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ki77a [65]
2 years ago
15

A mutation in the gene coding for a single-polypeptide enzyme results in the substitution of the amino acid serine, which has a

polar R group, by the amino acid phenylalanine, which has a nonpolar R group. When researchers test the catalysis of the normal enzyme and the mutated enzyme, they find that the mutated enzyme has much lower activity than the normal enzyme does.
Which of the following most likely explains how the amino acid substitution has resulted in decreased catalytic activity by the mutated enzyme?
Chemistry
1 answer:
Alik [6]2 years ago
4 0

Answer:

The correct answer is: The substitution altered the secondary and tertiary structure of the enzyme so that the mutated enzyme folds into a different shape than the normal enzyme does.

Explanation:

In the given condition, the substitution mutation of gene causes a replacement of serine by phenylalanine amino acids which causes a reduction in the activity of the enzyme.  Since serine is polar and has -OH or hydroxy group involves the information of binding of biological catalyst to the substrate.

The primary structure of a protein is significant which finalizes the number of amino acids their sequence. The mutation of protein also affects both secondary and tertiary structures as it disturbs the structure of the protein and affects the catalytic activity as well as the binding affinity of the substrate.

the substitution of serin by phenylalanine does not affect or influence the mass of enzyme.

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Give a possible explanation for the relative amounts of the isometric methyl nitrobenzoates formed in the nitration reaction. Co
vovikov84 [41]

Answer:

The electrophilic aromatic substitution reaction nitration is used to nitrate methyl benzoate and acetanilide with a nitronium ion. Crystallization was used to purify the product. The melting point was used to determine its purity and the regiochemistry of the products.

Explanation:

Methyl m-Nitrobenzoate is formed in this

reaction rather that ortho/para isomers

because of the ester group of your starting

product of methylbenzoate. The functional

group of ester is a electron withdrawing group

causing nitrobenzene (N02) to become in the

meta position. Thus N02 is a deactivating

group causing itself to be a meta director.

Basically you must look at the substituents

that are attached to your starting benzene ring

in order to figure out whether your reaction

with be ortho/para directors or meta

directors. If the substituents are electron

withdrawing groups then you will be left with

meta as your product but if your substituents

are electron donating groups then your

product will be ortho/para.

4 0
1 year ago
Give all possible, accurate, correct IUPAC names for the following compounds:
Viefleur [7K]

Answer:

A. 4-ethyl-hex-3,5-dien-2-ol.

B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol.

Explanation:

Hello there!

In this case, according to the given problems, it is possible to apply the IUPAC rules to obtain the following names:

A. 4-ethyl-hex-3,5-dien-2-ol because we have an ethyl radical at the fourth carbon and the beginning of the parent chain is on the Me (CH3) because it is closest to first OH.

B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol: because we start at the alcohol and have a chlorine atom on the second carbon, a methyl radical on the third carbon, a <em>tert</em>-butyl on the fifth carbon and the parent chain is benzene which is phenol as an alcohol.

Regards!

6 0
2 years ago
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
stira [4]

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

8 0
2 years ago
Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.
vazorg [7]
The answer to this question is D! The ball and stick model! Hope this helps :)
8 0
1 year ago
Read 2 more answers
If the composition of the reaction mixture at 400 k is [brcl] = 0.00415 m, [br2] = 0.00366 m, and [cl2] = 0.000672 m, what is th
Scorpion4ik [409]
Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl

According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]

and we have [Br2] = 0.00366 m  [Cl2]= 0.000672 m  [BrCl] = 0.00415 m

by substitution:
                          = [0.00415]^2 / ( [0.00366] * [0.000672])
             ∴   Q   = 7
7 0
2 years ago
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