A dipeptide bond connects two amino acids. Ser stands for Serine while His stands for Histidine. Isoelectric point is when all charges of the two amino acids cancel out. This can be estimated by finding the average of the isoelectric points (pI) of the two amino acids.
Serine: pI = 5.68
Histidine: pI = 7.59
Average pI = (5.68 + 7.59)/2 =<em> 6.635
</em><em />This is only a rough estimate because an arithmetic average will not coincide with the biochemistry of the amino acids. The true value of the isolectric point is determined through experimentation. However, it would be more ore less around 6.635.<em>
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Answer:
Diameter of a circle = 1.6 Cm
Step-by-step explanation:
Given:
Distance between first and third coins = 3.2 cm
Find:
Diameter of the coins = ?
Computation:
Number of circle = 3
In the given figure,distance between the first and third coins is 3.2 cm.
So,
Distance = 1st circle radius + 2nd circle diameter + 3rd circle radius
Distance = 1st circle radius + 2 (radius) + 3rd circle radius
Distance = R + 2R + R
3.2 Cm = 4 R
Radius = 0.8 Cm
Diameter of a circle = 2 × radius
Diameter of a circle = 2 × 0.8
Diameter of a circle = 1.6 Cm
∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find
where 
is the length of the altitude originating from vertex O, and so
where 
is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²
