Chapter 1

1. What cells do?(a) Form and build cells’ structure, and breakdown the unuseful structures(b) Form its own extracellular matrix and its own ‘glue’, such collagen and lamina basal,

CAM (cell-adhesion molecules)(c) Change shape and move, cytoskeleton(d) Sense and send information in the form of signal molecules that is affectes cells’ activity(e) Regulate the gene expression to meet the changing needs, e.g. ideal condition for

transciption in E.coli is in the presence of glucose, but it can also survive in lactose in a pinch.

(f) Grows and replicate, to reproduce the cells and inherited the gentic material for survival.(g) Cells die from aggravated assault or a programmed cell-suicide (necessary to avoid the

release of potentially toxic cell constituents)(h) Generate energy in cellular metabolism, cell respiration.

2. Microorganism and virus involvement in human life, might be useful or bring disease.

3. Cells and their part(a) Cell biology reveals the size, shape, location and movement of the cells(b) Biochemistry andbiophysics reveal the molecular structure and chemistry purified cell

constituents(c) Genetics reveals the consequences of damage genes(d) Genomics revelas differences in the structure and expression of the entire genom(e) Developmental biology reveals changes in the properties of cells as they specialized

4. Mutation might be good, bad, or indifferent, depend on the mutation. Can be inherited only if they are present in cells that potentially contribute to the formation of offspring.

Chapter 2

Review of the concepts

1. The gecko is a reptile with an amazing ability to climb smooth surfaces including glass. Recent discoveries indicate that geckos stick to smooth surfaces via van der Waals interaction between septae on their feet and the smooth surface. How is this method of stickiness advantageous over covalent interaction? Given that van der Waals forces are among the weakest molecular interactions, how can the gecko’s feet stick so effectively?

Anwer:

Karena ikatan van der Waals terjadi akibat perbedaan dipol, ketika setae pada kaki cicak mendekati dinding, interansi antara dipol setae dengan dinding menguat sehingga kaki cicak bisa menapak pada dinding. Kemudian ketika setae menjauhi permukaan dinding, makan ikatannya melemah dan terputus ketika mencapai jarak tertentu sehingga kaki bisa dipindahkan ke tempat lain saat cicak melangkah.

Lain halnya dengan ikatan kovalen yang kuat dan stabil, justru ikatan van der Waals akan memungkinkan cicak untuk melangkah pada dinding.

Kaki cicak dapat menempel dengan baik pada dinding karena setae dalam jumlah yang banyak memberikan interaksi dipol yang cukup kuat untuk menahan bobot cicak pada dinding.

2. The K+ channel is an example of a transmembrane protein (a protein that spans the phospholipid bilayer of the plasma membrane). What types of amino acids are likely to be found (a) lining the channel through which K+ passes, (b) in contact with the hydrophobic core of the phospholipid bilayer containing fatty acyl groups, (c) in the cytosolic domain of the protein, and (d) in the extracellular domain of the protein?

Answer:

(a) Negatively charges aminos acids(b) Hydrophobic, non-polar amino acids(c) Hydrophilic amino acids(d) Hydrophilic amino acids

3. V-M-Y-F-E-N: this is the single-letter amino acid abbreviation for a peptide. What is the net charges of this peptide at pH 7.0? An enzyme called a protein tyrosine kinase can attach phosphate to the hydroxyl groups of tyrosine. What is the net chages of the peptide at pH 7.0 after it has been phosphorylated by a tyrosine kinase? What is the likely source of the phosphate utilized of the kinase for this reaction?

Answer:

(a) V = 0, M = 0, Y = 0, F = 0, E = -1, N = 0 net charges = -1(b) Tyrosine (Y) = 0, tirosin kinase = enzim untuk menambahkan gugus fosfat ke asam amino.

Peptide V-M-Y-F-E-N-Phosphate net charges = -1 + (-1) = -2(c) ATP

4. Disulfide bonds help to stabilized the 3D stucture of proteins. What amino acids are involved in the formation of disulfide bonds? Does the formation of a disulfide bond increase or decrease entrophy (∆S)?

Answer:

(a) Cysteine(b) ∆G = ∆H – T.∆S

∆H = enthalpy, energi untuk membentuk ikatanJadi, kalau ikatan terbentuk ∆H naik, ∆S akan berkurang.

5. In the 1960s, the drug thalidomide was prescribe to pregnant women to treat morning sickness. However, thalidomide caused severe limb defects in the children of some women who took the drug, and its used for morning sickness was discontinued. It is now known that thalidmoide is administered as a mixture of two stereoisomeric compounds, one of which relieved morning morning sickness and the other of which was responsible for the birth defects. What are stereoisomers? Why might two such closely related compounds have such different physiologic effects?

Answer:

(a) Stereoisomers: isomer dari suatu atom, yang berbeda pada peletakan spasial atom atau molekul yang mengelilinginya.Khiral: atom C yang keempat elektron valensinya berikatan dengan gugus yang berbeda-beda.

(b) Karena setiap atom pada yang berada dalam struktur suatu molekul memiliki kemampan yang berbeda untuk berinteraksi dengan molekul yang lainnya, sehingga suatu senyawa dengan isomer dapat memiliki pengaruh yang berbeda pula untuk setiap isomernya.Because the arrangement of atoms within their structure differs, yielding their unique abilities to interact and chemically react with other molecules (pg. 33).

6. Name the compound show below.

Is this nucleotide a component of DNA, RNA or both?Name one other function of this compound.

Answer:

RNA. Function:

- Carries genetic information- mRNA transfer genetic information from DNA- tRNA coupled with amino acids to translate mRNA into protein- signal recognition particle

7. The chemical basis of blood-group specificity resides in the carbohydrates displayed on the surfac of red blood cells. Carbohydrates have the potential for great structural diversity. Indeed, the structural complexity of the oligoseccharides that can be formed from four sugars is greater than for oligopeptides from four amino acids. What properties od carbohydrate make this great structural diversity possible?

Answer:

Karena, ikatan peptida hanya memiliki satu model ikatan, sementara ikatan glikosida memiliki dua model,yaitu ikatan α (1,6) dan ikatan β (1,4)

8. Ammonia (NH3) is a weak base that under acidic condition become protonated to he ammonium in in the following reactoin:

NH3 + H+ NH4+

NH3 freely permeates biological membranes, including thise lysosomes. The lysosomes is a subcellular organelle with pH about 4.5-5.0; the pH of cytoplasm is ~ 7.0. What is the effect on the pH of the fluid content of lysosomes when cells are exposed to ammonia? Note: protonated ammonia does not diffuse freely accros membranes.

Answer:

pH dalam lisososm tidak terpengaruh oleh masuknya ion amonia, karena amonia bersifat buffer di mana ion proton dari amonia tersebut akan berinteraksi dengan larutan yang ada di dalam lisosom, sehingga ion positifnya tidak lepas ke lingkungan (tidak menambah keasaman).

9. Consider the binding reaction L + R LR, where L is a ligand and R is its receptor. When 1 x 10-3 M L is added to a solution containing 5 x 10-2 M R, 90% of the L binds to form LR. What is the Keq of this reaction? How will the Keq be affected by the addition of a protein that catalyzes this binding reaction? What is the Kd?

L + R LR

[L] = 1 x 10-3 M[R] = 5 x 10-2 M

90% L MKd = 10-9

Kd = ([L][R]) / ([LR])10-9 = ( (1 x 10-3) x (5 x 10-2) ) / [LR][LR] = 5 x 104

Keq = ([LR]) / ([L][R]) = 109

Jika ditambah dengan katalis, Keq tidak berubah.

10. The ∆Go’ for the reaction X + Y XY is – 1000 cal/mol. What is the ∆G at 25oC (298 K) starting with 0.01 M each X, Y and XY? Suggest two ways one could make this reaction energetically favorable.

Answer:

(a) ∆G = ∆Go’ + RT ln Q

∆G = ∆Go’ + RT ln ([P] / [R])

∆G = - 1000 cal/mol + (1.987 cal/degree.mol)(298 degree) ln (0.01/0.01.0.01)

∆G = - 1000 cal/mol + (529.162)(4.6)

∆G = - 1000 cal/mol + 2434.1452 = 1134.1452 cal/mol

(b) Increase the entrophy (∆S) enough or raise the temperature so that T.∆S term can overcome the positive ∆H

11. What is the ionization state of phosphoric acid in the cytoplasm? Why is phosphoric acid such a physiologically important compound?

Answer:

HPO4 - ↔ H+ + PO42-

Buffer mempertahankan keasaman di dalam sel agar kondisinya tetap ideal untuk menjalankan fungsi-fungsi sel.

12. According to the health experts, saturated fatty acids, which come from animal fats, are a major factor contributing to coronary heart disease. What distinguishes a saturated fatty acids from unsaturated fatty acids, and to what does the term saturated refer? Recently, trans unsaturated fatty acids, or trans fats, which raise total cholesterol levels in the body, have also been implicated in heart disease. How does the cis stereoisomer differ from the trans configuration, and what effect does the cis configuration have on the structure of the fatty acid chain?

Answer:

(a) Saturated = tidak ada ikatan rangkapUnsaturated = ada minmal 1 ikatan rangkapThe chain of carbon atoms are fully saturated with hydrogen atoms.

(b) Trans : berseberanganCis : bersebelahanWhen this oil is hydrogenated, it is not possible to control where the hydrogen atoms are added to the structure. If both hydrogen atoms are added to the same side of the structure, it is called a "Cis" fat. Cis fats exist naturally and, because the hydrogen atoms are crowded on one side of the molecule, they bend, allowing other chemicals and enzymes to bind to them (bisa diuraikan).If, however one hydrogen atom adds to one side of the structure and the other atom to the other side, it creates trans fats, like the one below. Trans fats do not exist naturally, with a very few exceptions. Because the structure is uncrowded, they do not bend and so other molecules and enzymes find it more difficult to bind to them. The shape of the molecule is therefore vital to its function, much in the same way as the shape of a key is important for the operation of a lock.

13. Chemical modifications to amino acids contribute to the diversity and function of proteins. For instance, Ƴ-carboxylation of specific amino acid is required to make some proteins

biologically active. What particular amino acid undergoes this modification, and what is the biological relevance? Warfarin, a derivative of coumarin, which is present in many plants, inhibits Ƴ-carboxylation of this amino acid and was used in the past as a rat poison. At present, it also used clinically in humans. What patients might prescribes warfarin and why?

Answer:

(a) Glutamate, form blood-clotting factors such as prothrombin(b) Warfarin = anticoagulant

Warfarin and related 4-hydroxycoumarin-containing molecules decrease blood coagulation by inhibiting vitamin K epoxide reductase, an enzyme that recycles oxidized vitamin K to its reduced form after it has participated in the carboxylation of several blood coagulation proteins, mainly prothrombin and factor VII. For this reason, drugs in this class are also referred to as vitamin K antagonists.[2] When administered, these drugs do not anticoagulate blood immediately. Instead, onset of their effect requires about a day before clotting factors being normally made by the liver have time to naturally disappear in metabolism, and the duration of action of a single dose of racemic warfarin is 2 to 5 days. Under normal pharmacological therapy the drugs are administered to decrease the action of the clotting factors they affect by 30 to 50%.[3]

Warfarin is used to prevent blood clots from forming or growing larger in your blood and blood vessels. It is prescribed for people with certain types of irregular heartbeat, people with prosthetic (replacement or mechanical) heart valves, and people who have suffered a heart attack. Warfarin is also used to treat or prevent venous thrombosis (swelling and blood clot in a vein) and pulmonary embolism (a blood clot in the lung). Warfarin is in a class of medications called anticoagulants ('blood thinners'). It works by decreasing the clotting ability of the blood.

Chapter 3

Protein function

1. Regulation, to activate or inactivate several systems, like transcription factor.2. Protein structure, builds structure, such as actin filament.

3. Movement, as motor, like actin an myosin in muscle cells.4. Catalysts, enzyme to increase the reaction rate.5. Transport, provide channels or transporter for molecullar traffic.6. Signaling, provide pathways for signaling.

1. The three-dimensional structure of a protein is determined by its primary, secondary and tertiary structures. Define the primary, secondary and tertiary structures. What are some of the common secondary structures? What are the forces that hold together the secondary structure and tertiary structures? What is the quarternary structure?

Answer:

(a) Primary structures: linear arrangements of amino acids.Secondary structures: the core elements of protein architecture, stable spatial arrangements of segment of a polipeptide chain held together by hidrogen bonds between backbone amide and carbonyl groups. The principal secondry structures are α-helix, β-turns and short U-shaped β-turns.Tertiary structures: the overall folding conformation of a polypeptide chain, the three-dimensional arrangements of all its amino acids residues which are stabilized only by hydrogen bonds, primarily stabilized by hydrophobic interaction between nonpolar side chains, together with hydrogen bonds between polar side chains and peptide bonds.

(b) α-helix, β-turns and short U-shaped β-turns(c) hidrogen bonds(d) multimeric protein consist of two or more polpeptide chains or subunit. 2. Proper folding of proteins is esential for biological activity. Describe the roles of molecular

chaperones and chaperonins in the folding of proteins.

Answer:

Chaperones binds and stabilized unfolded or partly folded proteins, thereby preventing these proteins from aggregating an being degraded.Molecular chaperonins form a small folding chamber into which an unfolded protein can be sequestered, giving it time and an appropriate environment to fold properly.

3. Proteins are degraded in cells. What is ubiquitin and what role does it play in tagging proteins for degradation? What is the role of proteosomes in protein degradation?

Answer:

(a) Ubiquitin marks cytosolic proteins for degrdation in proteosomes. Ubiqitin is 76-residues of polypeptide that determined which proteins are to be degraded by covalently bond. - The covalent addition of a single ubiquitin molecule (monoubiquitination) to a lysine

on a target protein- The addition of multiple single ubiquitins (multiubiquitination)- Linking the ubiquitin to the N-terminus of the target protein

- Polyubiquitination in which the ubiquitins are linked to one another via another via their Lys-63 residue instead of the Lys-48 position

(b) Proteosomes is anothe cells’ molecular machine. The numerous proteasomes dispersed throughout the cell cytosol proteolytically cleave ubiquitin tagged proteins in an ATP-dependent process that yeilds short (7- to 8-residue) peptides and intact ubiquitin molecules.

4. Enzymes can catalyze chemical reactions. How do enzymes increase the rate of a reaction? What constitutes the active site of an enzyme? For an enzyme-catalyzed reaction, what are

Km and Vmax? For enzyme X, the Km for substrate A is 0.4 mM and substrate B is 0.01 mM. Which substrate has a higher affinity for enzyme X?

Answer:

(a) By lowering the activation energy needed in the reaction.(b) Active sites contains two functionally important regions, one that recognizes and binds the

substrate (or substrates) and another that catalyzes the reaction after the substrate has been bound.

(c) Km is defines as the substrate concentration that yields a half-maximal reaction rate (½ Vmax) ans a measure of the affinity of an enzyme for its substrate while Vmax is the maximal velocity of reaction at saturating substate concentration.

(d) Km A = 0.4 mMKm B = 0.01 mMThe smaller Km, the more avid an enzyme can bind substrate from a dilute solution and the smaller the substrate concentration needed to reach half-maximal velocity.

5. Motor proteins, such as myosin, convert energy into a mechanical force. Describe the three general properties characteristics of motor proteins. Describe the biochemical events that occur during one cycle of movement of myosin relative to an actin filament.(a) 3 general properties characteristics of motor protein

- The ability to tracnduce a source of energy, either ATP or an ion gradient, into linear or rotary movement.

- The ability to bind and translocate along a cytoskeletal filament, nicleic acid strand, or protein complex.

- Net movement in a given direction(b) Operational model for the coupling of ATP hysrolisis to movement of myosin along an

actin filament.

6. The function of proteins can be regulated in a number of ways. What is cooperativity, and how does it influence protein function? Describe how protein phosphorylation and proteolytic cleavage can modulate protein function.

Answer:

(a) Cooperativity: the binding of one ligand molecule affects the binding of subsequent ligand molecules. This cooperativity permits many multisubunit proteins to respond more efficiently to small changes in ligand concentration than would otherwise be possible. In positive cooperativity, sequential binding is enhanced; in negative cooperativity, sequential binding is inhibited.

(b) Phosphorylation changes a protein’s charge and generally leads to a conformational change; these effects can significantly alter ligand binding by a protein, leading to an increase or decrease in its activity. Proteolytic cleavage irreversibly activates or inactivates some proteins.

7. A number of techniques can separate proteins on the basis of their differences in mass. Describe the use of two of these techniques, centrifugation and gel electrophoresis. The blood proteins tranferrins (MW 76 kDa) and lysozymes (MW 15 kDa) can be separated by rate zonal centrifugation or SDS polyacrylamide gel electrophoresis. Which of the two proteins will sediment faster during centrifugation? Which will migrate faster during electrophoresis?

Answer:

(a) Centrifugation is used for two basic purposes, as apreparative technique to separate one type or material from athers and as an analytical technique to measure physiological properties of macromolecules. This technique is used to separate particles and molecules that differ in mass or density.

Electrophoresis separates molecules on the basis of their charge, mass ratio. This technique separating molecules in a mixture under the influence of an applied electric field. Dissolved molecules in an electric field move or migrate, at a speed determined by their chage: mass ratio. For example, if two molecules have the same mass and shape, the one with the greater net charge will move faster toward an electrode.

(b) Transferrins (c) Lysozymes

8. Chromatography is an analytical method used to separate proteins. Describe the principles for separating proteins by gel filtration, ion-exchange, and affinity chromatography.

Answer:

(a) Gel filtration: protein separation besed on its size.

(b) In ion-exchange chromatography, the proteins are separated on the besis of differences in their charges.

(c) Affinity chromatography used the ability of proteins to bind specifically to other molecules. In this techniques, ligand molecules that bind to the protein of interest are covalently attached to the beads used to form the column. Ligans can be enzyme substartes or other small molecules that bind to specific proteins.

9. Various methods have been developed for detecting proteins. Describe how radioisotope and autoradiography can be used for labeling and detecting proteins. How does Western blotting detect proteins?

Answer:

(a) Radioisotopes labeled nucleotides of the molecules and autoradiography detected the the radioactivity by overlaiding it with a photographic emulsion sensitive to radiation. Autoradigraphy is commonly used in various assays for detecting specific isolated DNA or RNA sequences. Autoradiography is a semi0quantitative technique for detecting radioactively labeled molecules in cells, tissues, or electrophoresis gels.

(b) Western blotting, a powerful method for separating and detecting a protein in a mixtures

10. Physical methods are often used to determine protein conformation. Describe how x-ray crystallography, cryoelectron microscopy, and NMR spectroscopy can be used to determine the shape of proteins.

Answer:

(a) X-ray crystallography is used to determine the 3D structures of proteins. Beams of x-ray are passed through a protein crystal in which millions of protein molecules are precisely aligned in a rigid array characteristics. Atoms in the crystal scaterred the x-ray, produced a diffraction pattern to discrete spots when they are intercepted by photographic film.

Disadvantage: hard to crystallize the proteins, particularly large multisubunit proteins that requires a time-consuming trial-and-error effort to find just the right conditions.

Advantage: provide the most detail structures.

(b) Cryoelectron microscopy, protein sample is rapidly frozen in liquid helium to preserve its structure and then examined in the frozen, hydrated state. Pictures are recorded in film by using a low dose of electron to prevent radiation-induced damage to the structure. Recent edvances permit researchers to generate molecular models that compare with thise derived from x-ray crystallography.Most useful for large protein complexes which are difficult to crystallize

(c) NMR (nuclear magnetic resonance) spectroscopy is used to study 3D structures of small proteins containing about as masny as 200 amini acids. In this technique, a concentrated protein solution is placed in a magnetic field and the effetcs of different frequencies on the spin of different atoms are measured. From the magnitude of the effect, the distances between residues can be calculated; these distances are then used to generate a model of the 3D structure of the protein. The advantage: limited to proteins smaller than 20 kDa, acn be applied to protein domains.

11. Mass spectrometry is powerful tool in proteomics. What are four key features of a mass spectrometer? Describe briefly how MALDI and 2D-PAGE (two-dimensional polyacrilamide gel electrophoresis) could be used to identify a protein expressed in cancer calls but not in normal healthy cells.

Answer:

(a) Mass spectrophotometer, requires a method for ionizing the sample, usualy a mixture of peptides or proteins, accelerating the molecular ions, and then detecting the ions. In a laser desorption mass spectrophotometer, the protein sample is mixed with an organic acid and then dried on a metal target. Energy from a laser ionizes the proteins, and an electric field accelerates the ions down a tube to a detector. Alternatively, in an electrospray mass spectrometer. A fine mist containing the sample is ionized and then introduced into a separation chamber where the positively charged molecule are accelerated by an electric field. In both instruments, the time of flight is inversely proportional to its charge.

Ionization, acceleration, detection, molecular weight

(b) 2D-PAGE have been very useful in comparing the proteosomes in undifferentiated and differentiated cells or in normal and cancer cells because as many as 100 proteins can be resolved simultanously.

Analyze the data

Proteomics involves the global analysis of protein expressions. In one approach, all the proteins in control cells and treated cells are extacted and subsequently separated using two-dimensional gel electrophoresis. Typically, hundreds or thousands of protein spots are resolved and the steady-state levels of each protein are compared between control and treated cells. In the following exampe, only a few protein spots ar shown for simplicity. Proteins are separated in the first dimension on the basis of charge by isoelectric focusing (pH 4-10) and then separated by size by SDS polyacrylamide gel electrophoresis. Proteins are detected with a stain such as Coomassie blue and assigned numbers for idntification.

a. Cells are treated with a drug (“+ Drug”) or left untreated (“Control”) and then proteins are extracted and separated by two-dimensional gel electrophoresis. The stained gels are shown below. What do you conclude about the effect of the drug on the steady-state levels pf proteins 1-7?

b. You suspect that the drug may be inducing a protein kinase and so repeat the experiment in part a in the presence of 32P-labeled inorganic phosphate. In this experiment the two-dimensional gels are exposed to x-ray film to detect the presence of 32P-labeled proteins. The x-ray films are shown below. What do you conclude from this experimant about the effect of the drug on proteins 1-7?

c. To determine the cellular localization of proteins 1-7, the cells from part a were separated into nuclear and cytoplasmic fractions by differential centrifugation. Two-dimensional gels were run and the stained gels are shown below. What do you conclude about the cellular localization of proteins 1-7?

d. Summarize the overall proerties of proteins 1-7, combining the data from parts a, b, and c. Describe how you could determine the identitiy of any one of the proteins.