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Mariana [72]
3 years ago
14

Considering the following hydropathy plot for a protein with 4 transmembrane domain. This protein functions in the plasma membra

ne. The position of 3 positively charged residues near the first transmembrane helix is indicated by an asterisk on the plot. 1. Are the protein termini expected to be located inside (I) or outside (O) of the cell or display one terminus on the inside and the other one on the outside (I/O)
Mathematics
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

Step-by-step explanation:

Answer - O : 2

N-glycans to Asn-X-Ser/Thr sequence occurs on the lumenal side of the endoplasmic reticulum. So, it is present in outside of the cell.

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ZDAC = ZBAD.<br>What is the length of AB?<br>Round to one decimal place.​
goblinko [34]

Answer: Line AB = 5.2

Step-by-step explanation: We start with triangle ACD, with two sides given and angle A which shall be the reference angle can be calculated as,

SinA = opposite/hypotenuse

Where the opposite is 4.3 (line facing the reference angle) and the hypotenuse is 5.6 (line facing the right angle)

SinA = 4.3/5.6

SinA = 0.7679

By use of a calculator or a table of values

A = 50.17 degrees.

Having been told that both angles DAC and BAD are equal, then we move to triangle ADB where the reference angle is 50.17 (BAD) and the opposite is 4 (line facing the reference angle) and the unknown side is the hypotenuse (line AB).

Sin 50.17 = opposite/hypotenuse

Sin 50.17 = 4/AB

0.7679 = 4/AB

By cross multiplication we now have

AB = 4/0.7679

AB = 5.2090

Approximately to one decimal place,

AB = 5.2 units

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4 years ago
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I think it’s the Third one
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