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TiliK225 [7]
2 years ago
15

The terms motif (fold) and domain describe levels of protein organization more complicated than primary or secondary structure.

Differentiate between motifs and domains by moving each phrase to the appropriate bin. Note: If you answer any part of this question incorrectly, a single red X will appear indicating that one or more of the phrases are sorted incorrectly.
Motifs Domain Both
Choices are:
Stabilized by hydrophobic interactions
Stable globular units
depends on primary structure
clusters of seconday structure
Beta Alpha Beta unit
may be distinct functional units in a protein
unit of tertiary structure
repetetive supersecondary structure
may retain a 3D structure when seperated from rest of the protein
Chemistry
1 answer:
posledela2 years ago
8 0

Answer:

Each specific property of motif and domain is explained.

Explanation:

Domain;

  • May retain a 3D structure when separated from rest of the protein.          
  • Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
  • Stable globular units like pyruvate kinase
  • May be distinct functional units in a protein

Motif;

  • Repetetive supersecondary structure because they contain cluster of secondary structure.
  • Beta Alpha Beta unit is an example of motif
  • Clusters of secondary structure

Both Motif and Domain;

  • Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
  • Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.

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8 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

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6 0
2 years ago
Express the quantity 556.2 x 10^-12 in each units<br> A.) ms<br> B.) ns<br> C.) ps<br> D.) fs
zavuch27 [327]
The answer:
we should know the meaning of each abbreviation:
ms means  millisecond, its value is 10^-3 s 
ns means  means  nanosecond,   its value is 10^-9 s
ps means  picosecond, its value is 10^-12 s
fs means  femtosecond, its value is 1x 10^15 s

<span>Expressions of the quantity 556.2 x 10^-12 are</span>
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
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6 0
2 years ago
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s2008m [1.1K]
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5 0
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vivado [14]

Answer:

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Explanation:

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An alcohol as n-octyl alcohol has different polarity than an alkene as 1-octene.

Thus, using  thin layer chromatography is most easy to separate:

<em>B. n-octyl alcohol and 1-octene </em>

<em></em>

I hope it helps!

<em></em>

8 0
2 years ago
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