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devlian [24]
3 years ago
11

The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalyt

ic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Chemistry
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Please help me with this: Create 20 bullet points specifically about energy exchanges in Earth's systems. Also, it doesn't have
raketka [301]

The below is about the energy exchanges in earth systems.                                                                                                          

<u>Explanation</u>:

  • Energy exchanges in earth systems are of many types.  The earth systems are atmosphere, geosphere, stratosphere, hydrosphere, and biosphere. All these earth systems exchange energy with each other.
  • The earth gains energy reflected from the sky. It converts that energy back to space. That energy is equally given to all the planets in the sky.
  • Each planet will absorb that energy and radiate heat. This heat is absorbed by all the places on the earth. So this is the energy exchange in the earth systems.                                                                                
7 0
2 years ago
Read 2 more answers
The boiling point of a substance in City A is found to be 145°C the boiling point of th same substance in Xity B is 141°C which
DochEvi [55]
City B. Higher altitudes have lower boiling points due to lower atmospheric pressure at higher altitudes
4 0
3 years ago
The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release
KATRIN_1 [288]

The mass of melted gold to release the energy would be  3, 688. 8 Kg

<h3>How to determine the mass</h3>

The formula for quantity of energy is given thus;

Q = n × HF

Where n represents number of moles

HF  represents  heat of fusion

To find the number of moles, we have

235.0 = n × 12.550

number of moles = \frac{235}{12. 550} = 18. 725 moles

Note that molar mass of Gold is 197g/ mol

Let's note that;

Number of moles = mass/ molar mass

Mass = number of moles × molar mass

Mass = 18. 725 × 197

Mass = 3, 688. 8 Kg

Thus, the mass of melted gold to release the energy would be  3, 688. 8 Kg

Learn more about molar heat of fusion here:

brainly.com/question/15634085

#SPJ1

8 0
1 year ago
URGENT!!! A solution in which there is very little solute dissolved in a solvent.
ella [17]

Answer:

Is called a diluted solution

Explanation:

Having little solute makes the dissolving process in a large amount of solvent very easy to mix therefore it dilutes in the solvent

4 0
3 years ago
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