore, transfersomes can deform and pass through narrow constriction (from 5 to 10 times less than their own diameter) to underlying viable skin without measurable loss due to transdermal osmotic gradients which is developed by skin penetration barrier.

Transfersomes are an ideal choice for drug development of non-invasive therapeutic use. Based on the capability of increasing the transdermal flux, prolonging the release and improving the site specificity of bioactive molecules, transfersome has become a potential candidate for both systemic and topical delivery of drugs.

Transfersomes have high entrapment efficiency (e.g. close to 90% for lipophilic drugs);

Better penetration of intact vesicles due to its high deformability;

A carrier for both low and high molecular weight drugs;

Transfersomes protect the encapsulated drug from metabolic degradation;

Easy to scale up, simple procedures do not involve unnecessary pharmaceutically unaccepted additives;

Transfersomes are biocompatible and biodegradable as they consist of natural phospholipids.

Explanation:

https://www.creative-biostructure.com/Transfersomes-Production-614.htm

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A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density th

Volgvan

Answer:

0.153

Explanation:

We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.

So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.

ρVg = ρ'V'g + ρ"V"g

ρV = ρ'V'g + ρ"V"

Its new body volume = volume of water displaced, V = V' + V"

ρ(V' + V") = ρ'V' + ρ"V"

ρV' + ρV" = ρ'V' + ρ"V"

ρV' - ρ"V' = ρ'V" - ρV"

(ρ - ρ")V' = (ρ' - ρ)V"

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= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)

= (0.9988 g/cm³ ÷ 0.18 g/cm³)

V'/V" = 5.55

Since V = V' + V"

V' = V - V"

(V - V")/V" = 5.55

V/V" - V"/V" = 5.55

V/V" - 1 = 5.55

V/V" = 5.55 + 1

V/V" = 6.55

V"/V = 1/6.55

V"/V = 0.153

So, the fish must inflate its air sacs to 0.153 of its expanded body volume

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