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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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The human body is 0.0040% iron. How many milligrams of iron does a 165 pound person contain?​
Fofino [41]

Answer:

in a body of 165 pounds, the iron content is 2993.707 mg

Explanation:

  • % iron = 0.0040% = (mass iron / mass human) * 100

⇒ mass iron / mass human = 4 E-5

∴ mass human = 165 pounds

⇒ mass iron = 165 pounds * 4 E-5

⇒ mass iron = 6.6 E-3 pounds * ( 453.592 g/pound ) * ( 1000 mg/g )

⇒ mass iron = 2993.707 mg iron

8 0
2 years ago
Calculate the pH of a polyprotic acid given and sketch the titration curves for the following reaction:
victus00 [196]

Answer: (1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22

8 0
2 years ago
Urea contains two nitrogen atoms. what is the source of these two nitrogen atoms?
Oksi-84 [34.3K]
The source of these two nitrogen atoms are ammonia (NH₃) from <span>nitrogen compounds (mostly metabolism of amino acids) through which excess nitrogen is eliminated from organisms. This process is called urea cycle, which extracted </span>nitrogenous wastes.
The liver<span> forms it by combining two </span>ammonia<span> molecules</span><span> with a </span>carbon dioxide<span> </span><span>molecule.</span><span />
3 0
2 years ago
Lithium chloride forms three hydrates. They are LiCl.H2O, LiCl.2H2O and LiCl.3H2O.
Stels [109]

Answer:

The answer is LiCl.2H2O

Explanation:

Li=7

Cl=35.5

O=16

LiCl.H2O

7+35.5+16+2

60.5

%comp=60.5/78.5×100

22.9

LiCl.2H20

7+35.5+2(2+16)

42.5+36

78.5

%comp=36/78.5×100

45.9

LiCl.3H20

7+35.5+3(2+16)

42.5+54

96.5

54/96.5×100

56.0

7 0
2 years ago
Oxides of virtually every element are known. bromine, for example, forms several oxides when treated with ozone. suppose you all
FromTheMoon [43]
Answer is: empirical formula of the product is Br₂O₅.
Chemical reaction: x/2Br₂ + y/3O₃ → BrₓOy.
m(Br₂) = 1,250 g.
m(BrₓOy) = 1,876 g.
n(Br₂) = m(Br₂) ÷ M(Br₂).
n(Br₂) = 1,25 g ÷ 159,81 g/mol.
n(Br₂) = 0,0078 mol.
n(Br) = 2 · 0,0078 mol = 0,0156 mol.
m(O₃) = 1,876 g - 1,25 g = 0,626 g.
n(O₃) = 0,626 g ÷ 48 g/mol = 0,013 mol.
n(O) = 0,039 · 3 = 0,039 mol
n(Br) : n(O) = 0,0156 mol : 0,039 mol.
n(Br) : n(O) = 1 : 2,5.
5 0
2 years ago
Read 2 more answers
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