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Nezavi [6.7K]
1 year ago
6

you fuse a sequence that contains a short stretch of hydrophobic amino acids to the c-terminus of gfp.you monitor this protein b

y fluorescence microscopy to characterize the final localization.coursehero you predict the unmodified gfp protein will be found
Biology
1 answer:
Vesnalui [34]1 year ago
7 0

The sequence containing short stretch of hydrophobic amino acids fused  to the c-terminus of GFP, will be probably found outside the cell because it has no signal sequence.

GFP, called the Green Fluorescent Protein emits a green fluorescence when exposed to light. The protein was first isolated from a jellyfish <em>Aequorea victoria</em>. In laboratories, the protein is majorly used as a marker to track the movement of functioning of any substance.

Signal sequence are the short stretch oh polypeptides found attached to any protein to determine its target site. These sequences determine the pathway and the function that the protein will follow.

To know more about GFP, here

brainly.com/question/28333100

#SPJ4

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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
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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"

V'/V" = (ρ - ρ")/(ρ' - ρ)

= (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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