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Law Incorporation
1 month ago
11

A plot of velocity versus substrate concentration for a simple enzyme-catalyzed reaction produces a _____. This indicates that a

t some point, the enzyme is _____. Group of answer choices A) straight line; inhibited by product
B) hyperbolic curve; saturated with substrate
C) sigmoidal curve; inhibited by substrate
D) hyperbolic curve; activated by substrate
E) sigmoidal curve; saturated with substrate
Chemistry
1 answer:
castortr0y [3K]1 month ago
5 0

Answer:

B) Hyperbolic curve; substrate saturation

Explanation:

Enzymatic kinetics examines the rates of reactions catalyzed by enzymes. These studies offer insights into the mechanism of the catalytic reaction and enzyme specificity. Determining the reaction rate facilitated by an enzyme is generally straightforward, as purification or isolation of the enzyme is frequently unnecessary. Measurements are taken under optimal conditions for pH, temperature, and the presence of cofactors, utilizing saturating substrate concentrations. Under these circumstances, the observed reaction rate is the maximum velocity (Vmax). The rate can be measured by monitoring either product formation or substrate consumption.

Following the rate of product formation (or substrate consumption) over time yields the so-called reaction progress curve, or merely, reaction kinetics. This reacts as a hyperbolic curve

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How many known elements have atoms, in their ground-state, with valence electrons whose quantum numbers are n = 5 and l = 1?a. 3
Tems11 [2777]

Answer:

c. 6.

Explanation:

Based on the question's description, the elements must be from the p-block of the periodic table and in the fifth period. They should have valence electrons located in the 5p orbital.

The following elements from the p-block of period 5 meet these criteria: Sr, In, Sn, Sb, Te, and I.

Therefore, there are a total of six such elements.

7 0
2 months ago
Arrange the steps of glycogen degradation in their proper order. Hormonal signals trigger glycogen breakdown. Glucose 6‑phosphat
VMariaS [2998]

Answer: Please see answer below

Explanation:

The sequence for glycogen degradation is as follows:[

---> Hormonal signals initiate the breakdown of glycogen.

1. Glycogen undergoes debranching through the hydrolysis of α‑1,6 linkages.

2. Blocks of three glucosyl units are relocated by remodeling α‑1,4 linkages.

3. Glucose 1‑phosphate is derived from the non-reducing ends of glycogen and is transformed into glucose 6‑phosphate.

---> Glucose 6‑phosphate enters further metabolic pathways

Glycogen degradation consists of three stages:

(1) the release of glucose 1-phosphate from glycogen,

(2) transforming the glycogen structure for continued breakdown, and

(3) converting glucose 1-phosphate into glucose 6-phosphate for subsequent metabolism.

(https://www.ncbi.nlm.nih.gov/books/NBK21190)[[TAG_34]][[TAG_35]][[TAG_36]]

4 0
1 month ago
20 point, pls help. A 5.50 mole sample of a gas has a volume of 2.50 L. What would the volume be if the amount increased to 11.0
Anarel [2989]

Sagot:

Dapat mong suriin ito, isang mahusay na pagkakataon ito para sa iyo upang ipaalam sa mga kasapi ng iyong pamilya at magiging masaya akong tumulong sa iyo sa prosesong ito kung ikaw ay mayroong mga katanungan o alalahanin tungkol sa dad lowkey o kung kailangan mo ng tulong sa proyekto o kahit ano pa, sabihin mo lang sa akin kung ano ang kailangan mong ipagawa

5 0
1 month ago
For H3PO4, Ka1 = 7.3 x 10^-3, Ka2 = 6.2 x 10^-6, and Ka3 = 4.8 x 10^-13. A 0.10 M aqueous solution of Na3PO4 therefore would be
Anarel [2989]
The aqueous solution of Na3PO4 is described as "strongly basic."
3 0
2 months ago
A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut
Anarel [2989]

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
  • Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
2 months ago
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