Study Guide Key for CEM 109 Fall 2015 Remember you will need to show your work for full credit. n the real exam always work the problems you know best first. If you get hung up on a problem, you should move on and come back to it at the end. If you have time, check over your work. To test your speed, work this study guide as if it was the exam. ow long does it take to finish? 0.1. Draw a generalized structure for an amino acid (use may use R for the side chain ). amino group C C R R represents the side chain carboxyl group 0.2. Show how two amino acids join to form a peptide bond. + C C - R Predominant structure at neutral p 0.3. Why would it be very appropriate for the amino acid, aspartic acid, to lie on the outer surface of a cytoplasmic protein? A cytoplasmic protein is surrounded by water which is polar. Aspartic acid s side chain is polar and negatively charged at neutral p. It is hydrophilic. 0.4. What are the levels of structure of proteins? What kinds of forces/bonds maintain this structure? Primary (1 o ) structure of proteins is the amino acid sequence. It is maintained by covalent bonds called peptide bonds. Secondary (2 o ) structure is a regular repeating structure due to folding of the polypeptide chain. The main types are alpha-helix and beta sheet (either parallel or anti-parallel). Secondary structure is maintained by hydrogen bonds formed between a hydrogen (donor) attached to the nitrogen in the backbone of the chain and the non-bonding pair on the carbonyl oxygen (C=) in the backbone. Tertiary (3 o ) structure refers to the location of each atom of the protein relative to every other atom in three dimensional space. This structure is maintained by hydrophobic interactions, hydrogen bonding, covalent (disulfide, -S-S-) bonds, ionic bonds and London forces. Quaternary (4 o ) structure exists only for proteins that have more than one chain or subunit. It describes the way the subunits are arranged and bind to each other. It is maintained by the same forces as tertiary structure. 0.5. Given an organic molecule, be able to circle the chiral carbons. Example with chiral carbons circled: Br C C C C Cl Br
0.6. Determine the amino terminal end of each protein strand shown to the right. Determine which the strand reads amino group, α-carbon, carbonyl carbon. For the strand on the left, that direction is from the bottom to the top, so the amino end is towards the bottom of the page. For the strand on the right, the strand runs from top to bottom, so the terminal is at the top of the page (or further up.) 1. Describe the two conformations of hemoglobin and describe how they aid in transport of oxygen to your tissues. emoglobin (b) has two conformations: a form with a high affinity for oxygen (the main b form in the lungs) and a form with low affinity for oxygen (the main b form present in the extremities and at lower p). b shifts back and forth between these two forms as it circulates thru the body. In the lungs, after one 2 molecule binds to a b subunit, the affinity of the other three subunits for 2 increases. As each subunit binds 2, the attraction grows stronger (a cooperative effect), so in the lungs where 2 concentration is high, all of the subunits are bound to 2, meaning b can pick up the maximum load of 2. In the extremities, 2 concentration is low. As the first subunit releases 2, the affinity for 2 by the other subunits decreases. This allows the other subunits to more easily release 2. So in the tissue where 2 concentration is lowest and where 2 is needed most, the b molecule is able to deliver the maximum amount of oxygen. 2. Draw a reaction coordinate diagram for the hydrolysis of a dipeptide (like Gly-Ala). Then draw another line on the same diagram to describe the reaction when catalyzed by an efficient catalyst. transition state uncatalyzed reaction Eact uncatalyzed gly-ala + water catalyzed reaction G Eact catalyzed ΔG is negative gly + ala Reaction Progress The G values on my diagram should be G o. The graph was getting crowded so I didn t include the transition state for the catalyzed reaction, although it should be indicated. 3. What happens to the rate of an enzyme catalyzed reaction at low substrate concentrations (compared to the enzyme concentration) when you double the substrate concentration? Does the same apply when you are looking at high levels of substrate concentration? Explain. At low substrate concentrations (compared to the enzyme concentration), when you double the substrate concentration, you double the rate of the reaction, because at low substrate concentrations not all of the enzyme molecules are working. At high substrate concentrations, the rate generally does not increase much, or not at all, because at high concentrations, all of the enzyme molecule s active sites are saturated (constantly filled)
4 Describe the molecular basis of sickle cell anemia, one treatment, and indicate why the trait is found at relatively high levels in some populations. ormal human b has glutamic acid (Glu) at position 6 of its beta chain. A mutation in the DA results in the conversion of GAG (in the RA) to GUG (in the RA) in sickle cell hemoglobin (bs). GUG is the codon for valine (Val). The valine side chain is much less polar than that of Glu. When bs is in its low affinity form (deoxygenated), bs starts to polymerize into long chains that protect the valine side chain from water and precipitate (come out of solution to form a solid). bs (2)4 (aq) bs(s) + 4 2(aq) The long rods that form lead to a change in shape and loss of flexibility of the red blood cells. This in turn leads to restricted blood flow in the capillaries and rupturing of red blood cell membranes. Ultimately this causes the broader problems associated with sickle-cell crises (pain, liver problems, etc.). ne of the treatments for sickle-cell disease is hydroxyurea. This drug works by inducing the synthesis of the fetally produced gamma ( ) chain, that can partially replace the beta ( ) chain. This reduces the incidence of a sickle-cell crisis. Sickle-cell heterozygotes (individuals who have one wild type (wt) and one sickle cell b gene) are resistant to malaria. This resistance may account for the relatively high frequency of the sickle cell trait in individuals from areas where malaria is prevalent. 5. What is the Central Dogma of Molecular Biology? It shows the direction and flow of information. See top of next page. 6. What amino acid sequence would be coded from the mra template shown below? Make sure that you indicate the polarity of the peptide. Stop AUG GAC CCG UGC GUC GUU UCU GUG ACA UGA CAAC 3 (- terminus) Met -Asp- Pro - Cys Val Val Ser Val Thr (C- terminus) 7. Why is it important that DA be able to replicate itself millions of times without error? What feature of DA structure is particularly important with regard to avoiding errors? DA needs to be able to replicate itself without errors, because an error if not corrected results in a mutation. Mutations are rarely positive for the cell or organism, sometimes neutral and often have negative consequences. Mutations can result in the formation of ineffective proteins, can cause disease states, and cancer. The structural feature of DA that helps avoid mutations is that DA is double stranded with complementary bases. This means that (1) each strand contains the information in each sequence, and (2) copying errors can be detected early on as a mismatch that can be recognized and repaired before replication.
8. Give one sentence descriptions of the roles of mra, tra, and rra in the cell. mra, tra, and rra are three major classes of RA in the cell. Messenger RA (mra) carries genetic information from DA, located in the nucleus in eukaryotic cells, into the cytoplasm where the information can be translated into protein. Transfer RA (tra) is the direct interpreter of genetic information found in mra into the sequence of amino acids for a protein using its three base anti-codon located at one end and attached amino acid at the other. Ribosomal RA (rra) is the RA component of ribosomes and it is involved in the catalysis of linking one amino acid to the next. 9. What is the quaternary structure of proteins? What kinds of forces/bonds maintain this structure? Quaternary (4 o ) structure exists only for proteins that have more than one chain or subunit. It describes the way the subunits are arranged and bind to each other. It is maintained by the same forces as tertiary structure. 10. Describe where DA is located and where proteins are synthesized. What molecules and processes exist that allow the transfer of this information from one location to the other? DA is located in the nucleus. Proteins are synthesized on ribosomes in the cytoplasm. The information encoded in the DA is carried from the nucleus to the cytoplasm via messenger RA (mra). mra is transcribed from DA and processed in the nucleus and then passes thru the nuclear membrane into the cytoplasm and moves to the ribosomal machinery where it is translated to protein. The translation process uses tra s that have (1) an anti-codon on one end that can read an mra codon and (2) its designated amino acid on the other end. 11. Draw a DA molecule being replicated. Include the direction the DA is unwinding, the polarity of the parent and daughter strands and the direction the daughter strands are being leading strand synthesized. parental strands daughter strands lagging strand direction of unwinding and of replication 12. You are attending a family reunion, and the subject of mitochondrial DA comes up (as it often will at such gatherings). ow likely is it that you have the same mtda sequence as your cousin Luke? Luke is your mother s sister s son. Briefly explain your logic. 100% likely. Mitochondrial DA is inherited maternally in humans. Therefore both your mother and her sister inherited their mtda from their mother, and (barring mutation) both you and Luke will inherit that same mtda.
13 Draw a cartoon model of DA indicating how the three major pieces are connected. Indicate the location of hydrogen bonds. See drawing a right. TE: ydrogen bonds need to be dotted or dashed. I couldn t figure out how to do that on my picture P 4 P4 P 4 P4 P4 C A G T hydrogen bonding A C T P 4 14. Given the following DA sequence, what would the complementary RA sequence be? Indicate the polarity of the RA. G A T G G A C G T G C G T A A 3 3 U A C C U G C A C G C A U U 5 15. What is a mutation? Be familiar with different types of mutations. A mutation is a change in the DA that doesn t get repaired. nce a base change mutation is replicated, it is fixed as a mutation. ne type of mutation is a point or single base change. It can be neutral (resulting in a codon change for the same amino acid), a conservative change (resulting in a codon change for an amino acid whose side chain has similar polarity or charge), or non-conservative mutation (resulting in an amino acid whose side chain has very different polarity or charge). A point mutation could even result in a nonsense mutation, which is when a codon gets changed to one of the stop codons.) ther types of mutations that you should be familiar with include deletions, insertions and frameshift mutations. 16. If a researcher determined that DA from a new organism was 28.3% G, what would be the percentages of A, T and C in the DA of this creature? G and C are complementary, so the % of C must equal the % of G. So, G is 28.3% and C is 28.3%. Total is 56.6%, so that leaves 100-56.6 or 43.4% for A and T. The bases A and T are complementary, so the % of A must equal the % of T. So each percentage is half of 43.4%, or 21.7% So, A is 21.7% and T is 21.7%.
17. Is the nucleic acid strand shown on the left RA or DA? ow do you know? Draw an arrow that represents the 5 to 3 direction. The structure shown on the study guide is a strand 2 of DA. You can tell because the 2 carbon of the has an bonded to it, rather than an. ote: I have labeled some of the carbons in the s. 1' 4' 2 2 2' P - P - 18. What is the definition of lipids?. Lipids are biological compounds that are soluble in non-polar solvents. 19. What is a generalized structure of a fatty acid, of a triglyceride? A fatty acid consists of a carboxylic acid, the head, and a long, non-polar, carbon chain, the tail. C 3 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C 2 -C Glycerol fatty acid fatty acid tail head fatty acid A triglyceride consists of three long chained fatty acid molecules linked by a ester linkage to glycerol. 20. In general, how and why does unsaturation affect the melting points of fatty acids. Fatty acids have long carbon chains that are non-polar. The intermolecular forces involved are mainly London forces that depend on the molecules in question to lay close to each other. Unsaturated chains are fairly straight and are able to line up close to each other. The cis double bonded carbons in unsaturated molecules put a bend or kink in the chain. This reduces prevents the chains from lining up close to each other, reducing the London forces between the molecules. With reduced IMF, it takes less heat to melt unsaturated chains reducing the melting point. saturated unsaturated Answers to questions 17.5 and 21-23 will follow tomorrow.
17.5. Describe major steps in protein synthesis. a) Before protein synthesis starts, amino acids must be activated and bound, through their carboxyl group, to the appropriate tra. A tra bound to its amino acid is called a charged tra. b) During initiation, the two ribosomal subunits form around the 5 end of the mra and the appropriate charged tra binds to the first codon (usually AUG) of the message, in what is called the P site of the ribosome. Remember that the codon is on the mra and the anticodon on the tra must be complimentary to bind. c) Elongation. The next tra charged with the correct amino acid binds in an adjoining site (called the A site) next to the first charged tra. d) ow an enzyme (made of RA) breaks the bond between the first amino acid and its tra and forms a peptide bond joining the first and second amino acids. e) ow we have the translocation step. The first tra, having lost its amino acid, leaves the ribosome and at the same time the mra and remaining tra attached to the growing protein move relative to the ribosome so the tra is now in the P site. f) The A site is now open again and the next charged tra that matches the next codon can bind. Steps d, e, and f continue until a stop codon is encountered. g) Termination. There are no tra molecules with anticodons that match the stop codons. Instead there are special proteins called release factors that can bind in the A site and this starts the process where the protein chain is released along with the mra and the subunits of the ribosome come apart. Energy is provided along the way by hydrolyzing GTP. 21. What is the structure of glycerol? 22. What is a lipid bilayer? Why does it form in water? The lipid bilayer is a thin membrane made up of lipids that form a flat sheet, two molecules thick. The lipids have a polar head facing away from the bilayer and long non-polar tails forming the inside of the layer. This is the basic structure for cell membranes for all living things. They form in water because of the hydrophobic effect where non-polar structures are separated from water to increase entropy (randomness) and reduce the overall Gibbs free energy of the system. 23. In the context of the phrase fluid mosaic model of a membrane, what does fluid and mosaic refer to? Mosaic is used because the membrane is not composed of just lipid, but also contains protein molecules. Some of the lipid molecules and some of the proteins also have s attached. Altogether forms a mosaic structure. Fluid refers to the liquid nature of the membrane and the fact that the phospholipid and protein molecules can diffuse laterally through the layer.