This set of Structural Biology Multiple Choice Questions & Answers (MCQs) focuses on “Proteins Structure – Hill Constant”.
1. Hill constant was named after ___________
a) A. B. Hill
b) A. V. Hill
c) A. C. Hill
d) B. C. Hill
View Answer
Explanation: The Hill constant was described by an English scientist named Archibald Vivian Hill. Hill was a physiologist. He also contributed to the field of biophysics. He has secured Noble prize for his work on the mechanical work performed by the muscles and the heat produced.
2. Why does the Hill constant is used?
a) To determine the ligand concentration
b) To determine the receptor concentration
c) To determine the co-operativity
d) To determine the selectivity
View Answer
Explanation: It is used to determine the cooperative coefficient of a process. When there are two substances like ligand and a substrate in a given reaction mixture, it is necessary to know the competitiveness between a particular set of ligand and substrate.
3. In which experiment does the concept of Hill constant is used for the first time?
a) Binding of the antigen to the antibody
b) Binding of oxygen to hemoglobin
c) Binding of calcium to the calcium binding site
d) Binding of potassium to the Na-K pump
View Answer
Explanation: Hemoglobin takes up and carries 4 molecules of oxygen. Oxygen binds to the hemoglobin in a stepwise process. All the oxygen molecules do not bind to the hemoglobin at the same time. To understand the cooperativity in their binding, this topic is used.
4. Which of the following is an example of stepwise/cooperative binding?
a) Binding of sodium to the Na-K pump
b) Binding of potassium to the Na-K pump
c) Binding of oxygen to the hemoglobin
d) Binding of the antibody to the antigen
View Answer
Explanation: As you know, the oxygen binds to the heme part of the hemoglobin. Hemoglobin is made of 4 subunits with Iron atoms in each subunit. So the oxygen binds to the Fe atom of the first subunit and brings up the conformational change in the hemoglobin which promotes the binding of the second oxygen atom to the second subunit and so on.
5. Assume that there are 96 oxygen molecules in a system. What is the total number of hemoglobins required to carry those 96 molecules of oxygen?
a) 96
b) 90
c) 48
d) 24
View Answer
Explanation: One hemoglobin subunit carries one oxygen molecule. Single hemoglobin has 4 subunits hence one hemoglobin can carry 4 oxygen molecules. In the problem, it is given that there are 96 molecules in a particular system. So 96/4 = 24 hemoglobins are required to carry 96 molecules of oxygen.
6. What is the minimum number of ligands need to achieve a saturated system if there are 15 macromolecules (substrates) undergoing cooperative binding with 6 binding sites on each?
a) 60
b) 70
c) 90
d) 150
View Answer
Explanation: A saturation state can be obtained only when all the macromolecules are bound with the ligands at all ligand binding site. Therefore, if a molecule has 6 binding sites on it, then it needs 6 ligands to be saturated. Hence a system with 15 macromolecules need 15 × 6 = 90 ligands to be fully saturated.
7. Name the phenomenon where binding of a ligand to a macromolecule is affected by the previously bound ligands.
a) Selectivity
b) Cooperativity
c) Activity
d) Reactivity
View Answer
Explanation: Cooperativeness is the phenomenon where binding of a ligand to the substrate is affected by the previously bound ligand. For example, if a ligand 1 binds to one domain of a protein then it decides whether the ligand 2 should be allowed to bind to another domain or not.
8. Which of the following is not a hill equation / alternative formulation?
a) θ = [L]n / (Kd + [L]n)
b) θ = [L]n / ((KA)n + [L]n)
c) Log((1 – θ) / θ) = n log [L] – log Kd
d) Log(θ / (1 – θ)) = n log [L] – log Kd
View Answer
Explanation: The actual hill equation is θ = [L]n / (Kd + [L]n) and by substituting Kd = (KA)n we get θ = [L]n / ( (KA)n + [L]n). Then by reciprocating, rearranging, inverting and then taking log on both the sides, we get the third equation which is used for the calculation purpose, log(θ / (1 – θ)) = n log [L] – log Kd.
9. What does theta represent in the hill’s equation?
a) The fraction of occupied sites
b) The fraction of free sites
c) Total number of sites
d) Total number of free sites
View Answer
Explanation: For example, there will be several active sites in a macromolecule. These sites will be occupied by ligands like enzymes or ions. Hence the number of occupied sites divided by the total number of active sites gives the fraction theta.
10. Which notation is used to indicate the hill coefficient?
a) Kd
b) KA
c) n
d) L
View Answer
Explanation: The Hill coefficient is represented by the term ‘n’. This describes the effect of the substrate the presence of ligands that are previously bound to the substrate, on the binding of the free ligands.
11. What conclusion can you give if the Hill coefficient is equal to unity?
a) The system has reached saturation
b) The system has reached equilibrium
c) The system is non-cooperative
d) The ligand is incompatible
View Answer
Explanation: If the binding of a ligand is not influenced by the previously bound ligand then the system constitutes a non-cooperative system. If the Hill coefficient is equal to unity then it represents that ligand 1 does not affect the binding of ligand 2 to the macromolecule.
12. In a lac-operon system, the repressor protein (substrate) Binds to the operator site (ligand 1) and blocks the expression of the gene. Then allolactose(ligand 2) Binds to this repressor protein and makes the gene expression possible. For this situation, calculate the Hill coefficient.
a) 0
b) 1 and greater than 1
c) 1
d) 1 and less than 1
View Answer
Explanation: Lac-operon is the gene regulation system possessed by some of the prokaryotes. In the absence of glucose, they prefer lactose as their source of food. In this process, binding of repressor protein to the operator site or binding of the allolactose to the repressor protein is not affected by one another. Therefore, it is incorporative. Hence, the Hill constant is equal to 1.
13. Calculate the value of the Hill constant. ( Data : – Kd = 1, Ligand concentration = 4 and θ = 0.7) and mention the type of system.
a) 0.4 and positively cooperative
b) 0.4 and negatively cooperative
c) 0.6 and negatively cooperative
d) 0.6 and positively cooperative
View Answer
Explanation: We have the standard formula as, log (θ / (1 – θ)) = n log [L] – log Kd. Substituting the values Kd = 1, Ligand concentration, L = 4 and θ = 0.7, we get log (0.7 / (1.0.7)) = 0.368, log [L] = .6, log Kd = 0. Hence, Hill constant n = .368 / .6 = .6. Since ‘n’ has a value less than 1 it is negatively cooperative system.
14. If the Hill coefficient is positive then the system is positively cooperative.
a) True
b) False
View Answer
Explanation: The system cooperative only when the hill coefficient is greater than 1. In the question, it is given that hill coefficient is positive which means the Hill constant is greater than 0. Hence, the system is not positively cooperative. It can be negatively cooperative also for the values between 0 and 1.
15. In a system containing only macromolecules, the Hill coefficient is 1.
a) False
b) True
View Answer
Explanation: It is given that the system contains only macromolecules. This indicates that the ligands are absent in the system. Hence ligand concentration [L] will be zero. Hence by substituting this value in the Hill equation, we get, log(θ / (1 – θ)) = n log [0] – log Kd. The value of log [0] is not defined. Hence, the Hill coefficient (n) for such cases is not possible to determine.
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