Biochemistry Exercises: Water, Buffers, pH, Lipids, Amino Acids, Proteins, and Enzymes, Slides of Biochemistry

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BIOCHEMISTRY

Exercises

Unit 1 - 2 - 3 - 4

What you need to know:

• Structure and functions of water
• How to calculate pH of a strong acid, base and a weak acid
• Buffer, function of buffer Unit 1 Water - Buffers

1. Solubility of substances in water Predict which of the following substances are soluble in water. Explain you answer. Unit 1

The following substances are essential components of the human diet: Predict which of the following substances are soluble in water.

Tương tác ion Tương tác kỵ nước Tương tác van der Waals Liên kết hydro

2. Calculation of pH of pure water The ion product of water at 0 °C is 1. 14 × 10

• 15 M 2 , and at 100 °C it is about

5. 13 × 10

• 13 M 2 . What is the pH of pure water at 0 °C and 100 °C? Unit 1 Ion product of water: K W K W

= [H

] [OH

• ] [H

] = [OH

] = K W pH = - log [H

] At 0 o C: pH = 7. At 100 o C: pH = 6.

3. Identifying the Conjugate Base Which is the conjugate base in each of the pairs below? (a) RCOOH, RCOO − (b) RNH 2

, RNH

3

(c) H 2

PO

4 − , H 3

PO

4 (d) HPO 4 2 - , H 2

PO

4

(e) PO 4 3 - , HPO 4 2 - (f) H 2

CO

3

, HCO

3

(g) CO 3 2− , HCO 3 − Unit 1 HA  H

• A

acid or conjugate base or

protonated form deprotonated form

4. Acidity of Gastric HCl In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient’s stomach contained no ingested food or drink; thus assume that no buffers were present. What was the pH of the gastric juice? HCl + NaOH  NaCl + H 2 O n HCl = n NaOH = C M (NaOH) . V NaOH = 0.1 x 7.2. - 3 = 0.72. 10 - 3 mol => n HCl = 0.72. 10 - 3 mol => C M (HCl) = (0.72. 10 - 3 ) / (10. 10 - 3 ) = 0.072 M HCl → H + + Cl - →[H + ] = [HCl] = 0.072 M → pH = - log [H + ] Unit 1

6. Preparation of an Acetate Buffer Calculate the concentrations of acetic acid (p K a = 4.76) and sodium acetate necessary to prepare a 0.2 M buffer solution at pH 5.0. pH = pK a + log

[CH

3

COO

• ] [CH 3

COOH]

log

[CH 3 COO

• ] [CH 3

COOH]

= pH - pK a [CH 3 COO

• ] [CH 3

COOH]

[CH

3

COO

• ] and [CH 3

COOH]

Unit 1 0.126 M acetate and 0.074 M acetic acid^13

7. pH and Drug Absorption Aspirin is a weak acid with a p K a of 3.5. It is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane, the rate of which is determined by the polarity of the molecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones pass rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestine is about 6. Is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice. Unit 1

What you need to know:

• The definition and classification of carbohydrates
• The importance of carbohydrates
• The structure of carbohydrates
• The difference between various isomers Unit 2 Carbohydrates

H

Recognizing Epimers Identify the epimers of (a) D-Allose, (b) D-Gulose, and (c) D-Ribose at C-2, C-3, and C-4. Unit 2

Isomers which differ from each other only with regard to

• OH group on a single asymmetric carbon atom are called epimers. Monosaccharides Epimers