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mixas84 [53]
3 years ago
6

ADP binds to platelets in order to intiate the activation process. Two binding sites were identified on platelets, one with a Kd

of 350 nM and one with a Kd of 7.9 µM. Which of these is the low-affinity binding site? A) The binding site with a Kd of 350 nM is the low affinity site. B) The binding site with a Kd of 7.9 µM is the low affinity site. C) No way to tell
Physics
1 answer:
12345 [234]3 years ago
5 0

Answer:B) The binding site with a kid of7.9uM is the low affinity site.

Explanation:

The rate at which macro molecules like protein bind together is called affinity. The higher the affinity site the more readily these macro molecules fuses together. 7.9uM is a low affinity value than 359uM.

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Why construction is present near the bulb ​
vova2212 [387]

Answer:

The constriction causes the mercury column to break under tension, leaving a vacuum between the bottom of the column and that in the bulb, and the top of the column stays still at the position reached in the body - a "peak hold" system.

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Why does an energy transfer not always result in phase change?
mash [69]

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These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. "If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance

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3 years ago
Read 2 more answers
.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
3 years ago
Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid
ollegr [7]
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s

Average velocity of Y = 546.41 / 30 = 18.2 m/s
8 0
3 years ago
Read 2 more answers
Help me please, need more assistance
Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0
3 years ago
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