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Lilit [14]
3 years ago
6

In a particular enzyme, an alanine residue is located in a cleft where the substrate binds. A mutation that changes this residue

to a glycine has little effect on activity; however, another mutation, which changes the alanine to a glutamate residue, leads to a complete loss of activity. Provide a brief explanation for these observations.
Chemistry
1 answer:
TiliK225 [7]3 years ago
5 0

Explanation:

There are non-polar side chains also present in alanine. And, these non-polar side chains are involved in some hydrophobic interactions in active sideand then it changes into polar amino acid glutamate that will destroy the hydrophobic interaction.

Whereas in glycine there is only hydrogen present in the side chain therefore, it will not affect much.

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What is the chemical compound of nitrogen gas?
anzhelika [568]

Answer:

N2

Explanation:

The chemical compound of most gases is two times the element. In this case, it would be N2.

4 0
3 years ago
10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1
leva [86]

Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

6 0
3 years ago
How many grams in 4.5 moles of H2O?
Margarita [4]

the are 18.01528 grams

.....

3 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂     →      CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

                       C             :         CO₂

                        1             :          1

                      1.2           :          1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2  mol × 44 g/mol

Mass = 52.8 g

8 0
3 years ago
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