Select all statements that correctly describe hemoglobin and myoglobin structure. a. Molecular oxygen binds irreversibly to the Fe ( II ) Fe(II) atom in heme. b. Each hemoglobin or myoglobin molecule can bind four oxygen molecules. c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron. d. Each iron atom can form six coordination bonds. One of these bonds is formed between iron and oxygen. e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom. f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.

c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron.

e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom.

f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.

Explanation:

The differences between hemoglobin and myoglobin are most important at the level of quaternary structure. Hemoglobin is a tetramer composed of two each of two types of closely related subunits, alpha and beta. Myoglobin is a monomer (so it doesn't have a quaternary structure at all). Myoglobin binds oxygen more tightly than does hemoglobin. This difference in binding energy reflects the movement of oxygen from the bloodstream to the cells, from hemoglobin to myoglobin.

Myoglobin binds oxygen

The binding of O 2 to myoglobin is a simple equilibrium reaction:

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point